A mass m is fired horizontally, with a speed v_0, at a rigid rod of mass M and l

Question

A mass m is fired horizontally, with a speed v_0, at a rigid rod of mass M and length d. pivoted about a frictionless axle. O. The mass sticks to the rod after the collision, the figure shows the events just prior and just after the collision. What is the angular momentum of the system just before the collision about an axis through the the point O? Just after the collision, what is the moment of inertia of the system about an axis through O? If the angular speed of the system after the collision is w, what is the angular momentum of the system after the collision? Find a;, the angular speed, just after the collision, in terms of the the given quantities. What is the fractional change in the kinetic energy for the system as a result of the collision?

Explanation / Answer

(a)

initial angualr momentum of the system

Li = m vid/2

(b)

final angualr momentum is

Lf = ( 1/12 M d^2 + m( d/2)^2 ) w

(c)

from the conservastion of angualr momentum

Li = Lf

m vid/2= ( 1/12 M d^2 + m( d/2)^2 ) w

w = 6 mvi/ Md + 3md

(c)

the original energy is

1/2 m vi^2

the final energy

1/2 I w^2 = 1/2 ( 1/12 Md^2 + md^2/4) ( 36 m^2 vi^2/ ( Md + 3 md)^2

= 3 m^2 vi^2 d/2 ( Md + 3md)

the loss of energy is

1/2 mvi^2 - 3 m^2 vi^2 d/2 ( Md + 3md)= m M vi^2 d/ 2( Md+ 3md)

fractioal loss of energy

m M vi^2 d/ 2( Md+ 3md)/  3 m^2 vi^2 d/2 ( Md + 3md) = M/m

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