rotatioal motion question

a. Moment of inertia of disk A about the axis of rotation, IA

= 1/2 MAR^2

= 1/2 (7)(1.2)^2

= 5.04 kgm^2

By torque = IAa

5 = (5.04)a

Angular acceleration, a = 5/5.04 rads^-2

So, by w = u + at

w = 0 + (5/5.04)(4)

Angular velocity after 4 s, w = 3.97 rads^-1


b. Moment of inertia of the system after putting B on A, I

= 1/2 (MA + MB)R^2

= 1/2 (7 + 9)(1.2)^2

= 11.52 kgm^2

There is no external torque acting on the system

So, by the conservation of angular momentum

IAw = Iw2

Final angular velocity, w2 = (5.04)(3.97) / (11.52)

= 1.74 rads^-1

(a) Use impulse = change of angulat mometum
5 x 4 = (7x1.2^2/2)w
where w is the angular momentum of disk A
hence, w = 3.97 s^-1

(b) using conservation of angular momentum
Initial momentum = (1/2)(7x1.2^2)(3.97) kg-m2/s
Final mometum = 2 x (1/2)(16x1.2^2)w(2)^2

hence, (1/2)(7x1.2^2)(3.97) = 2 x (1/2)(16x1.2^2)w2
w2 = 7 x 3.97/(2x16) s^-1 = 0.81 s^-1

2009-12-04 19:45:57 補充：
sorry....the the final momentum should be (1/2)(16x1.2^2)w(2), the 2 should not be here. The final result is 1.62 s^-1