Solveeit Logo

Question

Question: Two discs of moment of inertia I<sub>1</sub> and I<sub>2</sub> and angular speeds \(\omega _ { 1 }\)...

Two discs of moment of inertia I1 and I2 and angular speeds ω1\omega _ { 1 } and ω2\omega _ { 2 } are rotating along collinear axes passing through their centre of mass and perpendicular to their plane. If the two are made to rotate together along the same axis the rotational KE of system will be

A

I1ω1+I2ω22(I1+I2)\frac { I _ { 1 } \omega _ { 1 } + I _ { 2 } \omega _ { 2 } } { 2 \left( I _ { 1 } + I _ { 2 } \right) }

B

(I1+I2)(ω1+ω2)22\frac { \left( I _ { 1 } + I _ { 2 } \right) \left( \omega _ { 1 } + \omega _ { 2 } \right) ^ { 2 } } { 2 }

C

(I1ω1+I2ω2)22(I1+I2)\frac { \left( I _ { 1 } \omega _ { 1 } + I _ { 2 } \omega _ { 2 } \right) ^ { 2 } } { 2 \left( I _ { 1 } + I _ { 2 } \right) }

D

None of these

Answer

(I1ω1+I2ω2)22(I1+I2)\frac { \left( I _ { 1 } \omega _ { 1 } + I _ { 2 } \omega _ { 2 } \right) ^ { 2 } } { 2 \left( I _ { 1 } + I _ { 2 } \right) }

Explanation

Solution

By the law of conservation of angular momentum

I1ω1+I2ω2=(I1+I2)ωI _ { 1 } \omega _ { 1 } + I _ { 2 } \omega _ { 2 } = \left( I _ { 1 } + I _ { 2 } \right) \omega

Angular velocity of system ω=I1ω1+I2ω2I1+I2\omega = \frac { I _ { 1 } \omega _ { 1 } + I _ { 2 } \omega _ { 2 } } { I _ { 1 } + I _ { 2 } }

Rotational kinetic energy =12(I1+I2)ω2= \frac { 1 } { 2 } \left( I _ { 1 } + I _ { 2 } \right) \omega ^ { 2 }

=12(I1+I2)(I1ω1+I2ω2I1+I2)2= \frac { 1 } { 2 } \left( I _ { 1 } + I _ { 2 } \right) \left( \frac { I _ { 1 } \omega _ { 1 } + I _ { 2 } \omega _ { 2 } } { I _ { 1 } + I _ { 2 } } \right) ^ { 2 } =(I1ω1+I2ω2)22(I1+I2)= \frac { \left( I _ { 1 } \omega _ { 1 } + I _ { 2 } \omega _ { 2 } \right) ^ { 2 } } { 2 \left( I _ { 1 } + I _ { 2 } \right) }.