A stationary stick of length L and mass M is pivoted around its center like a pr

Question

A stationary stick of length L and mass M is pivoted around its center like a propeller. A lump of mud also of mass M is thrown at the stick, near its end, with velocity v in a direction perpendicular to it as shown. The mud sticks and the stick+mud moves. What is the speed of the mud immediately after the collision? The rotational inertia for a stick rotated around its center as

shown is I=1/2ML2

A stationary stick of length L and mass M is pivoted around its center like a propeller. A lump of mud also of mass M is thrown at the stick, near its end, with velocity v in a direction perpendicular to it as shown. The mud sticks and the stick+mud moves. What is the speed of the mud immediately after the collision? The rotational inertia for a stick rotated around its center as shown is I=1/2ML2

Explanation / Answer

let x be the vertical distance of the mud to the lower end of the rod.
momentum is conserved.
so v*m=(M+m)*v_G
Assume that the angular speed of the system is w.
then the center of mass of the rod is moving with speed of wl/2 and the mud is wx.
we have that
M*wl/2+m*wx=m*v

V =M*wl/2+m*wx/m

also Conservation of angular momentum, the pivot point is the lower end of the rod.
m*v*x=Ml^2*w/3+m*x^2*w.
so we have
M*x*wl/2+m*wx^2=m*v*x .
so Ml^2*w/3=M*x*wl/2
so l=3x/2
so x=2l/3

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