A 3.80-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so

Question

A 3.80-kg cylindrical rod of length 2.00 m is suspended from a horizontal bar so it is free to swing about that end. A solid sphere of mass 0.25 kg is thrown horizontally with a speed

v1 = 13.0 m/s to hit the rod at the point A one-fifth of the way up from the bottom of the rod. The sphere bounces back horizontally with a speed v2 = 9.50 m/s,

while the rod swings to the right through an angle before swinging back toward its original position. What is the angular speed of the rod immediately after the collision?

Explanation / Answer

let,

mass of the cylindrical rod, m1=3.8 kg

length of the rod, l=2m

mass of the sphere, m2=0.25 kg

speed, v1=13 m/sec, v2=9.5 m/sec

moment of inertia, I=m1*l^2/3

I=3.8*2^2/3

I=5.07 kg.m^2

now,

by using law of conservation of angular momentum,

m2*v1*r=m2*v2*r+I*w

0.25*13*4*2/5+0.25*9.5*4*2/5=5.07*w

===> w=1.77 rad/sec

and

by using law of conservation energy,

1/2*I*w^2=m*g*h

1/2*5.07*1.77^2=3.8*9.8*h

=====> h=0.213 m

and

h=L*(1-cos(theta))

h=l/2*(1-cos(theta))

0.213=1*(1-cos(theta))

0.213=1-cos(theta))

====>

theta=38.12 degrees --------->

and

w=1.77 rad/sec

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