Rigid Bodies Equations of Motion General Plane Motion (Learn to solve any question)

Learn about dynamic rigid bodies and equations of motion concerning general plane motion with animated examples. We will use our usual f=ma equations along with relative acceleration to figure out the unknown forces.
🔹Rigid Bodies Relative Acceleration : • Rigid Bodies Relative ...
🔹Rigid Bodies Equations of Motion Translation: • Rigid Bodies and Equat...
🔹Rigid Bodies Equations of Motion Rotation: • Rigid Bodies Equations...
Intro(00:00)
The 2 kg slender bar is supported by cord BC (02:00)
A force of F = 10 N is applied to the 10 kg ring as shown (05:10)
The slender 12-kg bar has a clockwise angular velocity of (07:28)
Find more at www.questionsolutions.com
Book used: R. C. Hibbeler and K. B. Yap, Mechanics for engineers - dynamics. Singapore: Pearson Education, 2014.

[@arlow5497]

You not only post up a clean and clear process from start to finish but you're in the comments answering direct questions. . . Not all hero's wear capes. In fact, some teach dynamics.

[@QuestionSolutions]

Haha, that was funny! 😅 Many thanks. I wish you the best.

[@user-ml7nt2sq9f]

I agree

[@MrPJBarney]

No dislikes for a reason. Your videos are absolutely fantastic! Thank you so much. You have helping me through Dynamics right now. Today is the final exam, and I am absorbing so much knowledge and gathering so much understand. God bless you. Thank you. Jesus loves you.

[@MrPJBarney]

Video shared. *eight thumbs up* 10/10!

[@QuestionSolutions]

Great to hear! I hope you ace your exam. Best of luck :)

[@osmantaha4718]

These series are very very helpful, thanks for all of your efforts.

[@QuestionSolutions]

You are very welcome! Glad to hear they are helpful.

[@GenknownTutorial]

Love your videos Sir, especially the animations creating a very understandable visuals for the said problems, Btw Thank you Sir.

[@QuestionSolutions]

Thank you very much, I am glad it makes the problems easier to understand. :)

[@mohdhalimjaafar324]

thank you so much for this explanation. very helpful and 100% clear

[@QuestionSolutions]

I am glad to hear that, and you are very welcome :)

[@tseyincheng9854]

Far better than my tutor. Thank you for your sharing

[@QuestionSolutions]

Glad to hear it helped :)

[@11antonio11]

Thanks for all the help!

[@QuestionSolutions]

You're very welcome!

[@melekmnif1933]

This channel is a must watch if your are having trouble!!!

[@QuestionSolutions]

Many thanks!

[@mezonoor4022]

I put like even before watching the video :D thank you so much

[@QuestionSolutions]

Thank you very much! I appreciate every like and comment :)

[@alexanderblanco323]

I’ve watched every video, thank you

[@QuestionSolutions]

You rock! Thank you for taking the time to comment and for watching all the videos :)

[@ahmadaloufi4067]

In the first example ( 5:00 ) why is the acceleration of point B perpendicular to the rode? Is it because its normal acceleration pointing towards point c is zero ( released from rest ), and its tangential acceleration is perpendicular to the radius which is the rope??

[@melekmnif1933]

At 4:01 in the eq of ab you used a(g) aren't you supposed to use a(a) since you are using acceleration of point b relative to point a? But in this case a(g)=a(a) that's why right ?

[@QuestionSolutions]

This is a good catch on your part. To answer your question, no, I am actually using the acceleration of point B relative to the center of mass, point G. Unfortunately, I made a mistake and wrote "r(b/a)" when it should be "r(b/g)", however, you will see that right after that, in the next equation, it is in the correct form. You can use any point on a body to figure out relative acceleration, but usually, it'll either be a fixed point or the center of mass since we won't be given enough values to figure out anything else. Since we are using relative acceleration between points B and G, you will see that the position vector is from G to B, which is 0.15 m. 👍

[@feliceverbeek2838]

Hi! Love the videos. How would you proceed to solve a system with 5 unknowns and 5 equation? I find it difficult to start and to see which to substitute where. As in 5:10

[@QuestionSolutions]

This is a personal preference. For me, I like to use a matrices whenever it's more than 3 equations. If it's less than 3, I use substitution. Here is an easy example: www.mathsisfun.com/algebra/systems-linear-equations-matrices.html This method works for a very large number of simultaneous equations. The other method is to graph all the equations and see where they intersect.

[@Manuel-of7et]

I'm only confused on why the normal component of your relative acceleration equation is minus, I keep seeing it everywhere else as plus omega^2 *r. Which is correct? Are they both correct? Thanks!

[@QuestionSolutions]

It depends on how you take your position coordinates. So in this equation, the part you're asking about is the normal acceleration component of B with respect to A and the direction is always from B toward A. If you see a positive equation, look to see if it's r_(A/B) or r_(B/A). Which equation does your textbook use, positive or negative?

[@nelly-maps]

Have you ever thought of creating PDFs of the different section of work, and compiling it into its respective courses. That would be very helpful

[@QuestionSolutions]

I never really thought about it, maybe when I get free time, I will consider doing it. I just thought these videos are more of a supplementary resource along with coursework and textbooks. Thank you for the idea!

