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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?

  • $\begingroup$ 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

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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