Question

# A point object with mass 6 kg and a uniform rigid rod with mass 6 kg  and...

A point object with mass 6 kg and a uniform rigid rod with mass 6 kg  and length 11 meter are on a horizontal frictionless planar surface. Point object hits the rod vertically with velocity 8 m/s and sticks to the rod.

Part A

Calculate the angular velocity ω about the center of mass just after the point object with mass m sticks to the rod.

Part B

Calculate the ratio of the lost energy to the initial kinetic energy of the system?

here,

mass of object , m1 = 6 kg

the mass of rod , m2 = 6 kg

l = 11 m

initial vertical velocity , u = 8 m/s

a)

when the mass sticks to the rod

the center of mass of the system is (L/4) from the center of rod

using conservation of angular momentum

Li = Lf

(m * u * L/4) = (m * (L/4)^2 + m * L^2 /12 + m *(L/4)^2) * w

(8 * 11/4) = ((11/4)^2 + 11^2 /12 + (11/4)^2) * w

solving for w

the new angular velocity is 0.87 rad/s

b)

the ratio of the lost energy to the initial kinetic energy of the system, R = (KEi - KEf)/KEi

R = (0.5 * m * u^2 - 0.5 * (m * (L/4)^2 + m * L^2 /12 + m *(L/4)^2) * w^2 ) /(0.5 * m * u^2)

R = (8^2 - ( (11/4)^2 + 11^2 /12 + (11/4)^2) * 0.87^2 ) /(8^2)

R = 0.7

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