hollow, thin-walled sphere

2012-11-03 12:41 am

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

回答 (1)

用戶25975

2012-11-03 5:14 pm

✔ 最佳答案

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

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