標題:
hollow, thin-walled sphere
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發問:
A hollow, thin-walled sphere of mass 11.0kg and diameter 47.0cm 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 *(t) = At^2 + Bt^4 , where a has numerical value 1.20 and B has numerical value 1.60.1. What are the... 顯示更多 A hollow, thin-walled sphere of mass 11.0kg and diameter 47.0cm 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 *(t) = At^2 + Bt^4 , where a has numerical value 1.20 and B has numerical value 1.60. 1. What are the units of the constants A? (choose an answer) A. rad/s B. rad/s^2 C. rad/s^3 D. rad/s^4 2. What are the units of the constants B? (choose an answer) A. rad/s B. rad/s^2 C. rad/s^3 D. rad/s^4 3. At the time 3.00s , find the angular momentum of the sphere. unit = kg. m^2 /s 4. At the time 3.00s , find the net torque on the sphere. unit = N . m 更新: Q3 and 4 are incorrect. 更新 2: Q3) 72.9 更新 3: Q4) 71.0 N . m
最佳解答:
1, 2) As the angle is measured in radians, A and B should have units of rad/s2 and rad/s4 resp. 3) Radius of the sphere = 23.5 cm = 0.235 m So moment of inertia = 11 x 0.2352 = 0.6075 kg/m2 And the instantaneous angular speed is d(At2 + Bt4)/dt = (2At + 4Bt3) rad/s Hence at t = 3, angular speed = 6A + 108B = 180 rad/s So angular momentum = 0.6075 x 180 = 109.35 kg m2/s 4) Angular acceleration = d(2At + 4Bt3)/dt = (2A + 12Bt2) rad/s2 Hence at t = 3, angular acc. = 175.2 rad/s2 So net torque = 0.6075 x 175.2 = 106.43 N m 2012-11-04 00:19:01 補充： For Q3, 4, the formula of moment of inertia of a hollow spherical shell should be 2mr^2/3 So Q3 ans should be 109.35 x 2/3 = 72.9 kg m^2/s Q4 ans should be 106.43 x 2/3 = 71.0 N m
其他解答:63D0B758E2D502CC
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