- #1

Karol

- 1,380

- 22

## Homework Statement

A disc of radius R and mass M lays on a smooth table. a smooth channel is carved along the diameter and on both ends there are masses m. one of the masses is attached with a wire to a small, almost weightless motor that is in the center. the other mass remains fixed in place. the disc can rotate round it's center using a smooth nail.

At t=0 the disc rotates at ω

_{0}and the motor starts pulling the left mass with acceleration a

_{0}. what is the tension in the wire.

Now the nail is removed and at t=0 the disc again rotates at ω

_{0}and the motor running. find the distance s(t) of the center of mass of the system relative to the center.

What is the angular velocity of the disc as a function of time. use s(t).

What is the tension in the wire at t=0

## Homework Equations

Centrifugal force: ##F=m\frac{v^2}{R}##

Moment of inertia of a disc: ##I=\frac{1}{2}MR^2##

Distance with constant acceleration: ##s=\frac{1}{2}at^2##

## The Attempt at a Solution

The tension at t=0 with the nail is the centripetal force+force due to linear acceleration:

$$F=m(\omega_0^2R+a_0)$$

Moments round the left, moving, mass gives the C.O.M. distance s(t):

$$m\left( R+\frac{1}{2}at_0^2 \right)+M\left( \frac{1}{2}at_0^2 \right)=(M+m)\left( \frac{1}{2}at_0^2+s \right)$$

$$\rightarrow s(t)=\frac{a_0 t^2}{4}$$

The angular velocity ω without the nail is from conservation of angular momentum.

I consider, for the last term on the right side of the equation, the distance of each mass from the new C.O.M.

$$\frac{1}{2}M\omega_0^2 R^2+2mR^2\omega_0^2=\frac{1}{2}M\omega^2+2m\left( \frac{R+s(t)}{2} \right)^2$$

$$\omega_0^2 R^2 \left( \frac{M}{2}+2m \right)=\omega^2\left( \frac{M}{2}R^2+2m\left( \frac{R+s(t)}{2} \right)^2 \right)$$

I think the tension force at this situation is similar to the first when the nail was.