A 79g , 40-cm-long rod hangs vertically on a frictionless, horizontal axle passing through its center. A 15g ball of clay traveling horizontally at 2.0m/s hits and sticks to the very bottom tip of the rod.
To what maximum angle, measured from vertical, does the rod (with the attached ball of clay) rotate?
Apply consrvation of angular momentum
m*v*(L/2) = I*w
m*v*L/2 = (M*L^2/12 + m*(L/2)^2))*w
w = (m*v*L/2)/(M*L^2/12 + m*(L/2)^2))
= (15*2*0.4/2)/(79*0.4^2/12 + 15*0.2^2)
= 3.63 rad/s
now Apply Ebergy consrvation
0.5*I*w^2 = m*g*h
h = 0.5*I*w^2/(m*g)
= 0.5*((M*L^2/12 + m*(L/2)^2))*3.63^2/(m*g)
= 0.5*(79*0.4^2/12 + 15*0.2^2)*3.63^2/(15*9.8)
= 0.0741 m
= 7.41 cm
h = L*(1 - cos(thet))
h/L = 1 - cos(theta)
cos(theta) = 1 - h/L
theta =cos^-1(1 - h/L)
= cos^-1(1 - 7.41/40)
= 35.44 degrees <<<<<<<-------------Answer
A 79g , 40-cm-long rod hangs vertically on a frictionless, horizontal axle passing through its center....
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