Question

The angle turned by a body undergoing circular motion depends on time as $\theta =(2t^2-{\theta }_0{t}^3)\text{rad}$, where ${\theta }_0$ is constant. Then the angular acceleration of the body at $t=2\text{s}$ is

• A. $-12{\theta }_0{\text{rad/s}}^2$
• B. $2-6{\theta }_0{\text{rad/s}}^2$
• C. $4-12{\theta }_0{\text{rad/s}}^2$
• D. $4-3{\theta }_0{\text{rad/s}}^2$

Answer 1

The correct option is C $4-12{\theta }_0{\text{rad/s}}^2$

Angular acceleration of the body is given as
$\alpha =\frac{d^2\theta }{d{t}^2}$
We have, $\frac{d\theta }{dt}=\frac{d(2t^2-{\theta }_0{t}^3)}{dt}=4t-3{\theta }_0{t}^2$
Similarly, $\frac{d^2\theta }{d{t}^2}=\frac{d(4t-3{\theta }_0{t}^2)}{dt}=4-6{\theta }_{0}t$
Thus, at $t=2\text{s}$ we have
$\alpha =4-6{\theta }_0\times 2=4-12{\theta }_0{\text{rad/s}}^2$