Mass wraps around rod in circular motion

In summary, the problem is finding the time taken for a string of length l to completely wrap around a rod of radius R. The string is attached to a mass moving with constant tangential velocity v0, but increasing angular velocity due to acceleration from gravity. Using energy conservation, the equation mgx = 1/2 mv^2 + 1/2 Iw^2 can be used to find the time taken for the string to wrap around the rod. By finding an expression for dl/dt, where l is the length of the string, the time can be calculated.
  • #1
unscientific
1,734
13

Homework Statement


A string attached to the center of a rod with radius R has a mass attached at the other end, moving with speed v0. Because of acceleration due to gravity the mass moves down while undergoing circular motion, causing the string to wrap around the rod. Find an expression for the time taken for the string to completely wrap around the rod.




Homework Equations


v = rw


The Attempt at a Solution


-constant tangential velocity since acceleration due to gravity acts only downwards
-w = v/r, so as the string wraps around the rod, the angular velocity increases

w(t) = v0/r(t)

-w(t) is the same along all lengths of the string

-tried to figure out dr/dt, but to no avail
 

Attachments

  • mass rope rod.jpg
    mass rope rod.jpg
    2 KB · Views: 521
Last edited:
Physics news on Phys.org
  • #2
use energy conservation, where x is the distance moved downwards by the mass, so

mgx = 1/2 mv^2 + 1/2 Iw^2

and use, w = v/r
 
  • #3
supratim1 said:
use energy conservation, where x is the distance moved downwards by the mass, so

mgx = 1/2 mv^2 + 1/2 Iw^2

and use, w = v/r

hmm, i don't really understand how the term '1/2 Iw^2' comes about - the tangential velocity stays the same, all its doing is accelerating downwards

Loss in GPE = Gain in KE
mgx = 1/2 mvy^2 (in the y-direction)

I'm interested in finding out the time taken for a given string of length l to finish wrapping a rod of radius R entirely.

-at every point in time the mass is undergoing circular motion
-fixed tangential velocity, but increasing angular velocity
-in order to find the time taken to wrap around the rod, i must find an expression for dl/dt, where l is the length of the string. I imagine it to be some function of (r,v0 and t)
 
  • #4
it's okay, i solved it! :)
 
  • #5


I would approach this problem by first defining the variables and parameters involved. The radius of the rod, R, and the initial velocity of the mass, v0, are given. The mass itself can be represented by its position on the string, r(t), which changes as the string wraps around the rod.

Next, I would use Newton's laws of motion to analyze the forces acting on the mass. The only force acting on the mass is gravity, which causes it to accelerate downwards. This acceleration also causes a change in the angular velocity of the mass as it moves along the string.

To find the time taken for the string to completely wrap around the rod, I would use the equation v = rw, where v is the tangential velocity, r is the radius, and w is the angular velocity. Since the tangential velocity remains constant, we can set the initial tangential velocity, v0, equal to the tangential velocity at any point along the string, v(t).

Using this equation, we can solve for the angular velocity at any point along the string, w(t). Then, by integrating w(t) with respect to time, we can find the time taken for the string to completely wrap around the rod.

In conclusion, the expression for the time taken for the string to completely wrap around the rod would involve the initial velocity, radius of the rod, and the acceleration due to gravity. It would also involve integrating the angular velocity with respect to time.
 

1. What is circular motion?

Circular motion refers to the movement of an object in a circular path around a fixed point or axis. This type of motion can be seen in phenomena such as the Earth orbiting the sun, a spinning top, or a car turning a corner.

2. How does mass wrap around a rod in circular motion?

When an object with mass is attached to a rod and is moving in a circular path, it is experiencing centripetal force. This force, acting towards the center of the circle, keeps the object in its circular path. As the object moves, the direction of the force constantly changes, causing the mass to wrap around the rod.

3. What factors affect the circular motion of a mass on a rod?

The speed of the object, the radius of the circular path, and the mass of the object are all factors that can affect the circular motion of a mass on a rod. The greater the speed or radius, the greater the centripetal force needed to keep the object in its circular path. A larger mass will also require more force to maintain circular motion.

4. What happens if the centripetal force is not strong enough?

If the centripetal force is not strong enough to keep the object in its circular path, the object will move in a straight line tangent to the circle. This is known as the object "flying off" the circular path.

5. How is circular motion related to Newton's laws of motion?

Circular motion is an application of Newton's laws of motion. The first law states that an object will continue in its state of motion unless acted upon by an external force. In circular motion, the centripetal force acts as the external force to keep the object in its circular path. The second law relates the force, mass, and acceleration of an object, while the third law states that every action has an equal and opposite reaction, which can be seen in the direction of the centripetal force constantly changing in circular motion.

Similar threads

  • Introductory Physics Homework Help
Replies
19
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
660
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
947
  • Introductory Physics Homework Help
Replies
29
Views
1K
  • Introductory Physics Homework Help
2
Replies
38
Views
2K
  • Introductory Physics Homework Help
2
Replies
35
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top