A.)A merry-go-round has a mass of 317 kg and a radius of 3.36 m. The merry-
go round is spinning such that it completes one revolution every 3.58 s
with a 63.6 kg man standing at its center. (a) What is the moment of
inertia of the merry-go-round if we assume it is a uniform circular disc?
tries: 0
B.)What is the angular velocity of the merry-go-round?
C.)If the person walks straight to the outside of the merry-go-round (i. e.
along a radial line), what is its new angular velocity of the merry-go-
round when he reaches its edge?

A)

B) 1.76 rad/s

C) 1.26 rad/s

Explanation:

A)

The moment of inertia of a uniform circular disc rotating about its axis is given by

where

M is the mass of the disc

R is the radius of the disc

For the merry-go-round in this problem, we have:

M = 317 kg is the mass of the disc

R = 3.36 m is its radius

Substituting, we find the moment of inertia:

B)

The angular velocity of the merry-go-round is the rate of change of its angular displacement, and it is given by:

where

is the angular velocity

T is the period of revolution (the time taken for completing 1 revolution)

For the merry-go-round in this problem, it takes 3.58 s to complete one revolution so:

is the period of revolution

Therefore, the angular velocity is:

C)

When the man walks towards the edge, the angular momentum of the system, must be conserved, since there are no external torques acting on it.

The angurlar momentum is given by:

where

I is the moment of inertia

is the angular velocity

Since the angular momentum must remain constant, we can write:

where:

is the initial moment of inertia of the system

is the initial angular velocity

is the final moment of inertia, where:

m = 63.6 kg is the mass of the man

R = 3.36 m is the final distance of the man from the axis

Substituting,

And is the final angular velocity. Substituting,

he kicked it 10 feet straight up into the air? i think this works.

b.

explanation: