Answer:
the moment of inertia of the merry go round is 38.04 kg.m²
Explanation:
We are given;
Initial angular velocity; ω_1 = 37 rpm
Final angular velocity; ω_2 = 19 rpm
mass of child; m = 15.5 kg
distance from the centre; r = 1.55 m
Now, let the moment of inertia of the merry go round be I.
Using the principle of conservation of angular momentum, we have;
I_1 = I_2
Thus,
Iω_1 = I'ω_2
where I' is the moment of inertia of the merry go round and child which is given as I' = mr²
Thus,
I x 37 = ( I + mr²)19
37I = ( I + (15.5 x 1.55²))19
37I = 19I + 684.7125
37I - 19 I = 684.7125
18I = 684.7125
I = 684.7125/18
I = 38.04 kg.m²
Thus, the moment of inertia of the merry go round is 38.04 kg.m²
Answer:
[tex]52.44 kgm^2[/tex]
Explanation:
First convert from revolution per minute to rad per second knowing that a revolution is 2π rad and every minute has 60 seconds
[tex]\omega_1 = 37 rpm = 37*2\pi/60 = 3.875 rad/s[/tex]
[tex]\omega_2 = 19 rpm = 19*2\pi/60 = 1.99 rad/s[/tex]
According to the law of momentum conservation, the total angular before and after the child jump on must be the same:
[tex]\omega_1 I_w = \omega_2(I_w + I_c)[/tex] (1)
where [tex]I_w[/tex] is the moments of inertia of the merry-go-round, which we are looking for, and [tex]I_c[/tex] is the moments of inertia of the child, which can be calculated if we treat him as a point mass:
[tex]I_c = mr^2 = 34.5*1.2^2 = 49.68 kgm^2[/tex]
Therefore from eq. (1) we have:
[tex]3.875I_w = 1.99(I_w + 49.68)[/tex]
[tex]I_w(3.875 - 1.99) = 1.99*49.68 = 98.85 [/tex]
[tex]I_w = \frac{98.85}{3.875 - 1.99} = 52.44 kgm^2[/tex]
