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 (a) t = 1 s (b) t = 0.5 s (c) t = 0.25 s (d) t = 2 s

Answer: (a). Torque is zero when angular acceleration is zero. ω = (dθ)/(dt) = 6t² - 12t, α = (dω)/(dt) = 12t - 12 α = 0 ⇒ t = 1 s So the answer is (a) t = 1 s.

The instantaneous angular position of a point on a rotating wheel is… | NEET 2011 (AIPMT, Prelims)

Q72NEET 2011 · AIPMT · PRELIMSMedium

Rotational Kinematics & Dynamics

The instantaneous angular position of a point on a rotating wheel is given by the equation $\theta(t) = 2t^3 - 6t^2$

The torque on the wheel becomes zero at

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Answer: (a)

Torque is zero when angular acceleration is zero.
$\omega = \frac{d\theta}{dt} = 6t^2 - 12t$ , $\alpha = \frac{d\omega}{dt} = 12t - 12$
$\alpha = 0 \Rightarrow t = 1\text{ s}$
So the answer is (a) $t = 1$ s.

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