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Physics Problem

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A hollow, thin-walled sphere of mass 12.0kg and diameter 45.0cm 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 θ(t)=At2+Bt4, where A has numerical value 1.10 and B has numerical value 1.60.a. At the time 4.00s ,... 顯示更多 A hollow, thin-walled sphere of mass 12.0kg and diameter 45.0cm 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 θ(t)=At2+Bt4, where A has numerical value 1.10 and B has numerical value 1.60. a. At the time 4.00s , find the angular momentum of the sphere.(answer in kg?m2/s) b. At the time 4.00s , find the net torque on the sphere.(answer in N?m)

 

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最佳解答:

(a) Moment of inertia of sphere = (2/3).(12).(0.45/2)^2 kg.m^2 = 0.405 kg.m^2 Angular speed w = dθ/dt = 2At + 4Bt^3 Hence, angular momentum = 0.405 x (2At + 4Bt^3) At t= 4 s, angular momentum = 0.405 x ( 2 x 1.1 x 4 + 4 x 1.6 x 4^3) kg.m^2/s = 169 kg.m^2/s (b) Agular acceleration = dw/dt = 2A + 12Bt^2 Hence, net torque = moment of inertia x angular acceleration = 0.405 x (2 x 1.1 + 12 x 1.6 x 4^2) N.m = 125 N.m

其他解答:8758B59A7FA1EEA7
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