A uniform disc of mass M and radius R is supported vertically by a pivot at its periphery as shown. A particle of mass M is fixed to the rim and raised to the highest point above the centre. The system is released from rest and it can rotate about pivot freely. The angular speed of the system when it attached object is directly beneath the pivot, is

Question

A uniform disc of mass M and radius R is supported vertically by a pivot at its periphery as shown. A particle of mass M is fixed to the rim and raised to the highest point above the centre. The system is released from rest and it can rotate about pivot freely. The angular speed of the system when it attached object is directly beneath the pivot, is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-jrv13hdz74kp.png" /> (A) √(24 g/11 R) (B) √(8 g/11 R) (C) √(8 g/3 R) (D) √(14 g/3 R)

Tardigrade Admin · Accepted Answer

M g(2 R)+M g(4 R)=(1/2)((3/2) M R2+M(2 R)2) ω2 ⇒ 6 M g R=(11/4) M R2 ω2 ⇒ ω=√(24 g/11 R)

$\frac{24 g}{11 R}$

$\frac{8 g}{11 R}$

$\frac{8 g}{3 R}$

$\frac{14 g}{3 R}$

$M g (2 R) + M g (4 R) = \frac{1}{2} (\frac{3}{2} M R^{2} + M (2 R)^{2}) ω^{2}$
$\Rightarrow 6 M g R = \frac{11}{4} M R^{2} ω^{2}$
$\Rightarrow ω = \frac{24 g}{11 R}$

11R24g​​

11R8g​​

3R8g​​

3R14g​​

Mg(2R)+Mg(4R)=21​(23​MR2+M(2R)2)ω2 ⇒6MgR=411​MR2ω2 ⇒ω=11R24g​​

$\frac{24 g}{11 R}$

$\frac{8 g}{11 R}$

$\frac{8 g}{3 R}$

$\frac{14 g}{3 R}$

$M g (2 R) + M g (4 R) = \frac{1}{2} (\frac{3}{2} M R^{2} + M (2 R)^{2}) ω^{2}$
$\Rightarrow 6 M g R = \frac{11}{4} M R^{2} ω^{2}$
$\Rightarrow ω = \frac{24 g}{11 R}$