a ball on the end of a string is revolved at the uniform rate in a vertical circle of radius 72.0 cm. if its speed is 4 .00m/s and its mass is 0.300 kg, calculate the tension in the string when the ball ia (A) at top of its path and (B) at the bottom of its path.
2006-11-24 09:42:10 · 3 answers · asked by Luis M 1 in Science & Mathematics ➔ Physics
(A) centripetal force = weight of ball + Tension
(mv^2)/r = mg + T
T= (0.3x(4^2 )/0.72) - 0.3x9.81
T = 3.72 N
(B) centripetal force = Tension - Weight of ball
(mv^2)/r = T - mg
T= (0.3x(4^2 )/0.72) = 0.3x9.81
T= 9.61N
2006-11-24 17:00:05 · answer #1 · answered by kooshal 2 · 0⤊ 0⤋
For such a motion to take place there has to be a net force towards the centre of the circle to provide centripetal acceleartion, a equal (v.v)/r, which equals in this case to 16/0.72. So a net force F equal to ma given by
F = 0.3(16/.72) = 2/0.03 = 200/3 m/(s.s) must act on the ball.
When the ball is at the top, the external forces acting on it are the force with which earth attracts = mg, downawrds and tension T1 of the string also acting downawrds on the ball. So we have for net force F
F = mg + T1 = (.3)(9.8) + T1 = 200/3
T1 = 66.66 - 29.4 = 37.26 N
When the ball is at the bottom mg by earth keeps acting downward but thetenion in the string is now acting upward on the ball
F = T2 - mg = T2 - 29.4 = 66.66
T2 = 29.4 + 66.66 = 96.06 N
2006-11-24 10:52:09 · answer #2 · answered by Let'slearntothink 7 · 0⤊ 0⤋
that is what oppinions are for, ya know! it is a good thing to all be different. i personally think it's really interesting to hear everyone else's oppinions. we need to respect each other in the sence that we each have our own worldview but still be open to discussion. if you don't like that type of question then okay, i'm fine with that, but don't bother people with this type of question. go to a subject that's logical enough for you.
2016-05-22 23:05:07 · answer #3 · answered by ? 4 · 0⤊ 0⤋