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JEE Advanced 2007Paper 1
Exam
JEE Advanced
Year
2007
Paper
Paper 1
Subject
Physics
Chapter
Rotational Motion

JEE Advanced 2007 · Paper 1 · Physics · Rotational Motion

Paragraph: P_ 17 - 19 : Paragraph for Questions Nos. 17 to 19 Two...

Paragraph: P17 - 19\mathbf{P}_{17 \text{ - } 19} : Paragraph for Questions Nos. 17 to 19 Two discs AandandBaremountedcoaxiallyonaverticalaxle.ThediscshavemomentsofinertiaIand2I,respectivelyaboutthecommonaxis.DiscAare mounted coaxially on a vertical axle. The discs have moments of inertia I and 2I, respectively about the common axis. Disc A is imparted an initial angular velocity 2ω2 \omega using the entire potential energy of a spring compressed by a distance x1x_1. Disc BB is imparted an angular velocity ω\omega by a spring having the same spring constant and compressed by a distance x2x_2. Both the discs rotate in the clockwise direction.Question: When disc BB is brought in contact with disc A,theyacquireacommonangularvelocityintime, they acquire a common angular velocity in timet.Theaveragefrictionaltorqueononediscbytheotherduringthisperiodis. The average frictional torque on one disc by the other during this period is

Options

  1. A2Iω3t\frac{2 I \omega}{3 t}
  2. B9Iω2t\frac{9 I \omega}{2 t}
  3. C91ω4t\frac{91 \omega}{4 t}
  4. D3Iω2t\frac{3 I \omega}{2 t}

Correct answer

A. 2Iω3t\frac{2 I \omega}{3 t}

Step-by-step solution

Let ω′\omega^{\prime} be the common velocity. Then from conservation of angular momentum, we have (i+2I)ω′=I(2ω)+2I(ω)ω′=43ω\begin{aligned} (i+2 I) \omega^{\prime}& =I(2 \omega)+2 I(\omega)\\ \omega^{\prime}& =\frac{4}{3} \omega \end{aligned}From the equation, Angular impulse = change in angular momentum for any of the disc, we have \taut=I(2ω)−I(43ω)=2Iω3τ=2Iω3t\begin{aligned} \taut & =I(2 \omega)-I\left(\frac{4}{3} \omega\right)=\frac{2 I \omega}{3}\\ \tau & =\frac{2 I \omega}{3 t} \end{aligned}\therefore$ Option (a) is correct.
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