- #1
Jennings
- 36
- 0
1. Homework Statement
(Attached is an image depicting the situation)
A ball of mass m = 1.50kg is hung from a spring attached to a shaft. The length of the spring with the mass hanging from it is Lo = 50.0cm. The shaft then starts to rotate such that the spring stretches to a length L = 65.0cm. The spring constant of the spring is k=1450N/m.
A) At what angle theta does the rotating spring make with the vertical
B) How fast is the ball moving
C) How much work had to be done on the shaft to get the ball moving at the angle and speed given in parts a and b
Homework Equations
Springs : Fs = -kx
Centripital Force : Fc = mv²/r
The Attempt at a Solution
First thing I did was solve for the length of the spring at equilibrium (assuming the spring is massless). First I solve for the change in distance due to the mass attached.
mg = kx
mg/k = x = (1.50kg)(9.81m/s²)/(1450N/m)
x = 1.015cm
Length at Equilibrium
50.0cm - 1.105cm = 48.99cm
Now that I know the equilibrium length I will find the total force being applied to the spring since the length is known.
Change in length from equilibrium
65.0cm - 48.99cm = 16.01cm = .1601m
The total force acting on this spring is
F = kx = (1450N/m)(.1601m) = 232.1N
I don't even know if I'm on the right track with this. I have a free body diagram that I will post, but again I have no idea if I'm on the right track.
If I've made any mistakes please let me know.
Last edited by a moderator: