On a frictionless table, a glob of clay of mass 0.400 kg strikes a bar of mass 0

Question

On a frictionless table, a glob of clay of mass 0.400 kg strikes a bar of mass 0.540 kg perpendicularly at a point 0.170 m from the center of the bar and sticks to it.

a) If the bar is 0.660 m long and the clay is moving at 7.800 m/s before striking the bar, what is the final speed of the center of mass?'

b) At what angular speed does the bar/clay system rotate about its center of mass after the impact (in rad/s)?

On a frictionless table, a glob of clay of mass 0.400 kg strikes a bar of mass 0.540 kg perpendicularly at a point 0.170 m from the center of the bar and sticks to it. a) If the bar is 0.660 m long and the clay is moving at 7.800 m/s before striking the bar, what is the final speed of the center of mass?' b) At what angular speed does the bar/clay system rotate about its center of mass after the impact (in rad/s)?

Explanation / Answer

a)

Linear momentum conservation :

V= final linear velocity of COM of the system.

mv = (m+M)V

V = mv/(m+M) = 3.32 m/s...ans

b)

Angular momentum conservation :

mv*r = I.

r= 0.17

New COM is

x= m*0.17/(M+m) = 0.0723 m above the bar centre.

I = ML^2/12 + Mx^2 + m(0.17-x) = 0.0615

= mv*r/I = 8.623 rad/s....ans

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