Question

A child exerts a tangential 58.5 N force on the rim of a disk-shaped merry-go-round with a radius of 2.99 m.
If the merry-go-round starts at rest and acquires an angular speed of 0.1250 rev/s in 5.00 s, what is its mass?

Answer 1

w_o = initial angular velocity = 0 rad/s

w = final angular velocity = 0.1250 rev/s = 0.1250 (6.28) rad/s = 0.785 rad/s (Since 1 rev = 6.28 rad)

t = time taken = 5 sec

angular acceleration is given as

= (w - w_o)/t

inserting the values

= (0.785 - 0)/5

= 0.157 rad/s²

M = mass of merry-go-round = ?

R = radius of the merry-go-round = 2.99 m

I = moment of inertia of merry-go-round

F = tangential force on the rim = 58.5 N

Torque is given as

= F r eq-1

Torque is also given as

= I eq-2

using eq-1 and eq-2

I = F r

I (0.157) = (58.5) (2.99)

I = 1114.11 kgm²

moment of inertia of merry-go-round is given as

I = (0.5) MR²

1114.11 = (0.5) M (2.99)²

M = 249.24 kg