Answer:
72.75 kg m^2
Explanation:
initial angular velocity, ω = 35 rpm
final angular velocity, ω' = 19 rpm
mass of child, m = 15.5 kg
distance from the centre, d = 1.55 m
Let the moment of inertia of the merry go round is I.
Use the concept of conservation of angular momentum
I ω = I' ω'
where I' be the moment of inertia of merry go round and child
I x 35 = ( I + md^2) ω'
I x 35 = ( I + 25.5 x 1.55 x 1.55) x 19
35 I = 19 I + 1164
16 I = 1164
I = 72.75 kg m^2
Thus, the moment of inertia of the merry go round is 72.75 kg m^2.
This is a problem of conservation of momentum
Momentum before throwing the rock: m*V = 96.0 kg * 0.480 m/s = 46.08 N*s
A) man throws the rock forward
=>
rock:
m1 = 0.310 kg
V1 = 14.5 m/s, in the same direction of the sled with the man
sled and man:
m2 = 96 kg - 0.310 kg = 95.69 kg
v2 = ?
Conservation of momentum:
momentum before throw = momentum after throw
46.08N*s = 0.310kg*14.5m/s + 95.69kg*v2
=> v2 = [46.08 N*s - 0.310*14.5N*s ] / 95.69 kg = 0.434 m/s
B) man throws the rock backward
this changes the sign of the velocity, v2 = -14.5 m/s
46.08N*s = - 0.310kg*14.5m/s + 95.69kg*v2
v2 = [46.08 N*s + 0.310*14.5 N*s] / 95.69 k = 0.529 m/s
Distance , d = a+b
The unit of d is in meter and t is in seconds.
So the unit of a a must be meter.
Now we have unit of b
is meter.
So unit of b*
= meter
Unit of b = meter/
So unit of a = m and unit of b = m/
.
