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NEET 2011
Exam
NEET
Year
2011
Subject
Physics
Chapter
Rotational Motion

NEET 2011 · Physics · Rotational Motion

The instantaneous angular position of a point on a rotating wheel...

The instantaneous angular position of a point on a rotating wheel is given by the equation Q(t)=2t3−6t2Q(t)=2 t^3-6 t^2 The torque on the wheel becomes zero at

Options

  1. At=0.5 st=0.5 \mathrm{~s}
  2. Bt=0.25 st=0.25 \mathrm{~s}
  3. Ct=2 st=2 \mathrm{~s}
  4. Dt=1 st=1 \mathrm{~s}

Correct answer

D. t=1 st=1 \mathrm{~s}

Step-by-step solution

According to question, torque, τ=0\tau=0 Its means that, α=0α=d2θdt2\alpha=0\alpha=\frac{d^2 \theta}{d t^2}Given So θ(\theta(t)=2 t³-6 t²dθdt=6t2−12td2θdt2=12t−12(∵α=d2θdt2=0)12t−12=0t=1 s\begin{aligned}\frac{d \theta}{d t}& =6 t²-12 t \\ \frac{d^2 \theta}{d t^2}& =12 t-12 \quad\left(\because \alpha=\frac{d^2 \theta}{d t^2}=0\right) \\ 12 t-12 & =0 \\ t & =1 \mathrm{~s} \end{aligned}
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