Maximum torque on a rotating cylinder kept on a moving plank

  • #1
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Homework Statement


A cylinder of mass M and radius R is resting on a horizontal platform (which is parallel to the x-y plane) with its axis along the y- axis and free to rotate about its axis. The platform is given a motion in the x-direction given by x= Acos(ωt). There is no slipping between the cylinder and platform. The maximum torque acting on the cylinder during its motion is?

Homework Equations


τ = Iα = F x R
Frictional force = μmg


The Attempt at a Solution


Differentiating the given equation 2 times, I found out that maximum acceleration = Aω²
Torque will be maximum when acceleration is maximum to prevent slipping.
If an observer is placed on the plank,
the cylinder will appear at rest to him.
Thus to balance forces, frictional force= Pseudo force = mAω²
⇒μmg = mAω²
⇒μ = Aω²/g
Now to find torque due to friction -
We take a small ring on the disk (bottom part of the cylinder) of thickness dx and radius x,
Mass of this ring = M/πr² (2πr dx)
⇒dτ = M/πr² (2πx dx) μg x
Integrating x from 0 to R
τ = 2/3 μMgR
Putting the value of μ
τ = 2/3 MRAω²
But the answer is τ = 1/3 MRAω²
 
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Answers and Replies

  • #2
TSny
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Since the cylinder never slips, you are dealing with static friction. For static friction the formula ##f_s^{\rm max} = \mu N## represents the maximum static friction force that could exist. But you can't assume that the friction force in this problem will ever equal its maximum possible value. The coefficient of friction ##\mu## is irrelevant in this problem. You may assume that there is always enough friction to prevent slipping. So, at any instant of time, the friction force will adjust itself to whatever is needed to prevent slipping.

I don't follow what you are doing in your integration. You have rings "of thickness dx and radius x", yet you write 2πr dx. What does r represent? There is no need to integrate over the cylinder. Just treat the cylinder as a rigid body and apply Newton's laws. You can work in the frame of reference of the earth or the frame of reference of the platform (with pseudo-force).
 
  • #3
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You have rings "of thickness dx and radius x", yet you write 2πr dx.
Sorry that was a typing error. I corrected it.
Just treat the cylinder as a rigid body and apply Newton's laws.
I don't understand how.
Frictional force is acting forwards on the cylinder and backwards on the plank.
That means both have different accelerations.
 
  • #4
haruspex
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Frictional force is acting forwards on the cylinder and backwards on the plank.
That means both have different accelerations.
I don't see why that stops you following TSny's advice.
You know the horizontal acceleration of the point of the cylinder which is (instantaneously) the point of contact.
 
  • #5
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I don't see why that stops you following TSny's advice.
You know the horizontal acceleration of the point of the cylinder which is (instantaneously) the point of contact.
I am doing something wrong.
At the point of contact, at a particular time t
a + Rα = Aw²
Where a = acceleration of cylinder
From here a = Aw²/2
So,
τ = Iα = MR²/2 (Aw²/2R) = 1/4 (MRAω²)
I am not able to visualize this problem. It's because the cylinder is kept vertically. It's confusing me too much.
 
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  • #6
haruspex
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From here a = Aw²/2
Why?
I am not able to visualize this problem. It's because the cylinder is kept vertically. It's confusing me too much.
The axis of the cylinder is horizontal, but you seem to have understood that, so why do you say it is vertical?
 
  • #7
haruspex
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At the point of contact,
a + Rα = Aw²
Where a = acceleration of cylinder
These are functions of time. The trig function should be present.
 
  • #8
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These are functions of time. The trig function should be present.
Is it correct now?
 
  • #9
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Why?

The axis of the cylinder is horizontal, but you seem to have understood that, so why do you say it is vertical?
I misunderstood the question. Its clear now.
I the earlier post I assumed a = Rα
But that is clearly wrong.
I tried it again.
At a particular time t,
a + Rα = Aω² ---- (1)
Also, fR = τ = Iα
Where f = friction
⇒ fR = (MR²/2)α
⇒α = 2f/Mr ---- (2)
Also f = Ma ---- (3)
because f is the only force acting on the cylinder.
So from equations (1), (2) and (3),
f/M + 2f/M = Aω²
f/M = a = Aω²/3
⇒τ = MR²/2 (Aω²/3R) = 1/6 MRAω²
Its still wrong.
 