[@michielclaeys7784]

bro, you are a hero men, you saved my exam

[@QuestionSolutions]

Glad to hear that! I hope your exam went well, and keep up the good work.

[@pure_ligma2387]

thank you for these videos. Do you happen to have any overview or summary about all the topics in Dynamics?

[@QuestionSolutions]

I don't have an overview of all the subjects in dynamics, there are just too many topics, I only covered the stuff required for a first year course.

[@pure_ligma2387]

@@QuestionSolutions oh yeah i meant the first year material. Thanks for the response tho

[@QuestionSolutions]

@@pure_ligma2387 so everything in the Dynamics A-Z playlist covers pretty much everything needed for a normal first year course

[@cht5086]

Hi, at 6:52 the signs of moments at point C shouldn't be -(10cos45)(0.6) which creates a CW moment, and +(10sin45)(0.3464) CCW? Also for angular rotation, but the answer would be the same.

[@QuestionSolutions]

You can use either rotational direction as positive or negative. I use clockwise as positive. It's your personal choice, you will end up with the same answer.

[@rawadhasan9111]

For question 2, can we use sum of the forces in the y direction to find acceleration of G?

[@QuestionSolutions]

Yes, though you will need more equations since you will need to consider force N, which is eliminated when we take the moment about point C.

[@AlexiFrickEng]

hi there! Thanks sooo much for your videos they are amazing. i was just wondering at 8.50 you say we have 2 equations with three unknowns? Did you mean 4 unknowns? Also for the last question did you assume the angular accelerations direction? And if it come out negative we know its in the opposite direction?

[@QuestionSolutions]

Yes, yes and yes to the last question as well. :)

[@TheMounten1]

For the second question 6:29, why would it not be correct (not give the correct answer) if we were to equate the horizontal force in the x direction (horizontal component of 10N) with the horizontal force in the kinematic diagram? Or calculate the moment about point G?

[@QuestionSolutions]

So you can't equation the horizontal components because, first, it's a moment equation, but second, we need to account for the moment about the center of mass. The kinetic diagram doesn't deal with "forces" but more like moment/accelerations created throughout the whole object.

[@TechThinker-vw9ih]

Can you make some one shot video it will be really help full to understand the whole chapter at once...........By the way thank you for the explanation👍

[@QuestionSolutions]

Well, you can also watch the 3 videos that correspond to the chapter too 😅The whole playlist is in chronological order :)

[@SG-dw8jh]

my professor needs some help, what time is your office hours so I can send him 🤣

[@QuestionSolutions]

That was pretty funny! 🤣

[@christopherelkhoury7294]

Hello, great videos first of all. But do you have any tips on when to use the relative acceleration formula; since I am not sure when to use it. Thanks

[@QuestionSolutions]

If you know the acceleration of one point, you can usually find the acceleration of another point using the relative acceleration formulas. Over time, as you do more questions, you will have an intuitive idea on when to use what formulas.

[@christopherelkhoury7294]

@@QuestionSolutions Ok thanks!! By the way Do you have any video regarding relative acceleration in general plane? AKA 16.8 in the hibbeler book?

[@QuestionSolutions]

@@christopherelkhoury7294 You're very welcome. I do have a video on relative acceleration, but I don't know if it's the one you're referring to. kzbin.info/www/bejne/m5bCfaWLibt4a9E Do you mean rotational axes? If yes, then no, I didn't cover that. I only covered translating coordinate systems.

[@joedoyle4594]

At 4:14 , why do we only consider the tangential acceleration of point B? Would you not need to include the normal component as well?

[@joedoyle4594]

I've thought about it a bit longer, and i realise the normal component of ab is also w^2rbc, and w is zero at t=0. Is this reasoning correct?

[@QuestionSolutions]

@@joedoyle4594 That is correct.

[@sakn415]

For problem 2, may you explain what is Mk? Why in this problem we are including the acceleration 10(ag)(0.4) while in the first equation it was just I*alpha

[@QuestionSolutions]

Mk is the moment using the kinetic diagram. So the one on the right side. Also, what is the first equation you're referring to? Please use timestamps so I know where to look, thanks!

[@bilalors1486]

Hi, rods and bars take moment of inertia, in some questions it takes 1/3, in others it takes 1/12. I'm having a hard time logically making this up. In which cases is it 1/3 and in which cases is it 1/12 exactly? Thank you

[@QuestionSolutions]

Please see: cdn1.byjus.com/wp-content/uploads/2020/11/Moment-of-InertiaArtboard-1-copy-16-8.png

[@bilalors1486]

@@QuestionSolutions thanks a lot!

