zenterix
- 774
- 84
- Homework Statement
- The ends of thin threads tightly wound on the axle of radius ##r## of the Maxwell disc are attached to a horizontal bar. When the disc unwinds, the bar is raised to keep the disc at the same height. The mass of the disc with the axle is ##m##, the moment of inertia of the arrangement relative to its axis is ##I##.
- Relevant Equations
- Find the tension of each thread and the acceleration of the bar.
Here is a pictorial depiction of the problem
From Newton's 2nd law we have
$$2T-mg=0\implies T=\frac{mg}{2}$$
Then, considering the torques created by the threads we have
$$\vec{\tau}=I\vec{\alpha}=(-r\hat{k}+\frac{l}{2}\hat{i})\times T\hat{j}+(-r\hat{k}-\frac{l}{2}\hat{i})\times T\hat{j}$$
$$=2rT\hat{i}$$
$$\implies \alpha_z=\frac{2rT}{I}=\frac{rmg}{I}$$
Then
$$\omega_z=\frac{rmgt}{I}$$
The velocity of the axle at the point of contact of the thread with the axle is
$$\vec{v}=\vec{\omega}\times\vec{r}=\frac{rmgt}{I}\hat{i}\times (-r\hat{k})=\frac{r^2mgt}{I}\hat{j}$$
This is a problem from the book "Problems in General Physics" by Irodov. The back of the book answer says that
$$\omega_0=\frac{gmr^2}{I}$$
Why is there no factor of ##t## in the expression above?
As far as I can tell, because there is a constant torque there is a constant angular acceleration. Thus, angular velocity and tangential velocity (which coincides with the velocity of the bar being raised) are increasing with time.
My question is if I have made a mistake in my reasoning.
From Newton's 2nd law we have
$$2T-mg=0\implies T=\frac{mg}{2}$$
Then, considering the torques created by the threads we have
$$\vec{\tau}=I\vec{\alpha}=(-r\hat{k}+\frac{l}{2}\hat{i})\times T\hat{j}+(-r\hat{k}-\frac{l}{2}\hat{i})\times T\hat{j}$$
$$=2rT\hat{i}$$
$$\implies \alpha_z=\frac{2rT}{I}=\frac{rmg}{I}$$
Then
$$\omega_z=\frac{rmgt}{I}$$
The velocity of the axle at the point of contact of the thread with the axle is
$$\vec{v}=\vec{\omega}\times\vec{r}=\frac{rmgt}{I}\hat{i}\times (-r\hat{k})=\frac{r^2mgt}{I}\hat{j}$$
This is a problem from the book "Problems in General Physics" by Irodov. The back of the book answer says that
$$\omega_0=\frac{gmr^2}{I}$$
Why is there no factor of ##t## in the expression above?
As far as I can tell, because there is a constant torque there is a constant angular acceleration. Thus, angular velocity and tangential velocity (which coincides with the velocity of the bar being raised) are increasing with time.
My question is if I have made a mistake in my reasoning.