Solving Circular Motion Problems: Work and Tension in Springs

[TN17]

Homework Statement

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.

Homework Equations

F = mv²/ R
F = kx
W = Fxcos(theta)

The Attempt at a Solution

a)
F= mv²/ R = 1(1)² / 1 = 1 N

b)
According to Newton's third law, the force ON the string is equal to the force BY the string, so:
F= kx
1 = k(1-0.9)
10 = k

c)
If R' is the new radius, then:
F = mv²/ R'
and
F = kx = k(R' - 0.9)
SO
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

d)

Work for any circular motion = 0
because x and F are perpendicular to each other, so cos 90 = 0.

Is this correct?

 

[TN17]

Here is the image:
http://i1097.photobucket.com/albums/g349/Physics_/Spring.jpg

 

[Redbelly98]

(a), (b), and (c) look good.

For (d): I am not sure how to interpret this question. While it's true that no work is done while the mass is moving in a circle of fixed radius, there would be work done on it while the radius changes from 1.0 m to 1.23 m.

 

[TN17]

(a), (b), and (c) look good.

For (d): I am not sure how to interpret this question. While it's true that no work is done while the mass is moving in a circle of fixed radius, there would be work done on it while the radius changes from 1.0 m to 1.23 m.

That's true. =\
How would I find the work while the radius changes?

 

[Beer-monster]

The work done on the mass as the spring stretches would be stored as potential energy is the spring. The elastic energy is the spring is proportional to the extention squared.

 

[chatmike]

Do you go to MSS?

 

[aaku516]

omg, I go to mss too! xD