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LSU PHYS 2101 - Angular momentum

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1!Phys 2101 Gabriela González 2!No forces, no linear momentum: conservation of linear momentum. No torques, no angular momentum: conservation of angular momentum. total linear momentum total angular momentum €  F net=d P dt €  τ net=d L dt2!3! €  F net=d P dt=d M v com( )dt= Md v comdt= M a com € d L dt= τ net= I α = Id ω dt=d I ω ( )dt €  P = M v com L = I ω 4! €  L = I ω3!5!A 4.0 kg particle moves in an xy plane. At the instant when the particle's position and velocity are r = (2.0 i+ 4.0 j ) m and v= -4.0 j m/s, the force on the particle is F = -3.0 i N. At this instant, determine: (a) the particle's angular momentum about the origin, (b) the particle's angular momentum about the point x = 0, y = 4.0 m, (c) the torque acting on the particle about the origin, and (d) the torque acting on the particle about the point x = 0, y = 4.0 m. 6!A 4.0 kg particle moves in an xy plane. At the instant when the particle's position and velocity are r= (2.0 i+ 4.0 j ) m and v= -4.0 j m/s, the force on the particle is F = -3.0 i N. At this instant, determine: (a) the particle's angular momentum about the origin: L= r x p = (2,4,0)m x 4kg (0,-4,0)m/s = 32 kg m/s (0,0,-1) : into the page (b) the particle's angular momentum about the point x = 0, y = 4.0 m: L’ = r’ x p = (2, 0, 0) m x 4kg (0,-4,0)m/s = 32 kg m/s (0,0,-1) : the same (c) the torque acting on the particle about the origin: τ = r x F = (2, 4, 0)m x (-3,0,0)N = 12 Nm (0,0,1): out of the page (d) the torque acting on the particle about the point x = 0, y = 4.0 m: τ’ = r’ x F = (2,0,0)m x (-3,0,0)N = 0 : not the same! r v F x y4!7!A 30 kg child stands on the edge of a stationary merry-go-round of mass 100 kg and radius 2.0 m. The rotational inertia of the merry-go-round about its axis of rotation is 150 kg·m2. The child catches a ball of mass 1.0 kg thrown by a friend. Just before the ball is caught, it has a horizontal velocity of 12 m/s that makes an angle of 37° with a line tangent to the outer edge of the merry-go-round, as shown in the overhead view of the figure. What is the angular speed of the merry-go-round just after the ball is caught? 8!A 30 kg child stands on the edge of a stationary merry-go-round of mass 100 kg and radius 2.0 m. The rotational inertia of the merry-go-round about its axis of rotation is 150 kg·m2. The child catches a ball of mass 1.0 kg thrown by a friend. Just before the ball is caught, it has a horizontal velocity of 12 m/s that makes an angle of 37° with a line tangent to the outer edge of the merry-go-round, as shown in the overhead view of the figure. What is the angular speed of the merry-go-round just after the ball is caught?5!9!A 1.0 g bullet is fired into a 0.50 kg block that is mounted on the end of a 0.60 m nonuniform rod of mass 0.50 kg. The block–rod–bullet system then rotates about a fixed axis at point A. The rotational inertia of the rod alone about A is 0.060 kg·m2. Assume the block is small enough to treat as a particle on the end of the rod. (a) What is the rotational inertia of the block–rod–bullet system about point A? (b) If the angular speed of the system about A just after the bullet's impact is 4.5 rad/s, what is the speed of the bullet just before the impact? 10!A 1.0 g bullet is fired into a 0.50 kg block that is mounted on the end of a 0.60 m nonuniform rod of mass 0.50 kg. The block–rod–bullet system then rotates about a fixed axis at point A. The rotational inertia of the rod alone about A is 0.060 kg·m2. Assume the block is small enough to treat as a particle on the end of the rod. (a) What is the rotational inertia of the block–rod–bullet system about point A? (b) If the angular speed of the system about A just after the bullet's impact is 4.5 rad/s, what is the speed of the bullet just before the


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