Q79P 79.A uniform rod of length L res... [FREE SOLUTION]

91影视

University Physics with Modern Physics

Hugh D. Young, Roger A. Freedman

Physics

14th edition Edition 路 1596 pages

ISBN: 9780321973610

Chapter 1: Q79P (page 335)

79.A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod. (a) What is the final angular speed of the rod? (b) What is the ratio of the kinetic energy of the system after the collision to the kinetic energy of the bullet before the collision?

Short Answer

Expert verified

$(a) final angular speed of the rod is 蝇_{2} = \frac{6 v}{19 L} . (b) T h e r e q u i r e d r a t i o o f t h e k i n e t i c e n e r g y o f t h e s y s t e m i s \frac{E_{k_{2}}}{E_{k_{1}}} = \frac{3}{19}$

Step by step solution

Angular speed of the rod

(a)

Consider a rod of length with a bullet traveling parallel to the horizontal surface.

The final angular momentum is given by $L_{2} = I 蝇_{2} 鈥� 鈥� (1)$

Since, the total angular momentum is the angular momentum of both rod and bullet

localid="1667820163329" $that is I = I_{r} + I_{b} where I_{r} = \frac{1}{3} m_{r} L^{2} and I_{b} = \frac{1}{16} m_{r} L^{2} .$

localid="1667820233961" $Thus, the sum is I = \frac{1}{3} m_{r} L^{2} + \frac{1}{16} m_{r} L^{2} . Substitute I in (1) and simplify .$

$L_{2} = (\frac{1}{3} m_{r} L^{2} + \frac{1}{16} m_{r} L^{2}) 蝇_{2} = (\frac{1}{3} + \frac{1}{16}) m_{r} L^{2} 蝇_{2} = \frac{19}{98} m_{r} L^{2} 蝇_{2}$

localid="1667820376175" $Since, the initial angular momentum is L_{1} = \frac{1}{8} L m_{r} v By the law of angular momentum, \frac{1}{8} L m_{r} v = \frac{19}{98} m_{r} L^{2} 蝇_{2} which implies 蝇_{2} = \frac{6 v}{19 L} . Therefore, the final angular momentum is 蝇_{2} = \frac{6 v}{19 L}$

Kinetic energy of the system

$(b) The initial kinetic energy is given by E_{k_{1}} = \frac{1}{2} m_{b} v^{2} + \frac{1}{2} I_{r} 蝇_{1}^{2} At initial kinetic energy 蝇_{1} = 0 and m_{b} = \frac{m_{r}}{4} then, E_{k_{1}} = \frac{1}{2} \frac{m_{r}}{4} v^{2} .$

$The final kinetic energy is given by E_{k_{2}} = \frac{1}{2} I 蝇_{2}^{2} where I = \frac{19}{98} m_{r} L^{2} and 蝇_{2} = \frac{6 v}{19 L} then,$

$E_{k_{2}} = \frac{1}{2} I {蝇_{2}}^{2} = \frac{1}{2} (\frac{19}{48} m_{r} L^{2}) {(\frac{6 v}{19 L})}^{2} = \frac{1}{2} (\frac{19}{48} m_{r} L^{2}) (\frac{6 v}{19 L}) (\frac{6 v}{19 L}) = \frac{3}{152} m_{r} v^{2}$