Hw 5 Solutions

# D dt τ τ g a g g consequently the vector d dt g a

• Notes
• 8

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d dt τ τ = + = G A G G Consequently, the vector d dt G A has a magnitude 2 0 4 0 4 5 2 2 . . . + − = b g N m and is at an angle θ (in the xy plane, or a plane parallel to it) measured from the positive x axis, where θ = F H G I K J = − ° tan . . 1 4 0 2 0 63 , the negative sign indicating that the angle is measured clockwise as viewed “from above” (by a person on the + z axis). P6.- . If we write (for the general case) G r x y z = + + ± ± ± i j k, then (using Eq. 3-30) we find G G r v × is equal to yv zv zv xv xv yv z y x z y x + + d i b g d i ± ± ± . i j k (a) The angular momentum is given by the vector product G A G G = × mr v , where G r is the position vector of the particle, G v is its velocity, and m = 3.0 kg is its mass. Substituting

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(with SI units understood) x = 3, y = 8, z = 0, v x = 5, v y = –6 and v z = 0 into the above expression, we obtain ( ) 2 2 ˆ ˆ 3.0 [(3.0)( 6.0) (8.0)(5.0)]k ( 1.7 10 kg m s)k. = = − × G A (b) The torque is given by Eq. 11-14, G G G τ = × r F . We write G r x y = + ± ± i j and G F F x = ± i and obtain G τ = + × = − x y F yF x x ± ± ± ± i j i k e j e j since ± ± i i × = 0 and ± ± ± j i k. × = − Thus, we find ( )( ) ˆ ˆ 8.0m 7.0N k (56N m)k. τ = − = G (c) According to Newton’s second law G G A τ = d dt , so the rate of change of the angular momentum is 56 kg m 2 /s 2 , in the positive z direction P7. - (a) Since τ = dL / dt , the average torque acting during any interval t is given by τ avg = L L t f i d i , where L i is the initial angular momentum and L f is the final angular momentum. Thus 2 2 avg 0.800 kg m s 3.00 kg m s 1.47 N m 1.50s τ = = − , or avg | | 1.47 N m τ = . In this case the negative sign indicates that the direction of the torque is opposite the direction of the initial angular momentum, implicitly taken to be positive. (b) The angle turned is θ ω α = + 0 2 1 2 t t . If the angular acceleration α is uniform, then so is the torque and α = τ / I . Furthermore, ω 0 = L i / I , and we obtain ( ) ( ) ( )( ) 2 2 2 2 1 1 3.00kg m s 1.50s 1.467 N m 1.50s 2 2 20.4rad. 0.140kg m i Lt t I τ θ + + = = = (c) The work done on the wheel is ( )( ) 1.47 N m 20.4rad 29.9 J W τθ = = − = −
where more precise values are used in the calculation than what is shown here. An equally good method for finding W is Eq. 10-52, which, if desired, can be rewritten as ( ) 2 2 2 f i W L L I = .

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