A rod of mass m and length 2R can rotate about an axis passing through O in vertical plane. A disc of mass m and radius R is hinged to the other end P of the rod and can freely rotate about P. When di

Question

A rod of mass m and length 2R can rotate about an axis passing through O in vertical plane. A disc of mass m and radius R is hinged to the other end P of the rod and can freely rotate about P. When disc is at lowest point both rod and disc have angular velocity $ω$ . If the rod rotates by maximum angle $θ = 6 0^{\circ}$ with downward vertical, the $ω$ in terms of R and g will be (all hinges are smooth) :-

TestHub · Accepted Answer

The disc is free to rotate about P, so no torque acts on it about its centre. Its spin ω stays constant and its spin kinetic energy is the same at both positions, so it cancels. Only the disc's translation with P enters the energy equation

KE at the lowest point:

$\frac{1}{2} [\frac{m ( 2 R ) ^{2}}{3} + m (2 R)^{2}] ω^{2} = \frac{8}{3} m R^{2} ω^{2}$

Rise of disc centre: 2R(1 - cos60°) = R, rise of rod centre: R/2

$\frac{8}{3} m R^{2} ω^{2} = m g R + m g \frac{R}{2}$

$ω = \frac{9 g}{16 R}$

Answer: A

Physics - Rotation Question with Solution | TestHub

Options:

Doubts & Discussion