A bullet is shot through two rotating paper disks mounted a distance D= 1.50 m a

Question

A bullet is shot through two rotating paper disks mounted a distance D= 1.50 m apart along the same axle. Suppose that the bullet passes through the first disk at the moment that the disks begin accelerating from rest, due to a constant force of 2000 N applied tangentially to the axle by the engine. If the angular dispalcement betwenn the two bullet holes in the disks is measured to be 20 degrees, what is the speed of the bullet? Each paper disk has a mass of 100.0 g and a radius of 20 cm, and the axle has a mass of 10.0Kg and a radius of 2.00 cm

Explanation / Answer

I = 0.5*M*R^2 + 2*0.5*m*r^2

I = (0.5*10*0.02^2) + (2*0.5*0.1*0.2^2))= 0.006 kg m^2

torque = R*F = I*alfa

alfa = 6666.67 rad/s^2

speed of disk
v = s/t

(R+r)*alfa*t= (R+r)*theta /t

t^2 = theta/alfa = (20*3.14/180) / 6666.67

t = 7.23 *10^-3 s

speed of the bullet = d/t =1.5/(7.23*10^-3)= 207.34 m/s

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