Chapter 11. h = 5m. = mgh mv Iω 2. E f. = E i. v = 4 3 g(h h) = m / s2 (8m 5m) = 6.26m / s. ω = v r = 6.


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1 Chapter A solid cylinder of radius 10cm and mass 1kg starts from rest and rolls without slipping a distance of 6m down a house roof that is inclined at 30 degrees (a) What is the angular speed of the cylinder about its center as it leaves the house roof? (b) The roof s edge is 5.0m. How far horizontally from the roof s edge does the cylinder hit the level ground. This problem is very similar to the previous problem. We begin with energy conservation. Take the ground as zero height. H 5m + 6msin30 8m E i mgh h 5m E f mgh + 1 mv + 1 Iω E f E i mgh + 1 mv + 1 Iω mgh mgh + 1 mv + 1 (1 mr ) v R mgh 1 mv + 1 (1 m)v mg(h h) 1 v v g(h h) 3 4 v g(h h) v 4 3 g(h h) m / s (8m 5m) 6.6m / s ω v r 6.6m / s 0.1m 6.6rad / s This problem differs from the previous problem in that the cylinder is not launched horizontally. It has an initial velocity in the y direction as well as in the x direction... x i 0 x f? v ix 6.6cos30 5.4m / s v fx v ix a x 0 y i h y f 0 v iy 6.6sin m / s v fy? a y g t? Again we find the time
2 y f y i + v iy t + 1 a y t 0 h + v iy t 1 gt v iy ± t v iy 4 ( 1 g) h ( 3.13) ± ( 3.13) 4 ( 1 ( 1 9.8) 5 g) ( 1 9.8) 0.74s Again we find the horizontal distance x f x i + v ix t + 1 a x t 0 + v ix t + 0 x f 5.4m / s 0.74s 4.01m This problem is considerably more complicated because the projectile is not launched horizontally In the Figure, a solid brass ball of mass 0.80 g will roll smoothly along a looptheloop track when released from rest along the straight section. The circular loop has radius R 14.0cm and the ball has radius r R. (a) What is h if the ball is on the verge of leaving the track when it reaches the top of the loop if the ball is release at height h 6.00R, what are the (b) magnitude and (c) direction of the horizontal force component acting on the ball at point Q. h i mg Q R R When the ball is just about to come off the track, the centripetal force is entirely due to gravity. This allows us to solve for the velocity
3 mv R mg v Rg Now that we know the velocity, we can use energy conservation. The initial state is where we release the ball, the final state is at the top of the loop. E i mgh E f 1 mv + 1 I ω + mg(r) E i E f mgh mg(r) + 1 mv + 1 I ω mgh mg(r) + 1 mv + 1 ( 5 mr ) v mgh mg(r) + 1 mv + 1 ( 5 m)v r h g(r) + 1 v + 1 ( 5 )v g g(r) + 1 Rg + 1 ( 5 )Rg g (R) + 1 R R 7 10 R In the second part of this problem, we need the velocity so that we can compute the centripetal force again, since it is the net inward force. In this case, we know the initial height. The force at Q will be to the left.
4 E i E f mg(6r) mg(r) + 1 mv + 1 I ω mg(6r) mg(r) + 1 mv + 1 ( 5 mr ) v mg(5r) 1 mv + 1 ( 5 m)v v g(5r) ( 5 ) 10 7 g(5r) r 50 7 gr F mv R m 50 7 gr R 50 7 m g 11.8 A.0 kg particlelike object moves in a plane with a velocity components v x 30m / s and v y 60m / s as it passes through the point with (x,y) coordinates of (3.0, 4.0) m. Jus then, in unitvector notation, what is the angular momentum and (b) the point (.0, .0) m? We can draw what is happening in each case. The r is defined as coming from the axis and ending where the force is applied. In the first case
5 r pmv 45 r ( )î + ( ) ĵ 3.0m î 4.0m ĵ p mv 60kg m / s î + 10kg m / s ĵ L r p î ĵ ˆk det (3 10 ( 4) 60) ˆk 600kgm / s ˆk In the second case, the axis is at (,), so we need to recalculate r.
6 pmv 4 r 5 r (3.0 (.0))î + ( 4.0 (.0)) ĵ 5.0m î.0m ĵ p mv 60kg m / s î + 10kg m / s ĵ L r p î ĵ ˆk det 5 0 (5 10 ( ) 60) ˆk 70kgm / s ˆk A sanding disk with rotational inertia I kg m is attached to an electric drill whose motor delivers a torque of magnitude 16Nm about the central axis of the disk. About that axis and with the torque applied for 33 ms, what is the magnitude of the angular momentum and angular velocity of the disk. We begin by finding the angular velocity after 33 ms. We begin by finding the angular acceleration. τ I α α τ I 16Nm kg m rad / s
7 Now find the angular velocity ω i + α t s 440rad / s Now we can find the angular momentum. L I 0.58kg m / s The angular momentum of a flywheel having a rotational inertia of kg m about its central axis decreases from 3.0 to kg m /s in 1.50 s. (a) What is the magnitude of the average torque acting on the flywheel about its central axis during this period? (b) Assuming a constant angular acceleration, through what angle does the flywheel turn? (c) How much work is done on the wheel? (d) what is the average power of the flywheel? x DL Dt L f  L i 0.800kg m /s kg m /s 1.47Nm Dt 1.5s x Ia x 1.47Nm a I 0.140kg m rad/s L ~ i i 3.00kg m /s I 0.140kg m 1.48 rad/s i f i i + ~ it + 1 at rad/s $ 1.5s + 1 ( rad/s )(1.5s) 0.43 rad A man stands on a platform that is rotating (without friction) win an angular speed of 1. rev/s; his arms are outstretch and he hold a brick in each hand. The rotational inertia of the system consisting of the man, bricks and platform about the central vertical axis of the platform is 6.0 kg m. If by moving the bricks, the man decreases the rotational inertia of the system to.0 kg m, what are (a) the resulting angular speed of the platform and (b) the ratio of the new kinetic energy of the system to the original kinetic energy? (c) What source provided the added kinetic energy? This is an angular momentum conservation problem. Since all of the forces and torques are internal, the angular momentum cannot change. First we ll use conservation of angular momentum to find the final angular speed (a).
8 L f L i I f ~ f I i~ i 1. rev r radians ~ i $ 7.54 radians/s s rev ~ f I f (b) Ratio of Kinetic Energies I i 6 kg m ~i kg m $ 7.54 radians/s.6 radians/s KE R f KEi 1 I f ~ f 1 Ii~ i 1 $ ( kg m )(.6 radians/s) 3 1 (6 kg m )(7.54 radians/s) (c) Moving the bricks required work. The man s arms did work The rotational inertial of a collapsing spinning star changes to 1/3 its initial value. What is the ratio of the new rotational kinetic energy to the initial kinetic energy? I f I i ω i 3ω i I i I ω i i I f (I i / 3) ω i R 1 I f 1 I i ω i 1 (I i / 3)(3ω i ) 1 I i ω i 3
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