Question

A uniform circular disc of radius $50cm$ at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of $2.0rad{s}^{-2}$. Its net acceleration in ${ms}^{-2}$ at the end of $2.0s$ is approximately:

• A. $8.0$
• B. $7.0$
• C. $6.0$
• D. $3.0$

Answer 1

Answer: (A)

The correct option is A $8.0$
Given:

Radius, $R=0.5m$

Angular acceleration, $\alpha =2{rad/s}^2$

Time, $t=2s$

Assumption:

Acceleration asked is for a point on the rim.

After 2 s, Angular velocity is given by:

$\omega ={\omega }_o+\alpha t$

$\omega =0+2\times 2$

$\omega =4rad/s$

Centripetal Acceleration is given by, $a_c={\omega }^{2}R=4^2\times 0.5=8m/s^2$

Tangential Acceleration is given by, $a_t=R\alpha =0.5\times 2=1m/s^2$

Net acceleration, $a_{net}=\sqrt{a_c^2+a_t^2}$

$a_{net}=\sqrt{8^2+1^2}\approx 8m/s^2$