Cylinder lying on conveyor belt

haruspex

Homework Helper
Gold Member
2018 Award
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 realise 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.

JD_PM

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

haruspex

Homework Helper
Gold Member
2018 Award
Aren't you assuming constant acceleration with this statement?
No, that is a pefectly general equation.

kuruman

Homework Helper
Gold Member
Reading that made me realise 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.

"Cylinder lying on conveyor belt"

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving