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Unformatted text preview: 2300N 3m k g 80 kg 800N 1000N 1) Two people get on a balanced see saw that weighs 100 kg. The big person weighs 80 kg and the small person weighs 50 kg. The 80kg person is 3m from the fulcrum. a) Figure out how far the small person is from the fulcrum. Draw an extended force diagram. Ffulcrum on seesaw Στ=3*80*10
x*50*10=0 x=4.8m Fearth on seesaw Fbig on seesaw 50 kg 500N b) Calculate the force of the fulcrum. Draw a force diagram. ΣF=Ffulcrum
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500=0 Ffulcrum=2300N Fsmall on seesaw c) Imagine the 50 kg person is instead 3m away from the fulcrum. Determine the direcLon and magnitude of the net torque. Στ=3*80*10
3*50*10=900 Nm (out of the page) 2) If we take the bicep problem from the FNT, and rotate it as shown, but keep gravity poinLng down, it changes the problem. The forearm has a mass of 3kg. a) Draw complete force diagrams and extended force diagrams for the forearm. Στ = 180*30+30*15
Fbicep*5=0 Fbicep = 1170N ΣFy=Fupperarm+180+30
1170=0 Fupperarm=960N b) If the bicep exerts twice as much force, determine the magnitude and direcLon of the net torque on the forearm. 1170N 180N 30N 960N Στ = 180*30+.30*.15
2340*.05= 58.50 Nm (out of the page) c) Treat the forearm as a rectangular rod (see page 69 of your course notes) with length 30cm. If the torque remains constant for .1 sec, what is the angular velocity? Irod=1/3 m L2 = .5*3*.32 = .135 kg m2 ΔL = Στ Δt = 58.5 * .1 = 5.85 Δω=ΔL/I=5.85/.135=43.3 rad/s 3) A girl is riding on the edge of a merry go round. The merry go round weighs 100kg and has a radius of 2m (see page 69 of your course notes). The girl has a mass of 50kg. She is iniLally standing on the edge of the merry go round as seen in the picture. The merry go round has an angular velocity of 10 rad / s. a) Calculate the iniLal angular momentum of the merry go round and the girl. Idisk=1/2 m r2=.5*100*4= 200 Igirl = m r2 = 50 * 4 = 200 L=200*10+200*10 = 4000 kg m2/s b) She moves 1m towards the center of the merry go round. Calculate the new angular velocity and draw an angular momentum chart Igirl = m r2 = 50 * 12 = 50 L=4000 = I ω = (200+50) ω ω=16 rad / s Chart on the next page c) Was there a net torque when she moved in? Was there a net force? The center of mass is somewhere between the center of the mgr and the girl, so it is constantly changing direcLon as they rotate, which means there is a net force. That force is coming from the axle. But, since that force acts at the axis of rotaLon, there is no net torque. Li + ΔL = Lf Girl MGR Total 0 ...
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This note was uploaded on 01/05/2011 for the course PHY 7b taught by Professor Taylor during the Spring '08 term at UC Davis.
 Spring '08
 Taylor
 Angular Momentum, Momentum

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