[@szatyormester]

You can solve the second problem for point G too, that way you can even calculate the frictional and normal forces. I solved it using MATLAB. In direction x: -Ff+10*cos(45)=10*a In direction y: N-10*9.81+10*sin(45)=0 In direction z: alpha*10*0.4^2=0.4*sin(30)*10*cos(45)-0.4*cos(30)*10*sin(45)+Ff*0.4 Furthermore, we know that a=0.4*alpha, that makes it 4 variables and 4 equations. This way we get N=91.03 for the normal force and Ff=4.83 for the frictional force. The acceleration and angular acceleration are the same as in the video, a=0.224 and alpha=0.56

[@QuestionSolutions]

You can use many different methods to get the same answer, I only showcase one method. :)

[@HAGA7793]

At 10:14, on the moment equation, why did you show the y component on both sides of the equation. 12 * 9.81*0.75 is not the same as the y component of acceleration?

[@QuestionSolutions]

No, that acceleration is the acceleration due to gravity, on the right side is the acceleration at the center of mass of the object itself. :)

[@str3525]

Hi, I have a question. At 2:28, why is the 2(ag)y going upwards? isn't it supposed to be going downwards since the rod will go down and to the right

[@QuestionSolutions]

Notice how we got a negative value for AGy. So initially, we assumed it to be up, but we got a negative value, indicating that it is actually facing down. With pretty much any force, you can assume a direction, and at the end, you will end up with a positive or negative value. If it's positive, your assumption was correct, if it's negative, then it's opposite to your assumption. In either case, the magnitude value of your answer will be correct.

[@aashishkumarjha3993]

For the first question at 2:39, can we take moment about B . Taking moment about B will eliminate tension and we can find angular acceleration

[@QuestionSolutions]

You can solve these problems in multiple ways, so yes, if it works and you get the same answer, all is good. 👍

[@onuralpbuyukturkmen5140]

In the second question, when moment is taken about point G, all the forces N W and 10ag disappear and just F=10N and Igalpha are left. We can easily get alpha from this moment equation. However, it doesnt give the same alpha result as we take the moment about C. Can you please expalin why is this so?

[@QuestionSolutions]

Can you show your work? Hard to say without looking at the step you took 😅

[@onuralpbuyukturkmen5140]

@@QuestionSolutions Unfortunately I can not post a link of my solution as a comment (I dont know why). It is being deleted instantly.

[@QuestionSolutions]

@@onuralpbuyukturkmen5140 I am not sure why either, could you send me an email at contact @ questionsolutions.com

[@ayrin7160]

What's the explanation of this? In 2nd ques,I took moment about G and have different answer... Again tried taking horizontal summation of forces, it's again giving another answer.. This must not happen right? What to do?

[@DrDerivative]

in the final example, why does your position vector have both of its components as negative? shouldnt they be positive? Or atleast diffeent from the sense of the acceleration? aGx is to the left and you noteds it as negative in the relative motion analysis. and the position vector rb/g has its sense up and to the right so it should be opposite sign no?

[@QuestionSolutions]

For the future, please kindly use timestamps. So I assume you're referring to the position vector at 11:17? It goes from top to bottom, so if you look, you can see that you have to go down and to the left to go from the top point to the bottom point. We used right and up to be positive. So here, since to go from the top to bottom, you need to go down and to the left, both components will be negative. It will be even easier to see if you imagine a coordinate system at the ending point of the position coordinate. Also, you are trying to tie acceleration and position vectors together, when they are not related. A position coordinate is just a simple vector that goes from one point to another. So an object can be accelerating to the right but that has nothing to do with a position coordinate establishing some direction or a distance. Also, if you look at the animation, you can actually see that I animated the position coordinate from the top to down for this very reason, so to establish which way we are looking at the position coordinate. You wrote "vector rb/g" but we are using "rG/B" which means here, the position vector goes from b to g. I hope that helps.

[@elements-24]

At timestamp 4:18 you mentioned the absolute acceleration of point B was perpendicular to the cable. But in circular motion net acceleration is not along the tangential direction as this the sum of centripetal and tangential acceleration. Then why did you take this along tangent or perpendicular to the cable direction? Thanks

[@QuestionSolutions]

Here, I've drawn it for you: bit.ly/3lgRVae An easy way to visualize this is to actually think of point B as just a point, forget everything else. Then you can see how tangential and normal acceleration vectors can be drawn. I also think you're overthinking this, or you're missing/forgot or not understanding fundamentals when to comes to circular motion. I hope this makes it easier to visualize :) Best wishes with your studies!

[@ahri1844]

4:05 why is Agy up? should it not be down tangential to the arc G is going to take?

[@QuestionSolutions]

It's an assumption, so if you get a negative value, it's opposite to your assumption. If you got a positive value, then your assumption is correct.

[@HashemAljifri515]

Hello , first of all thank you for this useful tutorial . Now what if I don't know the direction of the acceleration at G ? Should I assume it to be right and up ?

[@QuestionSolutions]

You're welcome. Please use timestamps, I don't know where you're referring to. Thanks!

[@HashemAljifri515]

​@@QuestionSolutions2:22, the beam is accelerating to the right since it is going down I thought it should accelerate downward 🤔

[@QuestionSolutions]

@@HashemAljifri515 So it will swing upwards while tilting down, but that's really not that important. You can assume the direction whichever way you want. At the end, if you get a negative value, that means it's opposite to your assumption, and if you get a positive value, your assumption was correct. However, in both cases, you will get the same magnitude for the answer.

