Solving Circular Motion Problems: Work and Tension in Springs

In summary, we have a 1 kg ball attached to a spring that is attached to a fixed pivot. The ball moves in a circle of radius R in a frictionless plane with a velocity v. We calculate the tension in the spring at the point where it attaches to the ball, and the spring constant k based on the relaxed length of the spring. We then use this information to find the new radius of the ball's path when the velocity is doubled. We also note that no work is done on the mass while it is moving in a circle of fixed radius, but there would be work done while the radius changes. This work can be calculated using the potential energy stored in the spring, which is proportional to the extension squared.
  • #1
TN17
47
0

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?
 
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  • #3
(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.
 
  • #4
Redbelly98 said:
(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?
 
  • #5
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.
 
  • #6
Do you go to MSS?
 
  • #7
omg, I go to mss too! xD
 

FAQ: Solving Circular Motion Problems: Work and Tension in Springs

What is circular motion and how does it relate to work and tension in springs?

Circular motion is a type of motion where an object moves along a circular path. In this type of motion, the object's speed or velocity is constantly changing, as it moves in a curved path. Work and tension in springs are related to circular motion because they are both forces that act on an object to keep it moving in a circular path.

How do you calculate work in circular motion problems?

In circular motion problems, work can be calculated by multiplying the force applied to the object by the distance it moves along its circular path. This can be represented by the equation W = F*d*cosθ, where W is work, F is force, d is distance, and θ is the angle between the force and the direction of motion.

What is tension in a spring and how does it affect circular motion?

Tension in a spring is the force that is exerted by the spring when it is stretched or compressed. In circular motion, tension in a spring can affect the speed and direction of the object's motion. As the object moves along the circular path, the spring will stretch or compress, exerting a force on the object, which can change its velocity.

How can you use the concept of work in circular motion problems?

The concept of work in circular motion problems can help us understand how forces affect an object's motion in a circular path. By calculating the work done on the object, we can determine the amount of energy that is transferred to the object and how that affects its speed and direction of motion.

What are some common examples of circular motion problems involving work and tension in springs?

Some common examples of circular motion problems involving work and tension in springs include a ball attached to a string being swung in a circular path, a rollercoaster car going around a loop, or a satellite orbiting around a planet. In each of these examples, the forces of work and tension in springs are at play, affecting the object's circular motion.

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