A small mass M attached to a string slides in a circle (x) on a frictionless hor

Question

A small mass M attached to a string slides in a circle (x) on a frictionless horizontal table, with the force F providing the necessary tension (see figure). The force is then increased slowly and then maintained constant when M travels around in circle (y). The radius of circle (x) is twice the radius of circle (y).

Answer the following with:  true, false, greater than, less than, equal to

M's angular momentum at y is .... that at x.
As M moves from x to y, the work done by F is .... 0.
M's angular velocity at y is four times that at x.
M's kinetic energy at x is one quarter that at y.

While going from x to y, there is a torque on M

Explanation / Answer

The Conservation of Angular Momentum governs this situation. This is true because there is no external torque or force acting on this system, which consists of M, the string, and the object at the center exerting the centripetal force F. See reference 1, where the example of the skater is cited:

"The conservation of angular momentum explains the angular acceleration of an ice skater as she brings her arms and legs close to the vertical axis of rotation. By bringing part of mass of her body closer to the axis she decreases her body's moment of inertia. Because angular momentum is constant in the absence of external torques, the angular velocity (rotational speed) of the skater has to increase."

The situation with the rotating mass M on a decreasing radius is analogous to the skater's pulling arms and legs inward.

Calculations:
------------------
Angular momentum L and Moment of inertia I

Angular momentum is defined as:
(1) L = I * w,

I is the moment of inertia, which, for a point mass M rotating in circular motion, is given by (ref. 2):
(2) I = M * r^2

In our case,
(3) Ix = M * x^2 and
(4) Iy = M * y^2

It is given that x = 2y; therefore
(5) Iy = M * y^2 = M * (x / 2)^2 = (1/4) * M * x^2

which means that

(6) Iy = (1/4) * Ix <<==important result

Because of (1),
(7) Lx = Ix * wx and, from (6)
(8) Ly = Iy * wy = [(1/4) * Ix] * wy

However, since Conservation of L is in effect,
(9) Lx = Ly

so it must be true that
(10) wy = 4 * wx <<==important result


Kinetic energy:

The rotational kinetic energy KE is given by:
(11) KE = (1/2) * I * w^2

In our situation,
(12) KEx = (1/2) * Ix * wx^2 and
(13) KEy = (1/2) * Iy * wy^2

Combining with results (6) and (10) gives

(14) KEy = (1/2) * ((1/4) * Ix) * (4 * wx)^2

= 4^2 / 4 * KEx = 4*KEx <<==important result


ANSWERS:
-----------------
Using the results above, we are now ready to answer the several questions:

Q1. While going from x to y, there is a torque on M

A1. FALSE, since there is no force acting perpendicular to the radius--i.e., no F * r. The ONLY force acting is along the radius--i.e., directed toward the center--hence no torque.


Q2. M's angular momentum at x is .... that at y.

A2. EQUAL TO because of the Conservation of angular momentum.


Q3. M's kinetic energy at x is half that at y.

A3. FALSE; from (14), KEx = (1/4) * KEy


Q4. M's angular velocity at y is twice that at x.

A4. FALSE; from (10) wy = 4 * wx


Q5. As M moves from x to y, the work done by F is .... 0.

A5. GREATER THAN; in fact, from (14) we know how much work (energy) was expended:
KEy - KEx = 3*KEx.

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