A 0.10 kg solid rubber ball is attached to the end of an 0.80meter length of lig
ID: 1729048 • Letter: A
Question
A 0.10 kg solid rubber ball is attached to the end of an 0.80meter length of light thread. The ball is swung in a verticalcircle. Point P, the lowest point of the circle, is 0.20 metersabove the floor. The speed of the ball at the top of thecircle is 6.0 meters per second, and the total energy of the ballis kept constant. a) Determine the energy of the ball, using the floor as thezero point for gravitational potentional energy. b) Determine the speed of the ball at point P, thelowest point of the circle. c) dtermine the tension in the thread at (hint: use a FBD andconsider what force supplies the centripetal force here): i) the top of the circle ii) the bottom of the circle PLEASE HELP if you can! This is an extra credit homeworkquestion, and I am struggling for every point possible in thisclass. THANK YOU very much for your time! A 0.10 kg solid rubber ball is attached to the end of an 0.80meter length of light thread. The ball is swung in a verticalcircle. Point P, the lowest point of the circle, is 0.20 metersabove the floor. The speed of the ball at the top of thecircle is 6.0 meters per second, and the total energy of the ballis kept constant. a) Determine the energy of the ball, using the floor as thezero point for gravitational potentional energy. b) Determine the speed of the ball at point P, thelowest point of the circle. c) dtermine the tension in the thread at (hint: use a FBD andconsider what force supplies the centripetal force here): i) the top of the circle ii) the bottom of the circle PLEASE HELP if you can! This is an extra credit homeworkquestion, and I am struggling for every point possible in thisclass. THANK YOU very much for your time!Explanation / Answer
a) energy of the ball is kinetic +potential energy Et=1/2mv2+mgh =1/2*0.1*6*6+01*9.81*1 =2.781J b) gain in kinetic energy = loss in potential energy. speed of the block at the point P is, v2=36+9.81*0.8 v=6.621m/s. c) i) at the top position centripetal psudoforce is actingupward T=mv2/r-mgh. here h=1m, v=6m/s. ii) at the bottom position centripital psudoforce is actingdownward T=mv2/r+mgh. here h=0.2m, v=6.621m/s
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