Question

A cord of negligible mass is wrapped around the outer surface of a 5 kg disk of radius 0.5 m as shown. If the disk is released from rest, the angular velocity of the disk (in rad/s) after 3 seconds is
(acceleration due to gravity, g = 10 m/s${}^2$)

• A. 30
• B. 40
• C. 60
• D. 120

Answer 1

The correct option is B 40

Given data:
${\omega }_o=0$

$\therefore \omega ={\omega }_o+\alpha t=\alpha t$-------(1)

$\text{Torque,}T=I\alpha$
[P: Tension in the string, T : torque]

$\therefore P\times r=I\alpha$ --------(2)

Consider the downward motion of disk
Applying De Alemberts principle

$ma=mg-P$

$P=mg-ma=m[g-a]$ $[\because a=r\alpha ]$

$P=m(g-r\alpha )------\left(3\right)$

Substituting equation (3) in (2)

$m[g-r\alpha ]\times r=I\alpha$

$mgr-m{r}^2\alpha =I\alpha$

$\therefore \alpha =\frac{mgr}{1+m{r}^2}=\frac{mgr}{\frac{m{r}^2}{2}+m{r}^2}=\frac{2mgr}{3m{r}^2}$

$\alpha =\frac{2g}{3r}----------\left(4\right)$

Substitute (4) in (1)

$\omega =\alpha t=\frac{2gt}{3r}=\frac{2\times 10\times 3}{3\times 0.5}=40rad/sec$

(or) Another method


$I_P\alpha =mg.r$

$\frac{3}{2}m{r}^2\alpha =mgr$

$\alpha =\frac{2g}{3r}$

$\omega =\alpha t$

$\omega =\alpha t=\frac{2gt}{3r}=\frac{2\times 10\times 3}{3\times 0.5}=40rad/sec$