Angular momentum and conservation of angular momentum problem

In summary, the problem can be solved by using the conservation of angular momentum equation. The radius of the smallest possible circle on which the object can move is 0.85 m.
  • #1
mx2ko
7
0
A small 0.531-kg object moves on a frictionless horizontal table in a circular path of radius 0.85 m. The angular speed is 6.30 rad/s. The object is attached to a string of negligible mass that passes through a small hole in the table at the center of the circle. Someone under the table begins to pull the string downward to make the circle smaller. If the string will tolerate a tension of no more than 105 N, what is the radius of the smallest possible circle on which the object can move?



conservation of angular momentum
angular momentum = moment of inertia x angular velocity
moment of inertia = mass x radius squared
torque = force x lever arm
torque = moment of inertia x angular acceleration
maybe more equations I'm not sure.



I tried to use the conservation of linear momentum equation since you can find the momentum of the object when it is going around in the outer circle. I don't know what to do with the force or how to work it into the equation to solve for r of the little circle.
 
Physics news on Phys.org
  • #2
The answer can be found using conservation of angular momentum. The moment of inertia of a point mass is given by I = mr^2, where m is the mass of the object and r is the radius of the circle. The angular momentum of the object is L = Iω, where ω is the angular velocity. Therefore, the angular momentum of the object in the inner circle is equal to the angular momentum of the object in the outer circle. Rearranging this equation gives:L = mr1^2 * ω1 = mr2^2 * ω2where r1 and ω1 are the radius and angular velocity of the outer circle, and r2 and ω2 are the radius and angular velocity of the inner circle. Since the mass and angular velocity are known, we can solve for the radius of the inner circle:r2 = r1 * (ω1 / ω2)The angular velocity of the inner circle is limited by the tension in the string, which can be no greater than 105 N. Therefore, the angular velocity of the inner circle must be less than or equal to 6.30 rad/s. Plugging in the values, we get:r2 = 0.85 m * (6.30 rad/s / 6.30 rad/s) = 0.85 mTherefore, the radius of the smallest possible circle on which the object can move is 0.85 m.
 

1. What is angular momentum?

Angular momentum is a measure of an object's tendency to continue rotating or spinning about an axis. It is a vector quantity that depends on an object's mass, velocity, and distance from the axis of rotation.

2. How is angular momentum calculated?

Angular momentum is calculated as the product of an object's mass, velocity, and distance from the axis of rotation. In mathematical terms, it is represented as L = mvr, where L is angular momentum, m is mass, v is velocity, and r is the distance from the axis of rotation.

3. What is conservation of angular momentum?

Conservation of angular momentum states that the total angular momentum of a system remains constant when there is no external torque acting on the system. This means that if the net torque on a system is zero, then its angular momentum will remain constant.

4. How does angular momentum relate to rotational motion?

Angular momentum is directly related to rotational motion. Any object that is rotating has angular momentum, and the rate at which its angular momentum changes is equal to the net torque acting on the object. This is known as the angular momentum principle.

5. How is angular momentum conserved in real-life situations?

Angular momentum is conserved in many real-life situations, such as when a figure skater pulls their arms in to spin faster or when a diver tucks their body to rotate faster during a dive. Both of these actions decrease the object's moment of inertia, but the product of its mass and velocity increases to keep its angular momentum constant.

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
821
  • Introductory Physics Homework Help
Replies
23
Views
871
  • Introductory Physics Homework Help
10
Replies
335
Views
7K
  • Introductory Physics Homework Help
2
Replies
45
Views
2K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
737
Replies
13
Views
822
  • Introductory Physics Homework Help
Replies
18
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
22
Views
3K
Back
Top