Question

Joey, whose mass is m = 36 kg, stands at rest at the outer edge of a frictionless merry-go-round with the mass M = 300 kg and the radius R = 2.0 m. The merry-go-round is also at rest. Joey then begins to run around the perimeter of the merry-go-round, finally reaching a constant speed, measured relative to the ground, of 4.0 m/s. What is the final angular speed of the merry-go-round?

Answer 1

Answer:

Explanation:

m = 36 kg

M = 300 kg

R = 2 m

v = 4 m/s

Let the final angular speed of merry go round is ω.

Moment of inertia of the merry go round,

I = M x R² = 300 x 2 x 2 = 1200 kgm²

Moment of inertia of Joey

I' = m x R² = 36 x 2 x 2 = 144 kgm²

Use conservation of angular momentum

I x ω = I' x v/R

1200 x ω = 144 x 4 / 2

ω = 0.24 rad/s  

Answer 2

Answer:

$\omega=0.24\ rad.s^{-1}$

Explanation:

Given:

mass of person, $m=36\ kg$

mass of merry go-round, $M=300\ kg$

radius of merry go-round, $R=2\ m$

velocity of the person running, $v=4\ m.s^{-1}$

We consider merry go-round as a ring:

Now the moment of inertial of the ring is given as,

$I=M.R^2$

$I=300\times 2^2$

$I=1200\ kg.m^{-2}$

Moment of inertia of the person considering as a point mass:

$I_p=m.R^2$

$I_p=36\times 2^2$

$I_p=144\ kg.m^2$

Now according to the conservation of angular momentum:

$I.\omega=I_p.\omega_p$

where:

$\omega =$ angular velocity of the merry-go-round

$\omega_p=$ angular velocity of the person running

$1200\times \omega=144\times \frac{v}{R}$

$\omega=\frac{144}{1200} \times \frac{4}{2}$

$\omega=0.24\ rad.s^{-1}$