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:

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

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

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

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

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

Answer 2

In this case, the final angular speed of the merry-go-round is - 0.24 rad/s    

Given:

m = 36 kg  

M = 300 kg

R = 2 m

v = 4 m/s

Angular speed:

• rate of velocity at which an object or a particle is rotating around a center or a specific point in a given time period

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

1. Moment of inertia of the merry go round,  

I = M x R²

= 300 x 2 x 2

= 1200 kgm²

2. Moment of inertia of Joey  

I’ = m x R²

= 36 x 2 x 2

= 144 kgm²  

3. Use conservation of angular momentum

     

I x ω =

1200 x ω =

ω = 0.24 rad/s    

Thus, the final angular speed of the merry-go-round is - 0.24 rad/s.

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