Question

A merry-go-round with a rotational inertia of 600 kg m2 and a radius of 3. 0 m is initially at rest. A 20 kg boy approaches the merry-go-round along a path tangent to the rim of the at a speed of 5. 0 m/s. Determine the angular velocity of the system after the boy hops on the merry-go-round.

Answer 1

Hi there!

`\boxed{\omega = 0.38 rad/sec}`

We can use the conservation of angular momentum to solve.

`\large\boxed{L_i = L_f}`

Recall the equation for angular momentum:

`L = I\omega`

We can begin by writing out the scenario as a conservation of angular momentum:

`I_m\omega_m + I_b\omega_b = \omega_f(I_m + I_b)`

`I_m` = moment of inertia of the merry-go-round (kgm²)

`\omega_m` = angular velocity of merry go round (rad/sec)

`\omega_f` = final angular velocity of COMBINED objects (rad/sec)

`I_b` = moment of inertia of boy (kgm²)

`\omega_b`= angular velocity of the boy (rad/sec)

The only value not explicitly given is the moment of inertia of the boy.

Since he stands along the edge of the merry go round:

`I = MR^2`

We are given that he jumps on the merry-go-round at a speed of 5 m/s. Use the following relation:

`\omega = (v)/(r)`

`L_b = MR^2((v)/(R)) = MRv`

Plug in the given values:

`L_b = (20)(3)(5) = 300 kgm^2/s`

Now, we must solve for the boy's moment of inertia:

`I = MR^2\\I = 20(3^2) = 180 kgm^2`

Use the above equation for conservation of momentum:

`600(0) + 300 = \omega_f(180 + 600)\\\\300 = 780\omega_f\\\\\omega = \boxed{0.38 rad/sec}`