In a video game, a 5.0 kg bar of length 0.40 m hangs vertically, and traps some

Question

In a video game, a 5.0 kg bar of length 0.40 m hangs vertically, and traps some hostages to the right of the bar (the bar is prevented from swinging to the left/clockwise). The only way to rescue the hostages is to throw a 2 kg piece of clay (which will stick on the bar half way up its length), causing the bar to swing to the right/counter clockwise. How fast does the clay need to be traveling to get the bar/clay combination to swing such that it is perfectly horizontal at its highest point? Assume that the pivot is frictionless and that the clay ball is a point mass

Explanation / Answer

M = mass of bar = 5 kg

L = length = 0.4 m

I = moment of inertia of bar about its end = ML2/3 = 5 (0.4)2/3 = 0.267

v = speed of clay

r = L/2 = 0.4 /2 = 0.2 m

W = angular speed of bar - clay combination

Using conservation of angular momentum

initial angular momentum of clay + initial angular momentum of clay = angular momentum of clay bar combination

mvr + 0 = (I + mr2) W                       eq-1

using conservation of energy

Rotational KE of combination = PE

(0.5) (I + mr2) W2 = Mg(L/2) + mg(L/2)

(0.5) (0.267 + 2 (0.2)2) W2 = 5 (9.8) (0.2) + 2 (9.8) (0.2)

W = 8.89 rad/s

Using eq-1

mvr + 0 = (I + mr2) W  

2 (0.2) v + 0 = (0.267 + 2 (0.2)2) (8.89)

v = 7.71 m/s

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.