Question
A 60 g ball is tied to the end of a 50- cm-long string andswung in a vertical circle. The center of the circle, as shown inthe figure , is 150 cm above the floor. The ball is swung at theminimum speed necessary to make it over the top without the stringgoing slack. If the string is released at the instant theball is at the top of the loop, how far to the right does the ballhit the ground? A 60 g ball is tied to the end of a 50- cm-long string andswung in a vertical circle. The center of the circle, as shown inthe figure , is 150 cm above the floor. The ball is swung at theminimum speed necessary to make it over the top without the stringgoing slack. If the string is released at the instant theball is at the top of the loop, how far to the right does the ballhit the ground? If the string is released at the instant theball is at the top of the loop, how far to the right does the ballhit the ground?
Explanation / Answer
minimum speed necessary means that centrip force = mg . mv2 /r = mg v = gr = 9.80*0.50 = 2.2136 m/s . Now... the ball has this horizontal velocity when it isreleased and it is 150 + 50 = 200 cm = 2.00 m above the ground. . This means it takes t =2h/g = 2*2.00/9.80 = 0.63888seconds to hit the ground and it travels . distance = speed * time = 2.2136 *0.63888 = 1.414 metersto the right