A bullet of mass mB = 0.0127 kg is moving with a speed of 101 m/s when it collides with a rod of mass mR = 7.23 kg and length L = 1.09 m (shown in the figure). The rod is initially at rest, in a vertical position, and pivots about an axis going through its center of mass. The bullet imbeds itself in the rod at a distance L/4 from the pivot point. As a result, the bullet-rod system starts rotating.
a) Find the angular velocity, ω, of the bullet-rod system after the collision. You can neglect the width of the rod and can treat the bullet as a point mass.
b) How much kinetic energy is lost in the collision?
(Hint: Please enter a positive number for your answer! If ΔK < 0, then the amount of kinetic energy lost is a positive number.)
http://www.hizliupload.com/viewer.php?file=94332335472721891326.png
a) Find the angular velocity, ω, of the bullet-rod system after the collision. You can neglect the width of the rod and can treat the bullet as a point mass.
b) How much kinetic energy is lost in the collision?
(Hint: Please enter a positive number for your answer! If ΔK < 0, then the amount of kinetic energy lost is a positive number.)
http://www.hizliupload.com/viewer.php?file=94332335472721891326.png
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In this case the rod will not translate, it will only rotate. So, to determine ω, L (Angular Momentum) must be conserved about the hinge, since no external torque acts on that point.
Initially, L B (angular momentum of bullet about the hinge) = mB*v*perpendicular distance = (mB)vL/4.
Finally, L S (L system) = I(system)ω. [ I = moment of inertia]
I system = mR * L² /12 + mB * (L/4)² [assuming the rod to be uniform, its COM is at its centre. I about that point = m l² /12) = mR * L² /12 + mB * L² / 16 = L² [mR/12 + mB/16].
Conserving L, (mB) v L / 4 = L² [mR/12 + mB/16]ω
=> mB v = L (mR / 3 + mB /4) ω
=> ω = mB v / L (mR / 3 + mB /4) .
Putting values,
ω = 0.0127*101/1.09(7.23/3 + 0.0127/4) rad/s = 1.2827/2.63 = 0.488 rad/s (Answer).
K Initial = 1/2 mB v² = 0.5*0.0127*)(101)² = 64.77635 Joule.
K Final = 1/2 I ω² = 0.5 * 0.6 * (1.09)² * (0.488)² = 0.17 Joule.
So, Energy lost = 64.6 Joule (Answer)
Initially, L B (angular momentum of bullet about the hinge) = mB*v*perpendicular distance = (mB)vL/4.
Finally, L S (L system) = I(system)ω. [ I = moment of inertia]
I system = mR * L² /12 + mB * (L/4)² [assuming the rod to be uniform, its COM is at its centre. I about that point = m l² /12) = mR * L² /12 + mB * L² / 16 = L² [mR/12 + mB/16].
Conserving L, (mB) v L / 4 = L² [mR/12 + mB/16]ω
=> mB v = L (mR / 3 + mB /4) ω
=> ω = mB v / L (mR / 3 + mB /4) .
Putting values,
ω = 0.0127*101/1.09(7.23/3 + 0.0127/4) rad/s = 1.2827/2.63 = 0.488 rad/s (Answer).
K Initial = 1/2 mB v² = 0.5*0.0127*)(101)² = 64.77635 Joule.
K Final = 1/2 I ω² = 0.5 * 0.6 * (1.09)² * (0.488)² = 0.17 Joule.
So, Energy lost = 64.6 Joule (Answer)
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About a month ago, it was the same with me. But, i had a rigorous practice with the chapter as i prepared for the high school finals and now i think, i can manage all these. So, i would advice you to do the same. Learn where and when to conserve L, if needed take help from your teacher or the net.
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I understand you. all right, I will pay attention your advice and I will do very practice with the chapter. I have finals exams after about one mount. I hope, it is as required. very thanks for all in all.
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