A 77 g , 39-cm-long rod hangs vertically on a frictionless, horizontal axle pass

ID: 1502470 • Letter: A

Question

A 77 g , 39-cm-long rod hangs vertically on a frictionless, horizontal axle passing through its center. A 14 g ball of clay traveling horizontally at 2.1 m/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?

Explanation / Answer

The clay ball has momentum 0.014*2.1 = 0.0294 kg-m/s just before it hits the tip of the rod.

Its angular momentum with respect to the pivot point of the rod is 0.0294*0.39 = 11.466*10^-3 kg-m²/s.

The angular momentum of the stick and rod after collision must equal the initial angular momentum of the ball.

The angular momentum of a rotating system is I*, where I is the moment of inertia of the system.

The moment of inertia of the rod (mass Mr) plus ball (mass Mb) is

I = Mr*L²/3 + Mb*L² =(Mr/3 + Mb)*L²

I = 6.033*10^-3 kg-m²

Solve for the angular speed by equating the angular momentums:

I* =11.466*10^-3 kg-m²/s

6.033*10^-3 kg-m² * = 11.466*10^-3 kg-m²/s

=1.901 rad/s

The initial rotational energy of the system is 0.5*I*² = 0.01089 J

The potential energy of the system when it stops is elevation change of ball + elevation change of center of gravity of rod:

Mb*L*(1 - cos) + Mr*(L/2)*(1 - cos)

(1 - cos)*(Mb*L + Mr*L/2) =0.01089 J

cos = 1 - 0.01089/(Mb*L + Mr*L/2) = 0.4681

= 62.1º

NOTE: You cannot simply equate the initial kinetic energy of the ball with the system potential energy, since this is an inelastic collision and KE is not conserved.

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A 77 g , 39-cm-long rod hangs vertically on a frictionless, horizontal axle pass

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