A ball of mass 1 kg is attached to a spring. The spring is attached to a fixed pivot, P. The spring cannot bend. The ball moves in a circle of radius R in a horizontal plane with a velocity v. The spring is mass-less and the plane is frictionless.
(a) If R = 1.0 m and v = 1.0 m/s, what is the tension in the spring at the point where it attaches to m?
(b) If the relaxed length of the spring is 0.90 m, what is the spring constant k?
(c) If the ball and spring now rotate with v = 2.0 m/s, what is the new radius of the ball's path.
(d) How much work is done on the mass.
F = mv2/ R
F = kx
W = Fxcos(theta)
The Attempt at a Solution
F= mv2/ R = 1(1)2 / 1 = 1 N
According to Newton's third law, the force ON the string is equal to the force BY the string, so:
1 = k(1-0.9)
10 = k
If R' is the new radius, then:
F = mv2/ R'
F = kx = k(R' - 0.9)
mv2/ R' = k(R' - 0.9)
1(2)2/ R' = 10(R' - 0.9)
4/R' = 10(R' - 0.9)
4 = 10R'2 - 9R'
0 = 10R'2 - 9R' - 4
Quadratic formula gives me R'= 1.226 m
Work for any circular motion = 0
because x and F are perpendicular to each other, so cos 90 = 0.
Is this correct?