Exercise 10.40 A hollow, thin-walled sphere of mass 13.0 kg and diameter 47.0 cm

Question

Exercise 10.40

A hollow, thin-walled sphere of mass 13.0 kg and diameter 47.0 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 1.20 and B has numerical value 1.60.

Part A

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

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Part D

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

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Exercise 10.40

A hollow, thin-walled sphere of mass 13.0 kg and diameter 47.0 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 1.20 and B has numerical value 1.60.

Part A

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

Lz =   kgm2/s  

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Part D

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

z =   Nm  

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Explanation / Answer

angular momentum = I*w

I = hollow thin sphere = 2/3 * MR^2

I = 0.4786 kg.m^2

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

here A = 1.20 and B = 1.60

at t = 4

w = 2*1.20 * 4 + 4 * 1.60 * (4)^3 = 419.2

L = 0.4786 * 419.2

L = 200.629 kg .m^2/s

part b )

torque = t = dL/dt

t = 200.629/4 = 50.157 N-m

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