A hollow, thin-walled sphere of mass 14.0 kg and diameter 45.0 cm is rotating ab

Question

A hollow, thin-walled sphere of mass 14.0 kg and diameter 45.0 cm is rotating about an axle through its center. The angle (in radians) through which it turns as a function of time (in seconds) is given by (t)=At2+Bt4, where A has numerical value 1.10 and B has numerical value 1.50.

Part A

What are the units of the constant A?

Part B

What are the units of the constant B?

Part C

At the time 4.00 s , find the angular momentum of the sphere

Part D

At the time 4.00 s , find the net torque on the sphere

Explanation / Answer

Here ,

theta = A * t^2 + B * t^3

m = 14 Kg

radius ,r = 45/2 = 22.5 cm

moment of inertia , I = (2/3) * m * r^2

I = (2/3) * 14 * (0.225)^2

I = 0.4725 Kg.m^2

part A)

for the constant A

s^2 * A= rad

A = rad/s^2

the unit of A is rad/s^2

part B)

for the constant B

s^4 * B = rad

B = rad/s^4

the unit of constant B is rad/s^4

part c)

omega = d(theta)/dt

omega = d/dt(At^2+Bt^4)

omega = 2 * A* t + 4 * B * t^3

at t = 4 s

omega = 2 * 1.1 * 4 + 4 * 1.5 * 4^3

omega = 392.8 rad/s

angular momentum = 392.8 * 0.4725

angular momentum = 185.6 Kg.m^2/s

part d)

at t = 4 s

angular acceleration = d/dt(omega)

angular acceleration = 2 * A + 12 * B * t^2

at t = 4 s

angular acceleration = 2 * 1.1 + 4 * 1.5 * 4^2

angular acceleration = 98.2 rad/s^2

net torque = 98.2 * 0.4725

net torque = 46.4 N.m

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