1. A 0.l0-kilogram solid rubber ball is attached to the end ofa 0.80-meter lengt
ID: 1733639 • Letter: 1
Question
1. A 0.l0-kilogram solid rubber ball is attached to the end ofa 0.80-meter length of light thread. The ball is swung in avertical circle, as shown in the diagram above. Point P, the lowestpoint of the circle, is 0.20 meter above the floor. The speed ofthe ball at the top of the circle is 6.0 meters per second, and thetotal energy of the ball is kept constant.(a) Determine the total energy of the ball, using the floor asthe zero point for gravitational potential energy. (b) Determine the speed of the ball at point P, the lowestpoint of the circle. (c) Determine the tension in the thread at i. the top of the circle; ii. the bottom of the circle. (d) The ball only reaches the top of the circle once beforethe thread breaks when the ball is at the lowest point of thecircle. Determine the horizontal distance that the ball travelsbefore hitting the floor. 1. A 0.l0-kilogram solid rubber ball is attached to the end ofa 0.80-meter length of light thread. The ball is swung in avertical circle, as shown in the diagram above. Point P, the lowestpoint of the circle, is 0.20 meter above the floor. The speed ofthe ball at the top of the circle is 6.0 meters per second, and thetotal energy of the ball is kept constant.
(a) Determine the total energy of the ball, using the floor asthe zero point for gravitational potential energy. (b) Determine the speed of the ball at point P, the lowestpoint of the circle. (c) Determine the tension in the thread at i. the top of the circle; ii. the bottom of the circle. (d) The ball only reaches the top of the circle once beforethe thread breaks when the ball is at the lowest point of thecircle. Determine the horizontal distance that the ball travelsbefore hitting the floor.
Explanation / Answer
a. Total energy(TE) at bottom=Total energy(TE) attop=mgh+1/2*m*v2=0.1*g*1.8+1/2*0.1*62 b. (PE+KE)bot=(PE+KE)top mgh1+1/2*m*v12=mgh2+1/2*m*v22 0.1*g*0.2+1/2*0.1**v12=0.1*g*1.8+1/2*0.1*62 Solving we get the velocity at the bottom of it. c. At the bottom T1-mg=mv12/r T1=0.1*v12/0.8+0.1gN At the top it as T2+mg=mv22/r T2=0.1*v12/0.8-0.1g d. The vertical velocity with which it hits the floor is v2-u2=2gh v=2gh v=2g*1.8 m/s The time taken by it is v-u=gt The horizontal distance travelled by it as, H=6*t Hence we get by it. "Hope this helps!Best of luck for the rest of yourcoursework."Related Questions
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