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?

rad/s4

Part B

What are the units of the constant B?

rad/s4

Part C

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

186

Part D

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

(Only need answer for part D)

rad/s rad/s2 rad/s3

rad/s4

Part B

What are the units of the constant B?

rad/s rad/s2 rad/s3

rad/s4

Part C

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

Lz =

186

  kgm2/s

Part D

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

(Only need answer for part D)

Explanation / Answer

(t)=At^2+Bt^4

w = d/dt = 2At + 4Bt^3

a = dw/dt = 2A + 12Bt^2

I = 2/3* m*r^2 = 0.4725

at t=4s....

w = 2*1.1*4 + 4*1.5*4^3 = 392.8 rad/s

a =2*1.1 + 12*1.5*4^2 = 290.2 rad/s^2

part C:   Lz = I*w = 0.4725 * 392.8 = 186 kg-m/s^2

part D: T = I* a = 0.4725 * 290.2 = 137.12

torque = 137.12 Kg-m^2 /s^2

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