Question

A merry-go-round with a a radius of R = 1.84 m and moment of
inertia I = 206 kg-m^{2} is spinning with an initial
angular speed of ? = 1.54 rad/s in the counter clockwise direection
when viewed from above. A person with mass m = 66 kg and velocity v
= 4.5 m/s runs on a path tangent to the merry-go-round. Once at the
merry-go-round the person jumps on and holds on to the rim of the
merry-go-round.

1) What is the magnitude of the initial angular momentum of the merry-go-round?

2) What is the magnitude of the angular momentum of the person 2 meters before she jumps on the merry-go-round?

3) What is the magnitude of the angular momentum of the person just before she jumps on to the merry-go-round?

4) What is the angular speed of the merry-go-round after the person jumps on?

Answer #1

here,

R = 1.84 m , I0 = 206 kg.m^2

angular speed , w0 = 1.54 rad/s

v = 4.5 m/s

mass of person , m = 66 kg

a)

the magnitude of the initial angular momentum of the merry-go-round , Pi = I0 * w0

Pi = 317.24 kg.m^2/s

b)

the magnitude of the angular momentum of the person 2 meters before she jumps on the merry-go-round , Lp = m * v * (r + R)

Lp = 66 * 4.5 * 3.84 kg.m^2/s

Lp = 1140.5 kg.m^2 /s

c)

the magnitude of the angular momentum of the person just before she jumps on to the merry-go-round , La = m * v * (R)

Lp = 66 * 4.5 * 1.84 kg.m^2/s

Lp = 546.5 kg.m^2 /s

d)

let the angular speed of the merry-go-round after the person jumps on be w

using conservation of angular momentum

Li + Lp = (I + m * R^2) * w

317.24 + 546.5 = ( 206 + 66 * 1.84^2) * w

w = 2.01 rad/s

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