Centripedal Motion
Alright, I''m trying to review centripedal motion from my physics class last year, and I can''t seem to get this deceptively simple problem. Someone please tell me what I''m doing wrong.
I''ve got this problem that I found that says, "A ball is traveling at a uniform speed of 4.15 m/s with a mass of .3 kg and the radius of the circle in which it is traveling is .85m. Find the tension in the force of the rope keeping the ball in centripedal motion at the top of its circular travel and at the bottom."
For the top I did this:
(Defining downwards as the negative direction)
-FT1 - mg = (MV^2)/r
Yet, when I solve for this problem, I do not get the answer of 3.13N, I get around 9 N.
For the bottom I did this:
(again defining downwards as the negative direction)
FT2 - mg = (MV^2)/r;
For this answer I get around 9 N, which is the correct answer. However, I can''t understand why I get the answer for the first question wrong. I''m defining both problems with the same directions, so why do I get the same answer for both questions and not 3.13 N for the first question. If anyone could see this mistake, I''d be greatful. I''ve been looking for it around half an hour.
--BioX
The acceleration has magnitude given by M*V^2/r, but has different directions at different times.
-FT1 - mg = -(MV^2)/r (acceleration down, negative)
FT1 = -(-M*V^2/r + Mg) = -(-0.3*(4.15)^2/0.85 + 0.3*9.81) = 3.136
FT2 - mg = (MV^2)/r (acceleration up, positive)
FT2 = MV^2/r + Mg = 0.3*(4.15)^2/0.85 + 0.3*9.81 = 9.022
-FT1 - mg = -(MV^2)/r (acceleration down, negative)
FT1 = -(-M*V^2/r + Mg) = -(-0.3*(4.15)^2/0.85 + 0.3*9.81) = 3.136
FT2 - mg = (MV^2)/r (acceleration up, positive)
FT2 = MV^2/r + Mg = 0.3*(4.15)^2/0.85 + 0.3*9.81 = 9.022
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