[@sayanjitb]

For the first question, it can be clearly seen that center of the mass frame is non-inertial. But you wrote torque equation Tau=I*\alpha in this frame. Is this permissible? should we not write that equation only in an inertial frame? TIA

[@QuestionSolutions]

What time are you referring to?

[@sayanjitb]

@@QuestionSolutions At the timestamp (3:30), you explicitly showed forces acting on the center of mass. So evidently, it is a non-inertial frame. Then is it legit to write torque equation w.r.t this point, though?

[@QuestionSolutions]

@@sayanjitb Yes of course, you can write it about the center of mass, forces shown at the middle are simple mass times acceleration equations. I think you are overthinking the problem. Please check earlier videos on how to do this if you are confused.

[@guusjesmit1300]

why can't you take the moment around B? To cancel out the unkown force T. I thought you would be left with just gravity and the arm (0.15 meters). And that with the mass moment of inertia, the angular acceleration would be easy to determine. But it just doesn't seem to work in my calculations. Am i missing something? How would you do it? (second question with the beam)

[@QuestionSolutions]

If you're going to use a moment about point B, you have to recalculate the mass moment of inertia for a whole new point (B). Notice how we calculated the mass moment of inertia about point G (center), this makes the math simple with just a single moment equation about the center.

[@jiseongkwon66]

For the second problem at 7:03, why do we not use parallel axis theorem? I thought that since point c was away from the the center, we would need to specify the distance away as well.

[@QuestionSolutions]

I am unsure of your question. Parallel axis theorem to calculate the mass moment of inertia? Or for the use in our moment equation? The moment was calculated about point C, not at the center. The distances for each of the other forces were then written down in the equation at 7:03. So we have the 10 N force (2 components, both distances are shown), and on the other side we have the mass times acceleration, again a perpendicular distance of 0.4 m is shown. There are no other distances to specify. Sorry I couldn't help, I am unsure of what you're confused about 😅

[@jiseongkwon66]

I was wondering why we would use that formula for mass moment at the contact point. I thought that formula you used was only for the mass moment of inertia at the center, but since C is away from the center, wouldn’t we need to add m(d)^2 because C is “d” distance away?

[@jiseongkwon66]

I’m sorry I didn’t specify haha, I meant using parallel axis for mass moment of inertia

[@QuestionSolutions]

@@jiseongkwon66 No, you don't have to make it that complicated. You can just use the mass moment of inertia for a ring, so it's just I=mr^2. You only need to worry about the parallel axis theorem if you're going to calculate the mass moment of inertia yourself. Most of time, this isn't required since they are basic shapes and you can memorize (or look up) the equations for them. Also, even though we are taking the moment about point C, the moment about the center of mass, so "Iα" is about the center, not point C.

[@jiseongkwon66]

@@QuestionSolutions OHH I SEE NOW Thank you so much!! I appreciate you getting back to my question!

[@mohamedsamy5063]

12:16 Thank you for the video. I have a question, Why don't you use the parallel axis theorm and make the sumition of the T = (I + m*L^2)*alpha then it will be just 3 equations with 3 variables and can solve it directly ?

[@QuestionSolutions]

You're very welcome! To answer your question, you can solve these problems in a multitude of different ways. I only showcase one method, but you can definitely arrive at the same answer using a different path.

[@miquel07]

@@QuestionSolutions Hi! I've been using your videos throughout college. Is there any way you can make a video discussing Parallel Axis Theorem and Inertia given a radius of gyration? Amazing video!!

[@QuestionSolutions]

@@miquel07 I think I cover it partly here: kzbin.info/www/bejne/sJ7dhpd8e6h5ocU

[@miquel07]

@@QuestionSolutions This helped me so much! Thank you!!

[@arinzeanthony7447]

11:09 angular acceleration was negative why? While solving crank questions,I remember that angular velocity when in the clockwise direction is -ve and anti clockwise direction +be, but while we were solving for acceleration using this same formula we didn't negate angular acceleration Even when it was moving in the clockwise direction. Why negate it in this case?

[@QuestionSolutions]

Angular velocity vector is found using the right hand rule. So clockwise would mean the vector points away from you, and counterclockwise means the vector points towards you. Depending on how you establish your 3D coordinate system, you can consider either side positive. For all of these things, the directions are assumed, as long as you follow through with your directions for all the forces/moments, etc. If you end up with a negative answer, then your assumption was wrong, and it's opposite to your assumption. Please see: hyperphysics.phy-astr.gsu.edu/hbase/rotv.html for more details on vector directions.

[@ishaanmann7866]

for problem 1 at 2:24, why can't the movement of the rod be defined by n and t acceleration components. When I tried this, I said my an=0 (since w=0), and at= alpha*r. I then used the equation of Mg=IG*alpha and summed up my forces in the y direction. I ended up with 2 equations and two unknowns (Tbc, and alpha). It didn't work, since our answers weren't the same but I would appreciate it if you could explain why. If you look at problem 17-67 in hibbler, you'll notice they did a similar approach to the one I described, but for some reason that doesn't work on this question.

