A child at a local playground rolls from rest a 550-g dodgeball (hollow sphere)

Question

A child at a local playground rolls from rest a 550-g dodgeball (hollow sphere) without slipping down a slide to hit his 32-kg friend who is standing on the edge of a stationary 120-kg merry-go-round witha 1.2 m radius. If the ball leaves the bottom of the slide with a speed of 5.425 m/s and the friend catches the ball, how fast will the merry-go-round rotate? Show steps please.

Assumptions: When the ball leaves the bottom of the slide, its spinning motion due to rolling is negligible compared to its translational motion and its direction of motion is perpendicular to the vector pointing from the rotational axis of the merry-go-round to the friend. The merry-go-round is a soid disk. Enough static friction exists between the friend's shoes and the merry-go-round that the friend does not slip once the merry-go-round spins.

Explanation / Answer

we can use conservation of energy

let's start with calculating the KE of the ball = 1/2 mv^ 2= 0.5 ( 550x 10^-3) (5.425^2) =8.1 J

Rotational KE = 1/2 Iw^2

I = (120 ) (1.2^2) / 2 + 32( 1.2^2) = 132.48 kg m^2

8.1 = 1/2 ( 132.48) w^2

w= 0.349 or 0.35 rad/sec

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