Question

A 4.4 m diameter merry-go-round is rotating freely with an angular velocity of 0.80 rad/s. Its total moment of inertia is 1950 kg · m2. Four people standing on the ground, each of 50 kg mass, suddenly step onto the edge of the merry-go-round.
What is the angular velocity of the merry-go-round now?

Answer 1

Answer:

Angular velocity of merry go round is 0.64 $\frac{rad}{s}$.

Explanation:

Given :

Radius of merry go round $R = \frac{D}{2} = 2.2$ m

Angular velocity of merry go round $\omega = 0.8 \frac{rad}{s}$

Momentum of inertia $I = 1950$ $Kg m^{2}$

Mass of each people $M = 50$ Kg

From the formula of angular momentum conservation,

 $L = I\omega$

So initial angular momentum,

$L_{i} = 1950 \times 0.8 = 1560$

Now we calculate the MOI after people jump on merry go round,

$I_{p} =\frac{1}{2} MR^{2}$

But there are four people so we have to multiply with 4 and add merry go round MOI,

$I _{tot} = 1950 + 2\times 50 \times (2.2) ^{2}$

So total MOI after people jump on is,

$I _{tot} = 2434$ $Kg m^{2}$

From conservation of angular momentum,

$L_{i} = L_{f}$

$1560 = 2434 \times \omega_{f}$

So angular velocity is,

 $\omega _{f} = \frac{1560}{2434} = 0.64$ $\frac{rad}{s}$