[@QuestionSolutions]

It's really difficult for me to say without seeing the steps you took. Did you see your TA or professor during office hours to ask about your solution?

[@ishaanmann7866]

@@QuestionSolutions I didn't, but I guess my question is wouldn't the the rod undergo curvilinear motion when released from rest? And if it does, why don't we use normal and tangential components for acceleration

[@chicanery358]

At 4:20 , why is the acceleration at B perpendicular to the cord again? Is there no normal acceleration in the cord? Thanks!

[@QuestionSolutions]

Here, I've drawn it for you: bit.ly/3lgRVae An easy way to visualize this is to actually think of point B as just a point, forget everything else. Then you can see how tangential and normal acceleration vectors can be drawn.

[@chicanery358]

Thanks! May I clarify if the acceleration at B would still be wholly tangential to the circle if it had angular velocity?

[@rllsaa]

For the third question, why is assuming w=d/dt(theta) and alpha=d^2/dt^2(theta) provide different solutions? What is the fallacy?

[@rllsaa]

Edit: I made a error in my calculations. Works both ways.

[@QuestionSolutions]

@@rllsaa Glad to hear!

[@ruannnnnnnnn28]

do anyone know if the channel has videos on mass moment of inertia ? I'm writing exam tomorrow but thanks for these videos anyway , they are really helping me

[@QuestionSolutions]

No, currently there aren't videos on mass moment of inertia. In the future, I hope to create a video on that topic as well.

[@QuestionSolutions]

There is a video up now on mass moment of inertia. kzbin.info/www/bejne/sJ7dhpd8e6h5ocU

[@qihang2173]

At 2:23, where did the acceleration 2(aG)x and 2(aG)y come from? There are no external forces acting on it, isn't it?

[@QuestionSolutions]

In the kinetic diagram, it's just mass, so 2kg times acceleration broken into x and y components. It occurs at the center of mass because the object accelerates when you let go.

[@bjtheclasher5443]

​@@QuestionSolutions for problem 1. How did you get the final answer for (ag)y =-8.41m/s^2 and (ag)x=2.42m/s^2. thankyou

[@QuestionSolutions]

@@bjtheclasher5443 You can use wolfram alfa or desmos. Otherwise, you will have to use the substitution method to solve them which will take quite some time. Basically, you isolate for one variable and plug it into the next one. Then you isolate again, and so forth until you have one equation with a single variable. Then you can solve for that variable and solve the rest too.

[@minmax3160]

For question 2, wouldn't summing the moment at G also ignore Normal force and weight ?

[@QuestionSolutions]

Normal force and weight were not used when we summed the moment about C (6:33). Their lines of action goes through point C.

[@oriolaolamide9995]

@@QuestionSolutions Yes but the lines of action of Normal force and Weight also go through G so if we summed the moment about G shouldn't they also be ignored?

[@grantwiesehan9515]

@@oriolaolamide9995 thats what i was thinking and then you could ignore aGx if you go through G

[@Khadra2024]

at the first question, why the direction of aB to the down? I mean why it is not the same direction with tension?

[@QuestionSolutions]

So you have to visualize these problems. If you hang something with a rope, pull it to the side, its not going to go straight towards the point where the rope is hung from. It's actually going to go down in a circular path because gravity is pulling down on it. 👍

[@e-nes4042]

4.46 why is the direction of acceleration (aB) ?

[@QuestionSolutions]

It's tangent to the force vector.

[@howis-goin]

Sir, at 6:20, i thought the moment of inertia of a circle is Ig = 1/2mr^2, so how can it be Ig = mr^2?

[@QuestionSolutions]

This is a ring, not a circle, so you have to use the moment of inertia for a ring, which is mr^2. See: byjus.com/jee/moment-of-inertia-of-a-ring/

[@howis-goin]

@@QuestionSolutions oh, i never thought a ring would be different from a circle XD, thx you so much, anyway if there is a circle with the radius of r, and the radius of gyration of r2, so the moment of inertua of that circle would be I = 1/2.m.(r2)^2, right?

[@kilianrave7988]

Why is the friction ignored in the FBD of question 2?

[@QuestionSolutions]

Slipping doesn't occur, and also, if a coefficient of friction isn't given, you usually don't have to worry about friction on a question, unless what it's asking for is the friction. Just one more thing, on this problem, the moment was written about point C, so even if friction was there, it wouldn't have been used.👍

[@SadkAtaMzrak]

@@QuestionSolutions But if we ignore the friction force (not include in the FBD) and take moment about G, we get a differen result for the accelerations, does that tell us that there is or should be a friction force?

[@samiatabdulraheem1871]

Hi, please I have a question about the second question, why couldn’t we do the moment on point G instead of C. Wouldn’t that have been easier, I tried it but it was a different answer. Is there a reason why it wouldn’t work?

[@QuestionSolutions]

It should work, did you double check your distances properly?

[@blakejohnson4851]

at 12:00 shouldn't the -6i term actually be +6i? because you are multiplying -2^2(-3*cos(60º)) = +6

[@QuestionSolutions]

Yes, a typo, but it didn't matter since we only used the j components. Good catch.

