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Lecture_16

# Lecture_16 - Example Last time we showed that a solid disk...

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Example : Last time we showed that a solid disk of radius r and mass m will roll faster down an incline than a ring of radius r and mass m, since the moment of inertia of the disk is smaller than that of the ring. Use conservation of energy to calculate the anslational speed of the ring and disk at the bottom of the ramp and show that it is translational speed of the ring and disk at the bottom of the ramp and show that it is faster for the disk. Since gravity is the only conservative force that oes work on the objects by conservation of does work on the objects, by conservation of energy, the total mechanical energy at the top of the ramp for each object must equal the total mechanical energy at the bottom of the ramp. top E h bot E bot top E E = bot Rot Trans top Rot Trans PE KE KE PE KE KE bot bot top top + + = + + bot bot Rot Trans top KE KE PE + = 2 2 1 2 2 1 bot bot I mv mgh ω + = r v bot bot = 2 2 1 2 2 1 + = r v I mv mgh bot bot Plug this in above: Solve this for v bot :

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2 bot I mgh v = This is the speed that each object has at the bottom of the ramp. Notice that it depends on the moment of inertia, I . 2 r m + ing isk Ring 2 mr I = Disk 2 2 1 mr I = 2 2 ring 2 bot mr m mgh v + = 2 2 disk bot 2 mr m mgh v + = r gh v = ring bot 2 r gh v 3 4 bot disk = he disk has a greater translational speed at the bottom of the ramp and The disk has a greater translational speed at the bottom of the ramp, and thus arrives there first.
9.6 Angular Momentum To this point, we have looked at all the following concepts in terms of both translational and angular (rotational) motion: g( ) Translational Rotational Displacement x θ Velocity v ω Acceleration α cce e at o a Force F τ ass Mass m I Work d F d τ⋅ 2 Energy 2 2 1 mv 2 1 I Momentum mv I I is the rotational or Angular Momentum: I L =

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ω I L = Units? [Moment of Inertia] x [Angular velocity] [kg·m 2 ] x [rad/s] [kg·m 2 /s] Remember, when we studied linear momentum , we found that when the sum of , , the external forces was equal to zero, then the total momentum of the system was a constant of the motion, i.e. it was conserved. o f ext P P F = = 0 For rotational motion , we find a similar conservation law: If the sum of the external torques is equal to zero, then the total angular momentum of the system is a constant of the motion, i.e. it is conserved.
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Lecture_16 - Example Last time we showed that a solid disk...

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