A mass, m, moving with horizontal velocity, v, strikes and sticks to a thin rod

Question

A mass, m, moving with horizontal velocity, v, strikes and sticks to a thin rod of mass, M, and length, d that hangs vertically by a frictionless hinge (see Fig.). The moment of inertia of the rod about its centre of mass (COM), is given by I_COM = 1/12 Md^2. a) What is the angular velocity of the rod just after impact? b) What speed, v, must the mass, m, have in order for the rod to swing to the horizontal position, i.e. theta = 90 degree A mass, m, moving with horizontal velocity, v, strikes and sticks to a thin rod of mass, M, and length, d that hangs vertically by a frictionless hinge (see Fig.). The moment of inertia of the rod about its centre of mass (COM), is given by I_COM = 1/12 Md^2. a) What is the angular velocity of the rod just after impact? b) What speed, v, must the mass, m, have in order for the rod to swing to the horizontal position, i.e. theta = 90 degree

Explanation / Answer

initial angular momentum of the system Li = m*v*(2/3)d

after the mass sticks to the rod

final angular momentum Lf = [(1/12)*M*d^2 + m*((2/3)d)^2] *wf

from conservation of angular momentum

Lf = Li

[(1/3)*M*d^2 + m*((2/3)d)^2] *wf = m*v*(2/3)*d

((1/3)*M*d + m*(4/9)*d )*wf = m*v*(2/3)

(3Md + 4md) *wf = 6*m*v

wf = 6*m*v/(3Md + 4md)   <<<<--------answer

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after collision KE at the bottom = PE at 90 degree

(1/2)*I*wf^2 = (M*g*d/2) + (m*g*d*2/3)

wf = m*v*d*(2/3)/I

(1/2)*(mvd(2/3)^2)/I = (M + 2m/3)*d*g

(1/2)*m^2*v^2*d^2*4/(3Md + 4md) = (M+ 2m/3)*g*d

2m^2*v^2*d^2/(3Md + 16md) = (M + 2m/3)*g*d

2*m^2*v^2/(3M + 16m) = (M + 2m/3)*g

v^2 = (16m^2/3 + 9mM + 1.5M^2)*g/m^2

v^2 = ((16/3) + 9*M/m + 1.5*(M/m)^2)

v = sqrt(((16/3) + 9*M/m + 1.5*(M/m)^2))

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