A bullet with a mass m=4.83 g and initial speed of 330 m/s passes through a whee

Question

A bullet with a mass m=4.83 g and initial speed of 330 m/s passes through a wheel which is initially at rest as in the figure. The wheel is a solid disk with mass M=2.29 kg and radius R=18.6 cm. The wheel rotates freely about an axis through its center and out of the plane shown in the figure. The bullet passes through the wheel at a perpendicular distance 14.8 cm from the center. After passing through the wheel it has a speed of 48 m/s.

(a) What is the angular speed of the wheel just after the bullet leaves it? rad/s ( ± 0.02 rad/s)

(b) How much kinetic energy was lost in the collision? J ( ± 0.02 J)

Explanation / Answer

   mnass of bullet mb = 4.83 g, initial speed u = 330 m/s,final speed of bullet v = 48 m/s
    mass of wheel mw = 2.29 kg, of radius R = 18.6 cm.

   here conservationn of angular momentum , we know that the angular momentum L = I*W = mr^2*W= m*V*r ( v= r*W)

   now initial angular momentum of the bullet is = 0.00483 * 0.148 * 330 = 0.2359 kg m2/s
   final angular momentum of the bullet is = 0.00483 * 0.148 * 48= 0.03431 kg m2/s

   chang ein angular momentum of the bullet is = 0.2359 - 0.03431 = 0.20159 kg m2/s

   so the angular momentum of the wheel I*W = 0.2016,moment of inertia of the wehhel is I = 1/2 m_W R^2 = 0.5*2.29*0.186^2 = 0.03961 kg m2

a)   so angular speed of the wheel is W = 0.20159/0.03961 = 5.089 rad/s

b)   initial kinetic energy is = k.e of bullet = 1/2 mv^2 = 0.5*0.00483 * 330^2 = 263 J

   final kinetic energy is = k.e of bullet + k.e of wheel
  
               = 1/2 mvf^2 + 1/2 IW^2
= 0.5*0.00483 * 48^2 + 0.5*0.03961*5.089^2 = 6.0770 J

change in kinetic energy = 6.0770 - 263 = -256.923 J

   amount of kinetic energy lost is 256.923 J

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