321_PartUniversity Physics Solution

321_PartUniversity Physics Solution - 10-16 Chapter 10 2...

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10-16 Chapter 10 (b) 1 0; K = 22 11 2 AA B B Km vI ω =+= ( ) 2 4 1 3 (0.0100 kg)(0.200 m/s) [0.0500 kg][1.00 m] (0.120 rad/s) 3.2 10 J. += × The increase in kinetic energy comes from work done by the bug when it pushes against the bar in order to jump. EVALUATE: There is no external torque applied to the system and the total angular momentum of the system is constant. There are internal forces, forces the bug and bar exert on each other. The forces exert torques and change the angular momentum of the bug and the bar, but these changes are equal in magnitude and opposite in direction. These internal forces do positive work on the two objects and the kinetic energy of each object and of the system increases. 10.46. IDENTIFY: Apply conservation of angular momentum to the system of earth plus asteroid. SET UP: Take the axis to be the earth&s rotation axis. The asteroid may be treated as a point mass and it has zero angular momentum before the collision, since it is headed toward the center of the earth. For the earth, z z L I = and 2 2 5 IM R = ,where M is the mass of the earth and R is its radius. The length of a day is 2 r a d T π = , where is the earth&s angular rotation rate. EXECUTE: Conservation of angular momentum applied to the collision between the earth and asteroid gives 2 12 55 () MR mR MR =+ and 2 5 2 mM ⎛⎞ = ⎜⎟ ⎝⎠ . 21 1.250 TT = gives 1 1.250 = and 1.250 = . 2 0.250 ωω = . 2 5 (0.250) 0.100 M == . EVALUATE: If the asteroid hit the surface of the earth tangentially it could have some angular momentum with respect to the earth&s rotation axis, and could either speed up or slow down the earth&s rotation rate. 10.47. IDENTIFY: Apply conservation of angular momentum to the collision. SET UP: The system before and after the collision is sketched in Figure 10.47. Let counterclockwise rotation be positive. The bar has 2 1 2 3 Im L = . EXECUTE: (a) Conservation of angular momentum: 2 1 10 1 2 3 mvd m L =− + . 2 2 1 90.0 N (3.00 kg)(10.0 m s)(1.50 m) (3.00 kg)(6.00 m s)(1.50 m) (2.00 m) 39 . 8 0 ms + 588 rads . = . (b) There are no unbalanced torques about the pivot, so angular momentum is conserved. But the pivot exerts an unbalanced horizontal external force on the system, so the linear momentum is not conserved. EVALUATE: Kinetic energy is not conserved in the collision. Figure 10.47 10.48. IDENTIFY: dd t τ = L ! ! , so d L ! is in the direction of ! . SET UP: The direction of ! is given by the right-hand rule, as described in Figure 10.26 in the textbook. EXECUTE: The sketches are given in Figures 10.48a±d.
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Dynamics of Rotational Motion 10-17 EVALUATE: In figures (a) and (c) the precession is counterclockwise and in figures (b) and (d) it is clockwise. When the direction of either ω ! or τ ! reverses, the direction of precession reverses. Figure 10.48 10.49. IDENTIFY: The precession angular velocity is wr I Ω= , where is in rad/s. Also apply m = F a ! ! to the gyroscope. SET UP: The total mass of the gyroscope is rf 0.140 kg 0.0250 kg 0.165 kg mm += + = .
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321_PartUniversity Physics Solution - 10-16 Chapter 10 2...

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