9. A thin, uniform, metal bar, I (m) long, mass M(kg), is hanging vertically fro

Question

9. A thin, uniform, metal bar, I (m) long, mass M(kg), is hanging vertically from the ceiling by a frictionless pivot. Suddenly it is struck 3/5 / (m) below the ceiling by a small m-kg ball, initially traveling horizontally at v m/s. The ball rebounds in the opposite direction with a speed of u m/s. (ignore air resistance) A. Explain what happens to the angular momentum and mechanical energy of the system-before and after the collision. You need to explain why or why not conservation laws apply B. What angle does the bar make with the vertical when it stops swinging up?

Explanation / Answer

Since there is no exrernal force so angular momentum is conserved .

Conserving angular momentum

3mvl/5=(1/3)Ml^2*w-3mul/5

we got

w=(9m(v-u)/5Ml)

Now we can see that

Initial energy=0.5mv^2

final Energy=0.5mu^2+(0.5)*(1/3)*M*L^2*w^2

Initial energy is not equal to final energy so mechanical energy of system will be reduced while angular momentum will be conserved

B

Let the angle is theta

conserving Energy

0.5*(1/3)ML^2*w^2=Mgl(1-cos(theta))

1-cos(theta)=Lw^2/6g

theta=cos^(-1)(1-Lw^2/6g)

substitute value of w =(9m(v-u)/5Ml)

We will get angle in terms of v ,u .M and m .

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