Question

A rod of mass "10m" and length "L" is hanging through a pivot at one end. A lump of clay of mass "m" is thrown perpendicular to the rod hitting it at the opposite end of the pivot. The lump sticks to the rod causing the rod to swing in a vertical circle up to an angle of 90^o. Given [m, L] Determine:

a. The initial speed of the lump of clay.

b. The angular speed of the system just after the collision.

Answer 1

Let vo is the initial speed of clay.

let w is the angular speed of rod just after the collision,

Moment of Inertia of rod = M*L^2/12

= 10*m*L^2/12

mment of inertia about pivot = m*L^2

vertical vertical displacement of the center of mass of rod = L/2

vertical dispalcement of clay = L

after the collsion apply conservation of energy

final potentail energy = initial kinetic energy

m*g*L + M*g*(L/2) = 0.5*I*w^2

m*g*L + 10*m*g*(L/2) = 0.5*(10*m*L^2/12 + m*L^2)*w^2

6*m*g*L = (22/24)*m*L^2*w^2

6*g = (22/24)*L*w^2

w^2 = 6*g*L*24/22

w = sqrt(6*9.8*24/22)

= 8 rad/s <<<<<<<<<-------------------Answer for part B)

when the collsion takes place apply conservation of angular momentum

m*vo*L = I*w

m*vo*L = (10*m*L^2/12 + m*L^2)*w

m*vo*L = (22/12)*m*L^2*w

vo = (22/12)*L*w

vo = (22/12)*L*8

= 14.67*L   <<<<<<<<<-------------------Answer for part A)