Question

Question .29
M.M 4.00 N.M 1.00
A circular disc of mass 1 kg and radius 20 cm is
rolling down a string without slipping as shown in
figure. If acceleration due to gravity at that place is 9
m/s2, then the linear acceleration of the centre of
mass of the disc is​

Answer 1

Given:

Circular disc of mass 1 kg and radius 20 is rolling down a string. Acceleration at that place is 9m/s².

To find:

Acceleration of centre of mass of disc.

Calculation:

As per Free-Body diagram of disc;

$mg - T = ma \: \: \: \: .......(1)$

We can see that torque to the disc is provided by Tension in the wire ;

$\therefore \: \tau = I \times \alpha$

$= > \: T \times r = I \times \alpha$

$= > \: T \times r = I \times \dfrac{a}{r}$

$= > \: m(g - a)\times r = I \times \dfrac{a}{r}$

$= > \: m(g - a)\times r = \dfrac{m {r}^{2} }{2} \times \dfrac{a}{r}$

$= > \: \cancel{m}(g - a)\times \cancel{ {r}^{2} } = \dfrac{ \cancel{m {r}^{2}} }{2} \times a$

$= > 2g - 2a = a$

$= > 3a = 2g$

$= > a = \dfrac{2g}{3}$

$= > a = \dfrac{2 \times 9}{3}$

$= > a = 6 \: m {s}^{ - 2}$

So, final answer is

$\boxed{ \red{ \bold{ a = 6 \: m {s}^{ - 2} }}}$