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I tried to solve a physics problem that involves a cylinder on a truck that is accelerating. I'm supposed to solve the acceleration of the center of mass of the cylinder when the acceleration of the truck, and the radius and the mass of the cylinder are known. The cylinder rolls without slipping.

In the solution it is said that as the rolling occurs without slipping, the point on the cylinder that is in contact with the truck accelerates at the same rate as the truck. Why is this true? I can understand that their speed should be the same as they are momentarily in contact, but why their accelerations?

I've been taught that if an object is under forces, regardless of the positions of the forces wrt. the center of mass, the acceleration of the COM is the sum of these forces. So if one point on the cylinder accelerates at the same rate as the truck, why doesn't the COM of the cylinder accelerate also at the same rate as the truck?

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  • $\begingroup$ If an object rotates then not all points move with the same velocity. If the COM and the contact point have different velocities they also have different accelerations. $\endgroup$ – ja72 Apr 8 at 13:55
  • $\begingroup$ I'm curious. What was it about my answer that you had trouble accepting? $\endgroup$ – Chet Miller Apr 9 at 14:31
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The place to start on this problem is with the kinematics. Let v represent the velocity of the CM of the cylinder, and let $\omega$ represent the counterclockwise rate of rotation of the cylinder. Then the tangential velocity at the bottom of the cylinder is $$v_T=v+\omega R$$ where R is the radius of the cylinder. Since the cylinder does not slip relative to the truck bed, this tangential velocity of the cylinder must match the velocity of the truck bed at all times. From this it follows that the acceleration of the cylinder a, the angular acceleration of the cylinder $\alpha$, and the acceleration of the truck $a_T$ must be related by $$a+\alpha R=a_T$$

Now for the mechanics: If F represents the forward frictional force exerted by the truck bed on the cylinder, F is responsible for both the linear acceleration of the center of mass and also for the angular acceleration. Therefore, $$a=\frac{F}{m}$$and $$\alpha=\frac{FR}{\frac{1}{2}mR^2}$$Combining the above equations, we have $$\frac{3F}{m}=a_T$$So, the tangential force is: $$F=m\frac{a_T}{3}$$ From this, it follows that the acceleration of the center of mass is $$a=\frac{a_T}{3}$$

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If their velocities are the same, their accelerations will be the same. This comes from the relationship between acceleration and velocity: $$ \frac{d\mathbf{v}}{dt} = \mathbf{a} $$ If the same $ \mathbf{v} $ appears in this relationship for both the truck and the point on the cylinder at all moments in time, then the $ \mathbf{a} $ must be the same for both.

Given the linear acceleration of the point on the outside of the cylinder is $ \mathbf{a}$, the angular acceleration will be $ \mathbf{a}/r $. The only way this is possible is if the truck exerts a torque $$ \tau = rf = \frac{1}{2} m r^{2} a / r = \frac{1}{2}mra \implies f = \frac{1}{2} m a $$

That is, a frictional force of $ \frac{1}{2}ma $ is required to keep the cylinder from slipping. Accordingly, the acceleration of the cylinder is $ f = ma_{cyl} \implies a_{cyl} = \frac{1}{2} a $ (where the second $ a $ is the truck's acceleration).

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  • $\begingroup$ Another way of expressing this is to say that the truck exerts a torque on the cylinder, but not a force. This is incorrect. Torque is just a property of a force. You can't have a torque without a force. The acceleration of the COM arises from a net force on the object and doesn't depend on where the force is applied $\endgroup$ – Aaron Stevens Apr 8 at 2:34
  • $\begingroup$ Sorry for initial clumsy mistake. Now can you find the mistake in my edited post? (there must be one since I get a different answer from you) $\endgroup$ – Hunter Akins Apr 8 at 5:44
  • $\begingroup$ Your mistake is in assuming that the center of mass moves $2\pi r$ for every $2\pi$ radians it rotates. This is true for the distance it rolls along the truck bed, but not the distance it moves relative to the ground. You could argue that the cylinder is rolling "with slipping" relative to the ground (although there isn't any actual slipping due to the lack of contact with the ground). Furthermore, the acceleration of the COM needs to be in the same direction as the acceleration of the truck, since friction is acting in this direction $\endgroup$ – Aaron Stevens Apr 8 at 13:22
  • $\begingroup$ Mm, so if this is true for the distance it rolls with respect to the truck bed, and the truck bed is moving at v_truck, won't the cylinder stay stationary since it will exactly cancel out the trucks motion? $\endgroup$ – Hunter Akins Apr 8 at 16:00
  • $\begingroup$ No, because that would mean there is no net force acting on the cylinder, which is not true since we know friction is the only horizontal force acting on the cylinder $\endgroup$ – Aaron Stevens Apr 8 at 16:12

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