[@lumjiaming9960]

6:39 I want to about some question about the question 2. In question 2,why we cannot straight away use the equation Fx= max . Using this equation , Fcos 45= m ax .I cant get the ans 😢

[@QuestionSolutions]

If only it was that simple 😅 You have to account for the mass moment of inertia, along with all the forces created, including moments, when the force is applied to the object. There really is no way to get around this, it's a lot of work when it comes to general plane motion because it's not locked into an axis.

[@alexanderblanco323]

GOAT

[@QuestionSolutions]

Thank you :)

[@sasukekianhoong6130]

i just wonder why at 6:41 , we cannot write 10cos45 = ma and find a since it both of them are on horizontal axis ?

[@QuestionSolutions]

Sorry, what do you mean by find "a" by writing 10cos45 = ma? If you plug in 10 for mass, and solve for a, are you getting the same answer? I think I am having a hard time trying to understand your question. What is on the horizontal axis? 😅

[@sasukekianhoong6130]

@@QuestionSolutions I mean summation Fx=m(ag)x , so 10N multiply cos 45 = m(agx) I hope it clear now, sorry for the unclear question. I just wonder why I can't use the formula summation Fx= m (ag ) to find my acceleration

[@QuestionSolutions]

@@sasukekianhoong6130 So one problem is that you're not accounting for the mass moment of inertia of the object. Also, it's very hard to get an answer if the point we are considering is not the center of mass using just f=ma, it works when we think of particles, but cannot with rigid bodies. It's usually a good idea to keep mass moment of inertia in mind when we have to consider rigid objects.

[@mohanakrishnanr4868]

@@sasukekianhoong6130 it's because we have friction which is necessary to sustain pure rolling motion if you observe initially there is no angular velocity in the ring and as the motion proceeds angular velocity starts increasing thus there must be torque in the direction of motion which accounts for this angular acceleration so friction will be there that's why you cannot write equilibrium eqn like what you have stated.

[@Dpo-fm4dj]

@@QuestionSolutions but you used Sigma F = ma in the first question along y and x axis😅 what is the different now

[@NuNoNoob]

why cant we use F=ma in the x direction on question 2 to get the acceleration at G? Wouldnt it be 10cos45=ma(G)? and then you can solve for a(G)?

[@QuestionSolutions]

Did you try your method and get the same answer? Generally speaking, there are multiple ways to get to the same answer, I only showcase one method pertaining to a chapter. :)

[@sakn415]

didn't get the same answer with that method.

[@morgandamagon8804]

5:11 Sir, I'm having a problem solving the 5 unknowns, I hope you could provide me how to solve it😅

[@QuestionSolutions]

The easiest way is to use wolfram alpha or desmos to get the answer. If you don't have access to it, you will need to use the substitution method or a matrix to solve it. It will be time consuming though 😅

[@arinzeanthony7447]

2:53 please why is 2(ag)y upwards and not downwards? And also since torque is anticlockwise,why isn't the moment about CG in the kinetic diagram = -Ia? Instead it was positive?

[@QuestionSolutions]

For AG, it's just an assumption, if you get a positive value, your assumption was right, if you get a negative value, then it's opposite to your assumption. So here, we just assumed it to be upwards and to the right. For the moment, I picked counterclockwise to be positive, you can see the symbol at 3:32. You can pick clockwise to be positive, then your IG will be negative. Again, it's an assumption, all it means is that if you get a negative answer at the end, then it's opposite to your assumption.

[@arinzeanthony7447]

@@QuestionSolutions But if we assume that it's direction [ (ag)y ] is downwards, we'll end up getting a different answer. I understood the other ones which you explained apart from this particular one. Or did we assume that (ag)y is upwards because the hand is pulling it upwards? But that can't be because in the question,it says that it was released, that's why I think that (ag)y should be downwards. Please I really want to understand this, don't want it to seem like I'm disturbing,but can you please briefly explain why (ag)y is acting upwards and not downwards?

[@QuestionSolutions]

@@arinzeanthony7447 It really is just an assumption and I just did the whole question assuming it to be down. Then the only difference I got was that aGY = 8.41, so it's positive. See, we assumed it to be up, but we got a negative value, which means it's really downwards. When you assume it to be down, you get a positive value. Please redo the question, you will always end up with the same answer. Here is the version with the assumption that aGY is down: www.wolframalpha.com/input?i=tcos%2830%29%3D2x%2C+tsin%2830%29-%282*9.81%29%3D-2y%2C+%280.15t%29sin%2830%29%3D0.015a%2C+-bcos%2830%29%3D-y%2B0.15a%2C+bsin%2830%29%3Dx You can see that the "y" value is a positive 8.41. Most likely, when you assumed it to be down, you didn't account for the relative velocity equation where that y component needs to be negative now.

[@arinzeanthony7447]

@@QuestionSolutions Yes,I just did it again. ay=8.41m/s². The direction is indeed correct. You're simply the best, thank you very much. I really appreciate.

