Question

A rod of length Land negligible mass is attached to a uniform disk of mass Mand radius R (Figure 9.64). A string is wrapped around the disk, and you pull on the string with a constant force F . Two small balls each of mass mslide along the rod with negligible friction. The apparatus starts from rest, and when the center of the disk has moved a distance d, a length of string shas come off the disk, and the balls have collided with the ends of the rod and stuck there. The apparatus slides on a nearly frictionless table. Here is a view from above:

(a) At this instant, what is the speed vof the center of the disk? (b) At this instant the angular speed of the disk is$\mathit{\omega }$ . How much internal energy change has there been?

Short answer

a) $\sqrt{\frac{2\times F.d}{M+2m}}$

b) $F_s-\frac{1}{4}\left(M{R}^2+mL\right){\omega }^2$

Step-by-step solution

Identification of the given data

The given data can be listed below as-

• The mass of the uniform disk is, M .
• The length of the rod is, L.
• The radius of the uniform disk is, R .
• The mass of the two small balls is, m.
• The force needed to pull the string is, F .
• The distance that the center of the disk has moved is, d .
• The additional length of the string is, s .

Significance of the work-energy theorem for the string

This theorem illustrates that the change in the kinetic energy of an object is equal to the net work done on that object.

The expression for the work energy theorem is given as follows,

$W=∆K.E$

…(1)

Here, . W is the work done and $\mathit{∆}K\mathit{.}E$ is the change in kinetic energy.

Determination of the speed of the center of the apparatus

(a)

The expression for work done is given as follows,

$W=F.d$

Here, F is the force applied to the body and d is the displacement of the object due to the applied force.

The expression for the change in kinetic energy is as follows,

$∆K.E.=-\frac{1}{2}\left(M+2m\right)v^2$

Substitute all the values in equation (1).

$$\begin{gathered} F.d=\frac{1}{2}\times \left(M+2m\right)\times v^2 \\ v=\sqrt{\frac{2Fd}{M+2m}} \end{gathered}$$

Thus, the speed of the center of the apparatus is $v=\sqrt{\frac{2Fd}{M+2m}}$.

Determination of the the angular speed of the apparatus

b)

From the work-energy theorem, the external work done is expressed as,

$w_{ext}=∆K_{trans}+∆K_{rot}+∆E_{thermal}$

...(2)

Here, $w_{ext}$is the external work done which is equal to the twice the force needed to pull the string, $∆K_{trans}$is the translational kinetic energy, $∆K_{rot}$ is the rotational kinetic energy and $∆E_{thermal}$is the internal energy.

The expression of rotational kinetic energy is as follows,

$∆K_{rot}=\frac{1}{2}\left(\frac{1}{2}M{R}^2+2m{\left(\frac{L}{2}\right)}^2\right){\omega }^2$

Here, $\omega$is the angular speed.

Substitute all the values in equation (2).

$$\begin{gathered} F+F=F+\frac{1}{2}\left(\frac{1}{2}M{R}^2+2m{\left(\frac{L}{2}\right)}^2\right){\omega }^2+∆E_{internal} \\ ∆E_{internal}=F-\frac{1}{4}\left(M{R}^2+mL\right){\omega }^2 \end{gathered}$$

For$F=F_s$,

$∆E_{internal}=F_s-\frac{1}{4}\left(M{R}^2+m{L}^2\right){\omega }^2$

Thus, the change in the internal energy is $F_s-\frac{1}{4}\left(M{R}^2+m{L}^2\right){\omega }^2$.