System Of Particles And Rotational Motion Ques 35
35. The instantaneous angular position of a point on a rotating wheel is given by the equation $\theta(t)=2 t^{3}-6 t^{2}$. The torque on the wheel becomes zero at
[2011]
(a) $t=1 s$
(b) $t=0.5 s$
(c) $t=0.25 s$
(d) $t=2 s$
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Answer:
Correct Answer: 35.(a)
Solution:
- (a) $\theta(t)=2 t^{3}-6 t^{2}$
$ \Rightarrow \frac{d \theta}{dt}=6 t^{2}-12 t \Rightarrow \alpha=\frac{d^{2} \theta}{dt^{2}}=12 t-12=0 $
$ \therefore \quad t=1 sec . $
When angular acceleration $(\alpha)$ is zero then torque on the wheel becomes zero.
BETA
