A thin uniform circular disc of mass M and radius R is rotating in a horizontal plane ahout an axis passing through its centre and perpendicular to the plane with an angular velocity ω Another disc of same mass hut half the radius is gently placed over it coaxially. The angular speed o f the composite disc will be

Question

A thin uniform circular disc of mass M and radius R is rotating in a horizontal plane ahout an axis passing through its centre and perpendicular to the plane with an angular velocity ω Another disc of same mass hut half the radius is gently placed over it coaxially. The angular speed o f the composite disc will be (A) (5/4)ω (B) (4/5)ω (C) (2/5)ω (D) (5/2)ω

Tardigrade Admin · Accepted Answer

I1ω1=(I1+I2)ω2 [(MR2/2)]ω=[(MR2/2)+(M((R/2))2/2)]ω2 =[(MR2/2)+(MR2/8)]ω2=[(5MR2/8)] ω2=(4/5)ω

$\frac{5}{4} ω$

$\frac{4}{5} ω$

$\frac{2}{5} ω$

$\frac{5}{2} ω$

$I_{1} ω_{1} = (I_{1} + I_{2}) ω_{2}$
$[\frac{M R ^{2}}{2}] ω = [\frac{M R ^{2}}{2} + \frac{M ( \frac{R}{2} ) ^{2}}{2}] ω^{2}$
$= [\frac{M R ^{2}}{2} + \frac{M R ^{2}}{8}] ω_{2} = [\frac{5 M R ^{2}}{8}]$
$ω_{2} = \frac{4}{5} ω$

45​ω

54​ω

52​ω

25​ω

I1​ω1​=(I1​+I2​)ω2​ [2MR2​]ω=[2MR2​+2M(2R​)2​]ω2 =[2MR2​+8MR2​]ω2​=[85MR2​] ω2​=54​ω

$\frac{5}{4} ω$

$\frac{4}{5} ω$

$\frac{2}{5} ω$

$\frac{5}{2} ω$

$I_{1} ω_{1} = (I_{1} + I_{2}) ω_{2}$
$[\frac{M R ^{2}}{2}] ω = [\frac{M R ^{2}}{2} + \frac{M ( \frac{R}{2} ) ^{2}}{2}] ω^{2}$
$= [\frac{M R ^{2}}{2} + \frac{M R ^{2}}{8}] ω_{2} = [\frac{5 M R ^{2}}{8}]$
$ω_{2} = \frac{4}{5} ω$