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Chapter 11 Student Solutions Manual

# Chapter 11 Student Solutions Manual - Chapter 11 Student...

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Chapter 11 – Student Solutions Manual 5. By Eq. 10-52, the work required to stop the hoop is the negative of the initial kinetic energy of the hoop. The initial kinetic energy is K I mv = + 1 2 2 1 2 2 ω (Eq. 11-5), where I = mR 2 is its rotational inertia about the center of mass, m = 140 kg, and v = 0.150 m/s is the speed of its center of mass. Eq. 11-2 relates the angular speed to the speed of the center of mass: ω = v / R . Thus, ( )( 2 2 2 2 2 2 1 1 140 kg 0.150 m/s 2 2 v K mR mv mv R = + = = ) , which implies that the work required is – 3.15 J. 23. If we write then (using Eq. 3-30) we find G r x y z = + + ± ± ± i j k G G r F × is equal to yF zF zF xF xF yF z y x z y x + + d i b g d i ± ± i j ± k. (a) Plugging in, we find ( )( ) ( )( ) ˆ ˆ 3.0m 6.0N 4.0m 8.0N k (50N m) k. τ = = G (b) We use Eq. 3-27, where φ is the angle between | | sin , G G r F rF × = φ G r and . Now G F r x y = + = 2 2 50 . m and F F F x y = + = 2 2 10 N. Thus, rF = = 50 10 50 . m N N m, b gb g the same as the magnitude of the vector product calculated in part (a). This implies sin φ = 1 and φ = 90°. 29. (a) We use where G A G G = × mr v , G r is the position vector of the object, G v is its velocity vector, and m is its mass. Only the x and z components of the position and velocity vectors are nonzero, so Eq. 3-30 leads to G G r v xv zv z z × = − + b ± j. g Therefore, ( ) ( ) ( ) ( ) ( ) ( ) ( ) ˆ ˆ j 0.25 kg 2.0 m 5.0 m s 2.0 m 5.0 m s j 0. z x m xv zv = + = + − = G A (b) If we write then (using Eq. 3-30) we find G r x y z = + + ± ± ± i j k, G G r F × is equal to yF zF zF xF xF yF z y x z y x + + d i b g d i ± ± . i j ± k With x = 2.0, z = –2.0, F y = 4.0 and all other components zero (and SI units understood) the expression above yields

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