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  • #10
haruspex
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At a particular time t,
a + Rα = Aω² ---- (1)
No, as I pointed out before, the acceleration of the platform is not constant. The trig function of time should be there.
 
  • #11
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No, as I pointed out before, the acceleration of the platform is not constant. The trig function of time should be there.
I don't understand.
Should I replace a by acos(t) ?
 
  • #12
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No, as I pointed out before, the acceleration of the platform is not constant. The trig function of time should be there.
How do I know what trig function is it's acceleration?
 
  • #13
haruspex
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How do I know what trig function is it's acceleration?
You are given
x= Acos(ωt).
Derive the acceleration from that.
 
  • #14
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You are given

Derive the acceleration from that.
x = Acosωt --- (1)
a = -Aω² cosωt
But this will be the acceleration of the plank.
To convert it to acceleration of cylinder, I'll have to subtract distance travelled by cylinder on the plank from equation (1).
How should I do that?
 
  • #15
TSny
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Let

vc/g = the velocity in the x direction of the center of mass of the cylinder relative to the ground
vc/p = the velocity in the x direction of the center of mass of the cylinder relative to the plank
vp/g = the velocity in the x direction of the plank relative to the ground.

How are these related?
How can you use this relation to get the relationship between the accelerations ac/g, ac/p, and ap/g?
 
  • #16
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How are these related?
vc/p = vc/g - vp/g
I think the same will go for acceleration.
 
  • #17
TSny
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vc/p = vc/g - vp/g
I think the same will go for acceleration.
Yes
 
  • #18
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τ = MR²/2 (Aω²/3R) = 1/6 MRAω
I did this again.
But this time I took the value of α = (Aω² - a)1/R = 2Aω²/3R
τ = MR²/2 (2Aω²/3R) = 1/3 (MRAω²)
Am I right this time?
 
  • #19
TSny
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I did this again.
But this time I took the value of α = (Aω² - a)1/R = 2Aω²/3R
τ = MR²/2 (2Aω²/3R) = 1/3 (MRAω²)
Am I right this time?
You might be right, but your solution is broken up among several previous posts. I think it would be a good idea to put it all together neatly in one post. Also, why do you assume from the beginning that you can use Aω2 for the acceleration of the plank? In post #14 you have the acceleration of the plank as a function of time. Using this you can deduce the torque on the cylinder as a function of time and then see when the torque is maximum.
 
  • #20
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I think it would be a good idea to put it all together neatly in one post.
Equation given for plank -
x = A cos(ωt)
Differentiating with respect to t, twice
a = -Aω² cos(ωt)
Torque will be maximum when acceleration is maximum to prevent slipping.
Maximum value of a = Aω²
At a particular time t,
a + Rα = acceleration of plank
Because there is no slipping.
Here a = acceleration of cylinder
⇒a + Rα = Aω² ---- (1)
Using τ = Iα
fR = MR²/2 (α)
where f = friction
α = 2f/MR ---- (2)
Also f = Ma ---- (3)
Using equations (1), (2) and (3)
f/M + 2f/M = Aω²
⇒f/M = Aω²/3 = a ----(4)
Using equations (4) and (1),
α = (Aω² - Aω²/3)/R = 2Aω²/3R
Again, τ = Iα
τ = MR²/2 (2Aω²/3R)
= 1/3 (MRAω²)
Is this correct?
 
  • #21
TSny
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Yes, that looks good.
a = -Aω² cos(ωt)
Torque will be maximum when acceleration is maximum to prevent slipping.
OK, if this is clear to you before any calculations are done.

Or, you could use a = -Aω² cos(ωt) throughout the calculation to end up with τ =(1/3) MRAω²cos(ωt). So, you have the torque at any time and you can then clearly see that the maximum torque will be (1/3) MRAω² and that this occurs when the plank has its maximum acceleration.

Good work.
 
  • #22
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Or, you could use a = -Aω² cos(ωt) throughout the calculation to end up with τ =(1/3) MRAω²cos(ωt). So, you have the torque at any time and you can then clearly see that the maximum torque will be (1/3) MRAω² and that this occurs when the plank has its maximum acceleration.
Okay. Thank you Sir for helping me.
 
  • #23
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Thank you Haruspex Sir for helping once again.
 

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