Mints on a Rotating Turntable - Determining the Static Coefficient of Friction

[@subikksha4941]

I said 'wrong' in my mind when Bo mentioned centripetal force. That's the impact you've made sir, I'm having the physics in my mind! I could never thank you enough for this

[@FlippingPhysics]

I enjoyed this. thanks!

[@ronalds271]

Sir, you are one of the best physics teacher I have learnt from. Thank you for posting videos.

[@eggsamillion]

You are a funny man. Not in the strange way, just an entertaining way.

[@FlippingPhysics]

thanks. I entertain myself and hope others are as well.

[@trungpham7588]

From other sources, all I know is to understand the concept a bit then doing the exercises ... But there are a ton of small questions that I don't even know how to ask (Because of all of the assumptions in Physics, I believe), but, the way you taught, you clear it all. And once again, no other resources can do that. 🙏🙏🙏 Hat off to Flipping Physics.

[@FlippingPhysics]

So glad to be helping you learn!

[@andrewjustin256]

Mr. P why is that the case that it was the maximum starltic force and it is equal to or less than the coefficient of the static friction times the nornal force?

[@pat6595]

Why the Coefficient of S.T. should be greater or equal to 0.29? Can you explain further?

[@curiouscat5555]

I struggled with this too and had to listen to the explanation many times but I think I get it now: recall they say that the coefficient of static friction (u_s) x normal force = f_smax... But once f_smax is reached the mints will fly off the turntable because f_smax leads to f_k once reached (static friction "breaks" and an object must go into motion where it experiences kinetic friction). Since the mints are just chilling there, we know f_smax MUST be at least greater than tor equal to the greatest value here or else they would all be flying off the table with the highest value we calculated.

[@PrashantKumar-hj1pn]

Sir please make video on polarisation of light

[@gemacabero6482]

Thanks! I had a question. In problems like this I always get confused with the friction forces. Because I understand there is a friction force in the inwards direction, but is there a friction force in the z axis ( in the direction of the tangential acceleration) ?? Because if there is no friction on that direction I don't understand how the mints are not moving forwards or backwards as the disk rotates. Thank you!!

[@carultch]

Don't confuse the axis direction we assign for keeping track of angular quantities, with the direction forces actually act. The direction of angular kinematics vectors is more of a matter of bookkeeping and convention, than anything actually happening in that direction. In this situation, all vertical forces add up to zero. Those forces are the normal force and gravity. Normal force, will be as large as necessary to keep the mints from sinking through the turntable surface. The static friction force (aka traction) can either point radially inward, or tangential along a circular path, or some vector sum of the two that is in between. At constant speed, the tangential acceleration is zero, and the net force is radially inward. Static friction points radially inward. When tangential acceleration is concurrent with the speed, part of the static friction acts radially inward, and part of the static friction acts tangentially. As an example, given a 10 cm radius, and a speed of 10 cm/sec (i.e. 1 rad/sec), the centripetal acceleration is 0.1 m/s^2 and is radially inward. Suppose a mint is at the 12 O'clock position, when moving clockwise at this speed on a turntable that is simultaneously slowing down at a rate of 1 rad/sec^2. The corresponding tangential acceleration will also be 0.1 m/s^2, to the left. These two tangential accelerations add up as vectors, and become a total linear acceleration of 0.14 m/s^2, which points in the down-left diagonal direction. Friction would have to be large enough to supply the force for the total linear acceleration at this condition, otherwise the mint would drift.

[@Phoenix-ly5yz]

If radians is a placeholder, can 1.5pi rad/sec be written as 1.5pi/sec?

[@fedoraemperor568]

Not exactly. Your expression "pi/sec" is not correct. Pi is a mathematical constant equal to 3.1415..., not a unit. If you don't want your answer to include radians, you can format use an exponent of negative 1 (denoted ^-1) to identify that you want the reciprocal of seconds. So, the answer would become "1.5pi sec^-1". I hope this helps you understand better :)

[@ijheller]

At 4:24, Bo said "0.68" when he should have said "0.068." I acknowledge that there is no easy way to fix this, but just so you know.

[@FlippingPhysics]

Bummer. Yep, can't fix it now. Oh well. I am sure the internet will survive.

[@ethanbilderbeek7607]

isn't centripetal acceleration = v^2/r?

[@carultch]

Substitute v = omega*r, for tangential velocity in v^2/r. You will get omega^2*r

[@BanValsimot]

Should not force of friction act in the tangential direction? With no friction mints would not ffly out in outwards direction but in the tangential direction... :/ like water in the rotating bucket

[@FlippingPhysics]

The direction of the centripetal acceleration acting on an object moving at a constant angular velocity is derived here: kzbin.info/www/bejne/bHOZioWGjrt9a7M it is _inward_. That centripetal acceleration is caused by a centripetal force which is the net force in the in-direction. In this example that force is the force of friction.

[@BanValsimot]

Ah, thank you so much for the derivation video! I have observed mints as a passenger in a car that is turning. There is no inward force acting on a passenger, only an inward friction force acting on a tires. Car shell keeps passengers in a car and I thought tangential friction force mints on the table. But mints are tires in this case...:D. Thank you very much for your reply.

[@shiningstar3236]

Me staring the video for the whole hour trying to differentiate these people 😫 instead of listening

[@PrashantKumar-hj1pn]

Hello

[@FlippingPhysics]

hi

[@PrashantKumar-hj1pn]

Sir I am from India , your way is unique