[@QuestionSolutions]

@@arinzeanthony7447 Awesome. Remember, your directions are always assumptions. Sometimes, you can take a good guess, but they can be wrong. The only difference you will get is the positive/negative signs. So if it's negative, that means your assumption was wrong, and if it's positive, your assumption was right. 👍

[@arinzeanthony7447]

5:42 question 2. Why didn't we take moments about G? I know you said it'd make things easier but how? Also the horizontal component of 10 N force = 10cos45. Why can't we say. 10cos45=10a? It's confusing.

[@QuestionSolutions]

So you can take the moment about any point you want, it's up to you, and you will get the same answer. Sometimes, you might need more equations than at other points, but it's still doable. Also, for these types of problems, especially if you're studying to be an engineer, you should try your own methods out and see what you get, the problems you face, and how to solve them. So instead of wondering if you can, or can't, you can actually try taking the moment about G, then you can actually see what problems, if any, you might encounter, and it won't be confusing since you're trying it out on your own and gain experience in solving a question in more than one way. Also, please keep in mind that these problems can be solved in so many different ways, I just showcase 1 way of doing it. I usually try to take the path of least resistance, by eliminating as many unknowns as possible and using methods that save time. That doesn't mean this is the only way :)

[@armandojosecolon]

how do we get aB= 4.85 m/s^2?

[@QuestionSolutions]

Please give me a timestamp so I know where you're referring to. Thanks!

[@mahyahbintiidris9012]

Please Help, Q. For a body executing planar motion: (a) The linear velocity of all the points of the body is same. (b) The linear velocity of a point is function of its radius vector. (c) The linear velocity as well as linear acceleration is same for all points of the body. (d) None of the above. Please give the answer with detailed explanation if possible.

[@QuestionSolutions]

I don't solve questions like this since I'd have to solve everyone's questions and it would take too long 😅 It's better to ask these types of questions on a forum to get help :)

[@mahyahbintiidris9012]

@@QuestionSolutions Please suggest some forums that may help.

[@QuestionSolutions]

@@mahyahbintiidris9012 Google "physics help forums"

[@vikasrayudu7647]

why didnt u use parallel axis theorem for mass moment of intertia to shift Ig to Io

[@vikasrayudu7647]

in the last problem

[@vikasrayudu7647]

while taking moment about o

[@QuestionSolutions]

Moment equation is written about O, which means we don't care what happens there, since all the forces will get cancelled out (their lines of action will go through point O).

[@adelamini9873]

I'm confused as to how you know how to use the relative acceleration equation. as in how did you know that you have to go B to G and then how did you know you had to go B to A?

[@QuestionSolutions]

Please use timestamps so I know where to look 😅

[@adelamini9873]

@@QuestionSolutions at 10:34 and 11:38

[@QuestionSolutions]

@@adelamini9873 So the general rule is, what I want to find, goes on the left side. Here, it doesn't matter as much because you're just trying to compare one point to another to get a few equations to match the number of unknowns.

[@adelamini9873]

@@QuestionSolutions thank you so much for the speedy reply’s I really appreciate everything you do for your community

[@QuestionSolutions]

@@adelamini9873 You're very welcome! :)

[@RookiePresent]

At 11:26, why is the left side of the equation for a sub G -(aG)xi-(aG)yj and not -(12)(aG)xi-(12)(aG)yj? Is it just because the equation calls for accelerations only so mass isn't factored in?

[@QuestionSolutions]

Yes, that is correct. 👍 The relative acceleration equation is only with accelerations, so mass is not considered. Please see: kzbin.info/www/bejne/m5bCfaWLibt4a9E

[@bahiralilatif6639]

At 4:28, r_B/A turns into r_B/G why? Thanks

[@QuestionSolutions]

That's a typo, we are comparing point B to G, which means it should have been B/G. Good catch!

[@kanao1167]

For the second question why is there no friction force exist?

[@QuestionSolutions]

"Slipping does not occur" so friction doesn't' need to be accounted for.

[@moureeshvk3566]

how to solve a more than 4 eq in a calci

[@QuestionSolutions]

Most calculators can only solve 3, so the best option would be to use a matrix or use substitution. It'll take long, but there isn't another way to do it. If you're allowed to use wolfram alpha, that's your best method.

[@theadel8591]

Good video but i’m confused as to how you knew ab was normal to the cable ?? Even the justification you said (because it is rotation about fixed axis) seems to me unrelated to the fact ab is normal to the cable !!!

[@QuestionSolutions]

I am not sure how it's unrelated, that is actually the reason for it. This should also be in your textbook, and if you find it hard to believe, you should model this on the computer to see the acceleration vectors. 😅

[@user-cv7yb3kk5h]

This video is very helpful, but the speed is a little bit fast, not everyone is a genius!

[@QuestionSolutions]

Thanks!

