(a) 
The moment of inertia of a uniform-density disk is given by
where
M is the mass of the disk
R is its radius
In this problem,
M = 16 kg is the mass of the disk
R = 0.19 m is the radius
Substituting into the equation, we find
(b) 142.5 J
The rotational kinetic energy of the disk is given by
where
I is the moment of inertia
is the angular velocity
We know that the disk makes one complete rotation in T=0.2 s (so, this is the period). Therefore, its angular velocity is
And so, the rotational kinetic energy is
(c) 
The rotational angular momentum of the disk is given by
where
I is the moment of inertia
is the angular velocity
Substituting the values found in the previous parts of the problem, we find
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Explanation:
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Answer:
The correct answer is;
The magnitude of the force is 35.12 N
Explanation:
To solve the question, we note that the friction is zero and the force causes motion of a stationary mass
One of the equations of motion is required such as
v² = u² + 2× a× s
Where
v = Final velocity = 5.93 m/s
u = Initial velocity = 0 m/s , object at rest
a = acceleration
s = distance moved = 32 meters
But v = Distance/Time = 32 m /5.4 s = 5.93 m/s
Therefore
5.93² = 2×a×32
or a = 35.12/ 64 = 0.55 m/s²
Therefore Force F = Mass m × Acceleration a
Where mass m = 64 kg
Therefore F = 64 kg×0.55 m/s² = 35.12 N
