# Cylinder lying on conveyor belt

• JD_PM
In summary: I was thinking. I was trying to apply that conservation of energy equation and it wasn't working, and I was getting confused by the fact that I was adding scalar values, and now I realize I was adding components of a vector, and it just cascaded from there. I'll try to get back to this when I have a clearer mind. Thanks.In summary, the problem involves a bottle of water being placed on a moving conveyor belt and can be approached as a single cylinder with radius ##R##, mass ##M##, and moment of inertia ##I = M R^2##. The velocity of the belt is given as ##V(t)## and it is assumed that the cylinder has a uniform mass distribution
JD_PM said:
Mmm true but I think that is due to not considering an initial velocity on the cylinder. If we were to consider one different from zero, we would get an extra term that would make ##v_{CM}## different from zero even when ##V(t) = 0##
The problem specifies that the bottle (cylinder) and the belt are initially at rest.

haruspex said:
And not quite over. The answer you got in post #65 was correct, but it assumed constant acceleration. You need to get the same answer without that assumption.

In post #63, the signs on ω in the first two equations were inconsistent. In post #67 you have changed the sign on both occurrences of ω, so they are still inconsistent.

Once you have those two equations right, you don't need the third equation in those two posts. In fact, I am not sure how you arrived at it. Just eliminate ω between the first two.

Yeah the following equation is wrong (algebraic mistake):

$$v_{CM} = \frac{1}{2}[-(\frac{k^2}{R^2} + 1) R\omega + V(t)]$$

Then, combining the following two equations so as to solve for ##v_{CM}## one gets the same result than #65:

$$v_{CM} = V(t) - R \omega$$

$$v_{CM} = - \frac{k^2 \omega}{R}$$

$$v_{cm} = \frac{Vk^2}{k^2 + R^2}$$

JD_PM said:
Yeah the following equation is wrong (algebraic mistake):

$$v_{CM} = \frac{1}{2}[-(\frac{k^2}{R^2} + 1) R\omega + V(t)]$$

Then, combining the following two equations so as to solve for ##v_{CM}## one gets the same result than #65:

$$v_{CM} = V(t) - R \omega$$

$$v_{CM} = - \frac{k^2 \omega}{R}$$

$$v_{cm} = \frac{Vk^2}{k^2 + R^2}$$
Now you have arrived.

JD_PM
haruspex said:
Now you have arrived.

The problem has been solved.

However I am still interested in the approach @kuruman suggested on 36#; solving the problem from a non-inertial reference frame. I guess this approach is based on the second Newton's Law for rotation:

$$f - F = I \alpha$$

Where ##f## is the friction and ##F## is a fictitious force.

It's not much different from what you did. The acceleration of the belt is ##\frac{dV}{dt}##. In the non-inertial frame it manifests itself as a fictitious force ##ma## acting on the CM in the opposite direction as the acceleration of the belt. The torque equation about the point of contact on the belt gives ##MaR=(Mk^2+MR^2)\alpha## (note the use of he parallel axis theorem). Let ##a'_{cm}## be the acceleration of the CM in the non-inertial frame. Then ##a'_{cm}=\alpha/R##. Put this back in the torque equation, cancel the masses and get $$a'_{cm}=\frac{R^2}{k^2+R^2}a=\frac{R^2}{k^2+R^2}\left(-\frac{dV}{dt}\right).$$ The negative sign is introduced because ##a'_{cm}## and ##\frac{dV}{dt}## are in opposite directions. In the inertial frame, the acceleration of the CM is $$a_{cm}=a+a'_{cm}=\frac{dV}{dt}-\frac{R^2}{k^2+R^2}\frac{dV}{dt} =\frac{k^2}{k^2+R^2}\frac{dV}{dt}$$The velocity of the CM is$$v_{cm}(t)=\int a_{cm}dt=\frac{k^2}{k^2+R^2}\int \frac{dV}{dt}dt=\frac{k^2}{k^2+R^2}V(t).$$I am not convinced that this is a better way to approach the problem however.

JD_PM
kuruman said:
It's not much different from what you did. The acceleration of the belt is ##\frac{dV}{dt}##. In the non-inertial frame it manifests itself as a fictitious force ##ma## acting on the CM in the opposite direction as the acceleration of the belt. The torque equation about the point of contact on the belt gives ##MaR=(Mk^2+MR^2)\alpha## (note the use of he parallel axis theorem). Let ##a'_{cm}## be the acceleration of the CM in the non-inertial frame. Then ##a'_{cm}=\alpha/R##. Put this back in the torque equation, cancel the masses and get $$a'_{cm}=\frac{R^2}{k^2+R^2}a=\frac{R^2}{k^2+R^2}\left(-\frac{dV}{dt}\right).$$ The negative sign is introduced because ##a'_{cm}## and ##\frac{dV}{dt}## are in opposite directions. In the inertial frame, the acceleration of the CM is $$a_{cm}=a+a'_{cm}=\frac{dV}{dt}-\frac{R^2}{k^2+R^2}\frac{dV}{dt} =\frac{k^2}{k^2+R^2}\frac{dV}{dt}$$The velocity of the CM is$$v_{cm}(t)=\int a_{cm}dt=\frac{k^2}{k^2+R^2}\int \frac{dV}{dt}dt=\frac{k^2}{k^2+R^2}V(t).$$I am not convinced that this is a better way to approach the problem however.
Reading that made me realize that in non-inertial frames it is valid to deal with fictitious changes in momentum. If the frame has acceleration a=a(t) over an interval then conservation of momentum works by adding the "fictitious" term mΔv=m∫a.dt.
Something similar may be possible for angular momentum, and if such a rule were taken as standard you could avoid the integration step above.

kuruman said:
The acceleration of the belt is ##\frac{dV}{dt}##.
Aren't you assuming constant acceleration with this statement?

JD_PM said:
Aren't you assuming constant acceleration with this statement?
No, that is a pefectly general equation.

haruspex said:
Reading that made me realize that in non-inertial frames it is valid to deal with fictitious changes in momentum. If the frame has acceleration a=a(t) over an interval then conservation of momentum works by adding the "fictitious" term mΔv=m∫a.dt.
Something similar may be possible for angular momentum, and if such a rule were taken as standard you could avoid the integration step above.
Yes, having to do the integral is exactly why I am not convinced that my method is a better way to approach the problem. It has the advantage that it is based on a "snapshot" FBD and does not require justifying conservation of momentum over an interval Δt but, overall, I think it is lengthier.

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