[@ROTSTarge]

Honestly a vid helping me through the set of 5 eqs with 5 unks would be great X3 I'm rusty on Algebra

[@QuestionSolutions]

😅 I think there are lot of algebra videos on KZbin already. The easiest to use is probably a matrix, so if you search matrix equation solving, you should get lots of hits 👍

[@haseebahmed2581]

Im confused as to how u got the final solutions for Q1

[@QuestionSolutions]

Please use cymath.com or symbolab or wolframalpha if you'd like to see the steps for solving algebra problems. I think cymath is very good at showing all the steps :)

[@haseebahmed2581]

@@QuestionSolutions Im still struggling to understand. Any chance you could help? Thanks

[@QuestionSolutions]

@@haseebahmed2581 So what we have to do when we have 5 equations with 5 unknowns is try to solve them by either substitution or elimination. Imagine we have a simple scenario like this: 2x+9=3y, 3x+9=y. I am positive you would know how to solve this right? We would isolate for x or y in one equation and plug it into the next one? Then we can solve for it. It's the same exact thing, but now we have 5 equations. It's a lot more steps but the process is the same. Also, covert things like sin30 to decimal form. So sin30=0.5. Another tip is to write all the variables in simple letters, like x, y, z. So (a_G)_x = x and (a_G)_y = y. This makes it easier for us to see that they are just equations that need to be solved the same way we always do. To double check your answers, use this: www.wolframalpha.com/input?i=Tcos30%3D2x%2C+Tsin30-%282*9.81%29%3D2y%2C+%280.15%29Tsin30%3D0.015a%2C+-Bcos30%3Dy%2B0.15a%2C+Bsin30%3Dx Here, I have an example of 2 equations being solved using the substitution method: www.dropbox.com/s/fp3d9sepod9kbu9/Question%201%20solved.pdf?dl=0

[@launchpadmcquack98]

I think there is a way to solve the last problem with only three equations. radX = distance from center of gravity to IC, x component radY = distance from center of gravity to IC, y component alphaG = alphaIC = alpha aB = 2radX*alpha aA = 2radY*alpha (aG)x = radY*alpha (aG)y = radX*alpha This way you can eliminate (aG)y, (aG)x, aB, and aA by using the distance from the IC and solve with only 3 equations since you only need to find FNx, FNy and alpha. Let me know if I made a mistake.

[@QuestionSolutions]

Did you get the same answers? If yes, you didn't make a mistake. :)

[@darrylcarter3691]

That last problem was actually my favorite one

[@QuestionSolutions]

👍 That's good.

[@darrylcarter3691]

@@QuestionSolutions that last one can be helpful when finding force and moments of bars in a truss.

[@QuestionSolutions]

@@darrylcarter3691 The methods used for truss analysis are a lot different since it falls under statics, so non-moving parts. So you end up using equations of equilibrium instead of equations of motion. But all of these topics really come together and you can apply them to so many different cases. 👍

[@tinkywinky4554]

Sir in first question you wrote Mg=Igxalpha but in second question you wrote Mc=(Mk)c.I mean in second question u wrote again Igxalpha and you added 10Ag(0.4). I wonder why u added 10Ag(0.4)?

[@QuestionSolutions]

Please provide timestamps so I know where to look. I can answer your questions if you say at what time in the video you're asking about. Thank you :)

[@tinkywinky4554]

@@QuestionSolutions In 3:24 you just wrote Mg=Igxalpha. In 6:30 you wrote Mc=(Mk)c. In part of (Mk)c, you wrote again Igxalpha but then you added 10Ag(0.4). I wonder why you added 10Ag(0.4) to equation in 6:30 and you didnt in 3:24?

[@QuestionSolutions]

@@tinkywinky4554 Notice how at 6:30, we are not writing a moment equation about the center of mass, whereas at 3:20, we are writing it about the center of mass. Please see this video, the intro part at least: kzbin.info/www/bejne/fWXJYWCDfNmYsMU Thanks!

[@tinkywinky4554]

@@QuestionSolutions Thanks bro.

[@fariaalju4649]

Yep..that last one was gigantic..

[@QuestionSolutions]

It sure was!

[@Azzu974]

Any shortcuts for these problems? Its too lengthy

[@QuestionSolutions]

Unfortunately, I don't think so. General plane motion questions tend to take a lot of steps to solve since a lot of factors come into play. But I can tell you that the more questions you solve, the faster you will get.

[@e-nes4042]

last question very hard and long

[@QuestionSolutions]

yes 😅

[@AmCosmeaux]

yeah i’m fucked for my final exam lol

[@QuestionSolutions]

Try doing as many questions as possible. You got this!

[@momentofchange1306]

I love ur teaching but u are too fast. Pls make it more gradual for proper understanding. Thank u

[@QuestionSolutions]

I know it seems fast, but if you watch the playlist from the beginning and use the pause button, I promise it's not too fast. It just seems like it because I'm not taking 5 minutes to write down an equation, and it's pretyped. Also, the reason why I said these videos need to be watched in series is because the required material for this video is covered in the previous ones. Regardless, I will keep what you said in mind. I think a lot of the newer videos tend to be a bit slower.

[@wagehadawy]

سيدنا النبي.. صلوا عليه

[@QuestionSolutions]

Not sure what you are saying, sorry 😅

[@omaraliraqi3056]

@@QuestionSolutions He said Our prophet Muhammad pray on him

[@QuestionSolutions]

@@omaraliraqi3056 Thank you for the translation. Really appreciate it!

[@omaraliraqi3056]

@@QuestionSolutions All respect to you, Mr.