Ac = v^2/r
r = L cos phi
Horizontal:
FT cos phi = m Ac = m v^2/(L cos phi )
Vertical
FT sin phi = m g
solve this first for
FT = m g /sin phi
then
(m g /sin phi)cos phi = m v^2/(L cos phi)
note: m does not matter in the end.
now get v from circumference
2 pi r = v /1.8
v = 1.8*2*pi * L cos phi
v^2 = (1.8*2*pi * L)^2 cos^2 phi
so
( g /sin phi)cos phi = (1.8*2*pi * L)^2 cos^2 phi/(L cos phi)
note cos phi cancels
solve for sin phi and phi
go back and get FT
Estimate the force a person must exert on a string attached to a 0.160 kg ball to make the ball revolve in a circle when the length of the string is 0.600 m. The ball makes 1.80 revolutions per second. Do not ignore the weight of the ball. In particular, find the magnitude of FT, and the angle ϕ it makes with the horizontal. [Hint: Set the horizontal component of FT equal to maR; also, since there is no vertical motion, what can you say about the vertical component of FT?]
I have the force but how do you find the angle?
1 answer
1 answer