Work done shortening string of rotating object

[WestCoast]

Homework Statement

10kg is attached to the end of a string, which is thread through a vertical tube. The string and the tube always make a 90 degree angle. The distance from the tube to the mass is R and the velocity at which it rotates is V. How much work is done to shorten the distance from the tube to the mass to x?

__distance X________10kg
[l] string
[l]
[l] t
[l] u
[l] b
[l] e
[l]
.l
.l string

Homework Equations

work = torque * theta = r*F*sin(theta)

The Attempt at a Solution

string radius being shortened from X to x, so the displacement is (X-x)
does the velocity change?

 

[mfb]

work = torque * theta = r*F*sin(theta)

That is not a good approach. There is no torque - the string is always in radial direction.

does the velocity change?

Think about angular momentum.

 

[rcgldr]

The problem statement should mention that the 10kg mass is sliding on a frictionless surface and that there are no other losses of energy such as aerodynamic drag.

That is not a good approach. There is no torque - the string is always in radial direction.

Depends on how "radial" is defined in this case. During the time when the mass is pulled in by the string, it's path is not circular, and the direction of the string is not perpendicular to the momentary velocity of the mass.

Think about angular momentum

This is a good hint.

 

[mfb]

Depends on how "radial" is defined in this case.

With the obvious polar coordinate system.

and the direction of the string is not perpendicular to the momentary velocity of the mass.

That does not change the fact that we have zero torque.

 

[rcgldr]

The string is always in radial direction.

Depends on how "radial" is defined in this case.

With the obvious polar coordinate system.

The alternative would be the direction of the radius of curvature, which would always be perpendicular to the velocity of the mass. I'll wait to see if the original poster follows up on this thread.

 

[WestCoast]

we know angular momentum = (position vector relative to origin) x (linear momentum)
- linear momentum = 10kgV
-what is the position vector relative to origin?
-what angle does the position vector make with the linear momentum.
-- if < 180 angular momentum = 0.

 

[mfb]

we know angular momentum = (position vector relative to origin) x (linear momentum)
- linear momentum = 10kgV
-what is the position vector relative to origin?
-what angle does the position vector make with the linear momentum.
-- if < 180 angular momentum = 0.

To evaluate the product, is sufficient to know the angle between string and velocity. You can assume that the mass makes a nearly perfect circle around the end of the tube.

 

[rcgldr]

what angle does the position vector make with the linear momentum?

The angle is unknown, since the rate of pull on the string is unknown. The problem can be solved in terms of the initial radius R, and the final radius x, without knowing the rate at which this transition occurred. The angular momentum in this case would be:

ω = angular velocity
m = mass
r = radius
I = angular inertia = m r^2 (for a point mass moving in a circle)
v = linear velocity = ω r

angular momentum = ω I = ω m r^2 = v m r

The original poster hasn't replied to this thread, so he may have already solved it.