A thin uniform circular disc of mass M and radius R is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular velocity ω . Another disc of same dimensions but of mass M / 4 is placed gently on the first disc coaxially. The angular velocity of the system now is

Question

A thin uniform circular disc of mass M and radius R is rotating in a horizontal plane about an axis passing through its centre and perpendicular to its plane with an angular velocity ω . Another disc of same dimensions but of mass M / 4 is placed gently on the first disc coaxially. The angular velocity of the system now is (A) (2 ω/√2) (B) (4 ω/5) (C) (5 ω/4) (D) (3 ω/4)

Tardigrade Admin · Accepted Answer

As no torque is applied, therefore angular momentum is conserved. ∴ I1 ω1=I2 ω2 text or ω2=(I1 ω1/I2)=(((1/2) M R2) ω/((1)2 M R2+(1)2 (M/4) R2)=(4 ω/5)

$\frac{2 ω}{2}$

$\frac{4 ω}{5}$

$\frac{5 ω}{4}$

$\frac{3 ω}{4}$

As no torque is applied, therefore angular momentum is conserved.
$∴ I_{1} ω_{1} = I_{2} ω_{2}$
$or ω_{2} = \frac{I _{1} ω _{1}}{I _{2}} = \frac{( \frac{1}{2} M R ^{2} ) ω}{( \frac{1}{2} M R ^{2} + \frac{1}{2} \frac{M}{4} R ^{2} )} = \frac{4 ω}{5}$

2​2ω​

54ω​

45ω​

43ω​

As no torque is applied, therefore angular momentum is conserved. ∴I1​ω1​=I2​ω2​ or ω2​=I2​I1​ω1​​=(21​MR2+21​4M​R2)(21​MR2)ω​=54ω​

$\frac{2 ω}{2}$

$\frac{4 ω}{5}$

$\frac{5 ω}{4}$

$\frac{3 ω}{4}$

As no torque is applied, therefore angular momentum is conserved.
$∴ I_{1} ω_{1} = I_{2} ω_{2}$
$or ω_{2} = \frac{I _{1} ω _{1}}{I _{2}} = \frac{( \frac{1}{2} M R ^{2} ) ω}{( \frac{1}{2} M R ^{2} + \frac{1}{2} \frac{M}{4} R ^{2} )} = \frac{4 ω}{5}$