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vitfil [10]
3 years ago
11

A playground merry-go-round of radius R = 1.40 m has a moment of inertia I = 265 kg · m2 and is rotating at 11.0 rev/min about a

frictionless vertical axle. Facing the axle, a 27.0-kg child hops onto the merry-go-round and manages to sit down on the edge. What is the new angular speed of the merry-go-round?
Physics
1 answer:
Tom [10]3 years ago
6 0

Answer:

The value of new value of angular speed of merry go round.\omega_{2} = 0.96 \frac{rad}{sec}

Explanation:

Given data

r = 1.4 m

Moment of inertia I_{1} = 265 kg - m^{2}

N_{1} = 11 RPM

\omega_{1} = \frac{2 \pi N}{60}

\omega_{1} = \frac{2 \pi (11)}{60}

\omega_{1} = 1.15 \frac{rad}{sec}

From conservation of momentum principal

I_{1} \omega_{1}  = I_{2} \omega_{2} ------- (1)

I_{2} = m r^{2} + 265

I_{2} = 27 (1.4)^{2} + 265

I_{2} = 317.92 \ kg m^{2}

Put all the values in equation  (1)

265 × 1.15 = 317.92 × \omega_{2}

\omega_{2} = 0.96 \frac{rad}{sec}

This is the value of new value of angular speed of merry go round.

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We were told to calculate the speed of the ball,

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