Question

Show your capability!???Very easy problem.

Answer 1

• The moment of inertia of the rod about the perpendicular axis passing through its mid point

• `I_(rod)=1/12xxM_(rod)xxL^2`,

• where mass of the rod `M_(rod)=0.75kg`

• and length of the rod `L=0.4m`

• So `I_(rod)=1/12xx0.75xx0.4^2=0.01kgm^2`

• Now the moment of inertia of each ring of mass `m=1kg` at initial position `r_1=0.1m` from center of rotation `O`

• `I_1=mr_1^2=1xx0.1^2=0.01kgm^2`

• And the moment of inertia of each ring of mass `m=1kg` at final position `r_2=0.2m` from center of rotation `O`

• `I_2=mr_2^2=1xx0.2^2=0.04kgm^2`

• Given that the initial angular velocity of the system for initial position of the ring be `omega_1=30` rad/s

• Let the angular velocity of the system in final position of the ring be `omega_2`

• Hence applying conservation of angular momentum we can write

• `(I_(rod)+2I_2)omega_2=(I_(rod)+2I_1)omega_1`

`=>omega_2=((I_(rod)+2I_1)omega_1)/(I_(rod)+2I_2)`

`=>omega_2=((0.01+2xx0.01)xx30)/(0.01+2xx0.4)` rad/s

`=>omega_2=10` rad/s

Answer 2

See below.

Explanation:

The kinetic energy is conserved so

`E_1=1/2J_1 omega_1^2=E_2=1/2J_2omega_2^2`

Here

`J_i =1/2(J_r+2r_i^2m)` with

`r_1 = 0.10`
`r_2=0.20`
`omega_1 = 30`
`m = 1` and

`J_r=(m_r L^2)/12 = (0.75 xx 0.40^2)/12`

so

`omega_2 = (sqrt(J_1/J_2))omega_1 = (sqrt((J_r+2r_1^2m)/(J_r+2r_2^2m)))omega_1 = 17.3205` [rad/sec]