First slide

Dynamics of rotational motion about fixed axis

Question

The instantaneous angular position of a point on a rotating wheel is given by the equation $θ$ (t) = 2t³ $-$ 6t². The torque on the wheel becomes zero at [CBSE AIPMT 2011]

Moderate

A

t = 0.5 s
B

t = 0.25 s
C

t = 2 s
D

t = 1 s

Solution

According to question,
torque, $τ = I α = 0$
$\Rightarrow α = 0 or α = \frac{d^{2} θ}{d t^{2}} = 0$
Given, $θ (t) = 2 t^{3} - 6 t^{2}$
So, $\frac{d θ}{d t} = 6 t^{2} - 12 t$
$\Rightarrow \frac{d^{2} θ}{d t} = 12 t - 12$ $(∵ α = \frac{d^{2} θ}{d t^{2}} = 0)$
$12 t - 12 = 0 \Rightarrow t = 1 s$

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