#8 A horizontal, solid disk is free to rotate on an axle through its center, but

Question

#8 A horizontal, solid disk is free to rotate on an axle through its center, but is not allowed to translate. Two small rockets on the edge each apply a force, F= 1.88N, as shown, for 3.00 seconds. The diameter of the disk is .500 m. Its mass is 2.25 kg. The disk is initially rotating with a period of 4.75 seconds. Use I=(l/2)MR^2 for the moment of inertia of the disk. A) Using the idea that Et=Ia, find the period of the disk after 3.00 s. B) Using the idea that TNET delta t=delta L, find the period of the disk after 3.00 s.

Explanation / Answer

Radius of the disk, R = d/2 = 0.5/2 = 0.25 m

I = (1/2)*M*R^2

= (1/2)*2.25*0.25^2

= 0.0703 kg.m^2

initail angular speed, w1 = 2*pi/T1

= 2*pi/4.75

= 1.32 rad/s

A) Tnet = I*alfa

alfa = Tnet/I

= 2*F*R/I

= 2*1.88*0.25/0.0703

= 13.37 rad/s

now Apply, w2 = w1 + alfa*t

= 1.32 + 13.37*3

= 41.4 rad/s

so, time periode, T2 = 2*pi/w2

= 2*pi/41.4

= 0.15 s

B)) we know, Net torque acting, Tnet = dL/dt

Tnet = I*(w2 - w1)/t

2*F*R = I*(w2 - w1)/t

2*F*R*t = I*(w2 - w1)

w2 = w1 + 2*F*R*t/I

= 1.32 + 2*1.88*0.25*3/0.0703

= 41.4 rad/s

so, time periode, T2 = 2*pi/w2

= 2*pi/41.4

= 0.15 s

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