Suppose a 0250 kg ball Is thrown at 160 ms to a motionless p

Suppose a 0.250 kg ball Is thrown at 16.0 m/s to a motionless person standing on ice who catches It with an outstretched arm as shown In Figure 9.31. (a) Calculate the final linear velocity of the person, given his mass Is 80.0 kg. 0.05 m/s (b) What is his angular velocity if each arm has a 5.00 kg mass? You may treat his arms as uniform rods of length 0.9 m and the rest of his body as a uniform cylinder of radius 0.170 m. Neglect the effect of the ball on his rotational inertia and on his center of mass, so that it remains in his geometrical center. 2.9 rad/s (c) Compare the Initial and final total kinetic energy. 0.66875 X (Initial energy / final energy)

Solution

mass of the rest of the body=80-5-5=70 kg
moment of inertia of uniform cylinder about its central axis=0.5*mass*radius^2=0.5*(70)*0.17^2=1.0115 kg.m^2

moment of inertia of a uniform rod about an axis passing through its edge and perpendicular to the rod=mass*length^2/3

=5*0.9^2/3=1.35 kg.m^2

but the axis of rotation here is 0.17 m (radius of the cylinder assumption of the body) from the edge of the uniform rod

hence using parallel axis method for calculation of moment of inertia,

moment of inertia of each hand=1.35+mass*0.17^2

=1.35+5*0.17^2

=1.4945 kg/m^2

hence total moment of inertia of the system=moment of inertia of the body +2*moment of inertia of the hand

=1.0115+2*1.4945

=4 kg.m^2

part a:

as momentum is conserved,

final momentum=initial momentum

==>(80+0.025)*v=.25*16

==>v=0.0498 m/s=0.05 m/s

part b: angular velocity=tangential veloicty/radius

=0.05/(0.9+0.17)

=0.04673 rad/sec

part c:

initial kinetic energy=0.5*mass of the ball*speed^2

=0.5*0.25*16^2=32 J

final kinetic energy=0.5*moment of inertia*angular speed^2+0.5*mass*linear speed^2

=0.5*4*0.04673^2+0.5*80*0.05^2

=0.10437 J

hence initial energy/final energy=32/0.10437=306.6