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

ID: 1347281 • Letter: A

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).


true false greater than less than equal to  M's angular momentum at y is .... that at x.
true false greater than less than equal to  M's kinetic energy at y is four times that at x.
true false greater than less than equal to  M's angular velocity at y is four times that at x.
true false greater than less than equal to  While going from x to y, there is a torque on M
true false greater than less than equal to  As M moves from x to y, the work done by F is .... 0.


Explanation / Answer

part a )

Angular momentum = moment of inertia * angular velocity

Moment of inertia = mass * radius^2

Angular velocity = radians/ sec

Angular momentum = mass * radius^2 * w

The radius at x is 2 times radius at y.

The angular velocity at x is 1/2 the angular at y.

Angular momentum at x = mass * (2 * ry)^2 * 1/2 * w y

Angular momentum at y = mass * ry^2 * w y

Angular momentum at x/angular momentum at y

= mass * (2 * ry)^2 * 1/2 * w y / mass * ry^2 * w y = r

Angular momentum at x / angular momentum at y = 4/2 = 2

less than

part b )

Fx = mass * vx^2/r1

Fy = mass * vy^2/(1/2 * r1)

vy^2 / vx^2 = ½

KE = 1/2 * mass * velocity^2

KEy/KEx= 1/2

False it 2 time

part c )

wy = 4*wx

true

part d )

true

tension in the string is causing the mass to rotate around the center of the circle

part e )

greater than 0

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