Question

Tape is wound around a disk shaped object which is at rest On a frictionless surface. Assume the disk is homogeneous, has mass M and radius R. The free end of tape is pulled horizontally with a force F. Find an expression for horizontal distance d covered by the disk as a function of length l of the tape unwound. Ignore the mass the tape as compared with the mass of the disk.

Answer 1

Moment of inertia of the solid disk:

$$I=\frac{MR^2}2.$$

When we pull the tape, it creates a torque:

$$\tau=FR.$$

This torque creates angular acceleration:

$$\alpha=\frac{\tau}{I}=\frac{2F}{MR}$$

This angular acceleration provides tangential acceleration:

$$a_\tau=\alpha R=\frac{2F}{M}.$$

As the tape unwinds, the disk rotates for some angle $\theta$. If the length of the tape unwound is L, the angle corresponds to

$$\theta=\frac{L}{\pi}.$$

For the disk to unwind for angle $\theta$ (from bottom to top are the surface, disk, and tape), it will take a time of

$$t=\sqrt{\frac{2\theta}{\alpha}}=\sqrt{\frac{2L}{\alpha\pi}}=\sqrt{\frac{LMR}{F\pi}}.$$

Meanwhile under the influence of linear acceleration

$$a_l=\frac{F}{M}$$

the disk will travel distance d for the same time:

$$t=\sqrt{\frac{2d}{a_l}}=\sqrt{\frac{2dM}{F}}.$$

Equate time expressed in terms of d and L:

$$\sqrt{\frac{2dM}{F}}=\sqrt{\frac{LMR}{F\pi}},\\\space\\ d(L)=\frac{R}{2\pi}L.$$