 #1
zenterix
 61
 20
 Homework Statement:
 Consider a conveyor belt moving with a constant velocity ##\vec{v}_{g,c}## relative to the ground reference frame. A ball of mass ##m## and radius ##R## is placed on the conveyor belt initially at rest relative to the ground frame. What happens to the ball? Assume there is friction due to contact forces between ball and conveyor belt.
 Relevant Equations:

Because there is a relative velocity between the conveyor belt and the ball, there is kinetic friction ##f_k## (see picture below).
##f_k## accelerates the ball and causes rotation due to torque applied at contact point about center of mass (CM).
Torque equation
$$\vec{\tau}_{cm,ball}=Rf_k\hat{k}=I_{cm}\alpha_{cm}\hat{k}$$
$$\implies \vec{\alpha}_{cm}=\frac{Rf_k}{I_{cm}}\hat{k}$$
##2^{nd}## Law Equation
$$f_k\hat{i}=ma_{g,cm}\hat{i}$$
$$\implies \vec{a}_{g,cm}=\frac{f_k}{m}\hat{i}$$
We can integrate ##\vec{\alpha}_{cm}## and ##\vec{a}_{g,cm}## to obtain angular and linear velocities
$$\vec{\omega}_{cm}=\frac{Rf_k t}{I_{cm}}\hat{k}$$
$$\vec{v}_{g,cm}=\frac{f_k}{m}t \hat{k}$$
Here is my depiction of the initial state:
Note that the presence of ##f_k## means the ball is initially slipping. We also know that the linear and angular speeds of the ball are increasing in time. At some point, the ball should stop slipping.
The condition for no slipping is that the speed of the ball relative to the conveyor belt be the same as the arc length speed of rotation, ie
$$\vec{v}_{c,cm}=\vec{r}_{c,cm} \times \omega_{cm}=R\hat{j} \times \omega_{cm}\hat{k}$$
$$\implies \vec{v}_{c,cm}=R\omega_{cm}\hat{i}$$
Notice the negative sign means that when not slipping, the ball's CM is still moving to the left relative to the conveyor belt.
But we also know
$$\vec{v}_{g,cm}=\vec{v}_{g,c}+\vec{v}_{c,cm}$$
$$\implies \vec{v}_{c,cm}=(\frac{f_k}{m}tv_{g,c})\hat{i}$$
We see that initially, the relative speed of the ball's CM to the conveyor belt is just the negative of the speed of the conveyor belt, ##v_{g,c}##, but because the ball is accelerating, this relative speed becomes less negative in time. At some point it must reach the relative speed associated with no slipping, ##R\omega_{cm}##.
Recall that
$$\vec{\omega}_{cm}=\frac{Rf_k t}{I_{cm}}\hat{k}$$
So when we equate our two expressions for ##\vec{v}_{c,cm}## we have
$$\frac{f_k}{m}tv_{g,c} = \frac{Rf_k t}{I_{cm}}$$
$$\implies t _{ns}= \frac{v_{g,c}}{\frac{f_k}{m}+\frac{f_k R^2}{I_{cm}}}$$
At ##t_{ns}## the point of contact between ball and conveyor belt is at rest and there is no friction any longer (kinetic nor static).
This represents a final state of static equilibrium.
Is this analysis correct?
Note that the presence of ##f_k## means the ball is initially slipping. We also know that the linear and angular speeds of the ball are increasing in time. At some point, the ball should stop slipping.
The condition for no slipping is that the speed of the ball relative to the conveyor belt be the same as the arc length speed of rotation, ie
$$\vec{v}_{c,cm}=\vec{r}_{c,cm} \times \omega_{cm}=R\hat{j} \times \omega_{cm}\hat{k}$$
$$\implies \vec{v}_{c,cm}=R\omega_{cm}\hat{i}$$
Notice the negative sign means that when not slipping, the ball's CM is still moving to the left relative to the conveyor belt.
But we also know
$$\vec{v}_{g,cm}=\vec{v}_{g,c}+\vec{v}_{c,cm}$$
$$\implies \vec{v}_{c,cm}=(\frac{f_k}{m}tv_{g,c})\hat{i}$$
We see that initially, the relative speed of the ball's CM to the conveyor belt is just the negative of the speed of the conveyor belt, ##v_{g,c}##, but because the ball is accelerating, this relative speed becomes less negative in time. At some point it must reach the relative speed associated with no slipping, ##R\omega_{cm}##.
Recall that
$$\vec{\omega}_{cm}=\frac{Rf_k t}{I_{cm}}\hat{k}$$
So when we equate our two expressions for ##\vec{v}_{c,cm}## we have
$$\frac{f_k}{m}tv_{g,c} = \frac{Rf_k t}{I_{cm}}$$
$$\implies t _{ns}= \frac{v_{g,c}}{\frac{f_k}{m}+\frac{f_k R^2}{I_{cm}}}$$
At ##t_{ns}## the point of contact between ball and conveyor belt is at rest and there is no friction any longer (kinetic nor static).
This represents a final state of static equilibrium.
Is this analysis correct?