A horizontal disc rotating freely about a vertical axis through its centre makes 90 revolutions per minute. A small piece of wax of mass m falls vertically on the disc and sticks to it at a distance r from the axis. If the number of revolutions per minute reduce to 60, then the moment of inertia of the disc is

Question

A horizontal disc rotating freely about a vertical axis through its centre makes 90 revolutions per minute. A small piece of wax of mass m falls vertically on the disc and sticks to it at a distance r from the axis. If the number of revolutions per minute reduce to 60, then the moment of inertia of the disc is (A) mr 2 (B) (3/2) m r2 (C) 2 mr 2 (D) 3 mr 2

Tardigrade Admin · Accepted Answer

w1=(2 π 90/60)=3 π rps w2=(2 π 60/60)=2 π rps Using angular momentum conservation I(1.5)=(I+m2)(1) (I/2)=m r2 I=2 mr 2

$m r^{2}$

$\frac{3}{2} m r^{2}$

$2 m r^{2}$

$3 m r^{2}$

$w_{1} = \frac{2 π 90}{60} = 3 π r p s$
$w_{2} = \frac{2 π 60}{60} = 2 π r p s$
Using angular momentum conservation
$I (1.5) = (I + m^{2}) (1)$
$\frac{I}{2} = m r^{2}$
$I = 2 m r^{2}$

mr2

23​mr2

2mr2

3mr2

w1​=602π90​=3πrps w2​=602π60​=2πrps Using angular momentum conservation I(1.5)=(I+m2)(1) 2I​=mr2 I=2mr2

$m r^{2}$

$\frac{3}{2} m r^{2}$

$2 m r^{2}$

$3 m r^{2}$

$w_{1} = \frac{2 π 90}{60} = 3 π r p s$
$w_{2} = \frac{2 π 60}{60} = 2 π r p s$
Using angular momentum conservation
$I (1.5) = (I + m^{2}) (1)$
$\frac{I}{2} = m r^{2}$
$I = 2 m r^{2}$