Question

A disc of mass M radius R moves under gravity while unwinding string under non slip condition. Mark the correct statement(s):

$\square$ Acceleration of the disc is $\frac{2g}{3}$

$\square$ Rate of ($\overrightarrow{L}$) change of angular momentum of disc $\frac{d\overrightarrow{L}}{dt}$ about a point along string is in $-\hat{k}$ direction.

$\square$ Angular momentum of disc is conserved about centre of mass.

$\square$ Rate of change of angular momentum $\frac{d\overrightarrow{L}}{dt}$ of disc about center of mass is in $+\hat{k}$ direction.

• A. Acceleration of the disc is $\frac{2g}{3}$
• B. Rate of ($\overrightarrow{L}$) change of angular momentum of disc $\frac{d\overrightarrow{L}}{dt}$ about a point along string is in $-\hat{k}$ direction.
• C. Angular momentum of disc is conserved about centre of mass.
• D. Rate of change of angular momentum $\frac{d\overrightarrow{L}}{dt}$ of disc about center of mass is in $+\hat{k}$ direction.

Answer 1

Answer: (A), (B)

Acceleration of the disc is $\frac{2g}{3}$ Rate of ($\overrightarrow{L}$) change of angular momentum of disc $\frac{d\overrightarrow{L}}{dt}$ about a point along string is in $-\hat{k}$ direction.

Answer 2

Solution Explanation:

1. Acceleration:
   For a disk of mass M and radius R, using Newton’s second law and the no‐slip condition (a = αR) together with the moment of inertia $I_{CM}=\frac{1}{2}MR^2$, one obtains:

$$mg-T = Ma \quad \text{and} \quad T R = I_{CM}\alpha = \frac{1}{2}MR^2\frac{a}{R}.$$

This gives $T=\frac{1}{2}Ma$ and, substituting in the first equation,

$$mg-\frac{1}{2}Ma = Ma \quad\Rightarrow\quad mg=\frac{3}{2}Ma,\quad\text{so}\quad a=\frac{2}{3}g.$$

Thus the first statement “Acceleration of the disc is $\frac{2g}{3}$” is correct.

2. Angular momentum about a point on the string:
   Choose a fixed point on the (vertical) string. By using the parallel‐axis theorem, the moment of inertia about this point is

$$I_A=I_{CM}+MR^2=\frac{1}{2}MR^2+MR^2=\frac{3}{2}MR^2.$$

The no–slip condition gives $\alpha=a/R=\frac{2g}{3R}$. Hence the rate of change of angular momentum about that point is

$$\frac{dL}{dt}=I_A\alpha=\frac{3}{2}MR^2 \cdot \frac{2g}{3R}=MgR.$$

Since the disk rotates clockwise (when viewed in the plane of motion) its angular acceleration (and therefore $\frac{dL}{dt}$) points into the page, which we denote as $-\hat{k}$. Thus statement 2 is correct.

3. Angular momentum about center of mass:
   There is a nonzero torque about the center of mass (due to the tension force acting at the rim) so the angular momentum about the CM is not conserved. Hence statement 3 is false.

4. Direction of $\frac{d\overrightarrow{L}}{dt}$ about CM:
   As just noted, the torque (and thus $\frac{dL}{dt}$) about the center of mass is in the same sense as the angular acceleration (clockwise), i.e. $-\hat{k}$, not $+\hat{k}$. Hence statement 4 is false.

Answer:
Only the first two statements are correct.