Question

5. An insect of mass 20 g crawls from the centre to the outside edge of a rotating disc of
mass 200g and radius 20 cm. The disk was initially rotating at 22.0 rads−1
. What will be
its final angular velocity? What is the change in the kinetic energy of the system?

Answer 1

Given:

M=200g=200/1000=0.2kg

R=20cm=20/100=0.2m

m=20g=20/1000=0.02kg

ω1=22rad/s

Using Law of conservation of angular momentum: J1ω1=J2 ω2

Let us calculate Initial angular momentum:  J1

J1=Mr²/2

=0.2 X (0.2)²/2=0.2x0.2x0.2/2=4x 10⁻³ Kgm²

Let us calculate final angular momentum : J2

J2=Mr²/2  + mr²=4x10⁻³ + 0.02 x( 0.2)²

=4.8x 10⁻³ kgm²

Now,

ω2=(J1/J2)ω1

=(4x10⁻³/4.8x 10⁻³) x 22⇒18.3 rad/s

Change in K.E

Δω=J2ω2²/2- J1ω1²/2
=[(4.8 x1 10⁻³ x 18.3x18.3)/2 ] -- [(4x 10⁻³ x 22 x  22)/2]

=-164.3 joules

∴Final angular velocity is 18.3 rad/s and  change in the kinetic energy of the system is -164.3 J