A 0.40 kg ball on a string is whirled on a vertical circle at a constant speed. When the ball is at the three o'clock position, the tension is 16 N. What are the tensions in the string when the ball is at the twelve o'clock and at the six o'clock positions.
2007-10-17 12:36:47 · 3 answers · asked by Singh J 1 in Science & Mathematics ➔ Physics
Yes I can.
2007-10-17 12:41:20 · answer #1 · answered by Holden 5 · 0⤊ 0⤋
At the 3:00 position, the tension, T3, is equal to the centripetal force. The ball needs a 16 N net force towards the center regardless of the position on the circle.
At 12:00 and 6:00, the ball's weight enters in. At the 12:00 position the ball's weight provides some of the centripetal force needed. To find T12, subtract W = m*g from T3. At the 6:00 position, to get a 16 N net force on the ball, the tension needs to support the ball's weight in addition to providing the centripetal force. T6 = W + T3.
2007-10-17 13:38:37 · answer #2 · answered by sojsail 7 · 0⤊ 0⤋
It's been a while since i did these problems but i think this is how it goes...
at three oclock the string does not help reduce force of the gravity so 16N is the tension of the string when the string is not affected by gravity
so at 6 oclock the string has to pull harder because of gravity and at 12 oclock the string pulls less
just figure out the force of gravity on the mass (9.8 * mass) and add for 6 oclock and subtract for 12 oclock.
i think this is how to do it
2007-10-17 13:05:22 · answer #3 · answered by Sunky 1 · 0⤊ 0⤋