A small round solid object beta equals 3/4 of mass m and radius r is released fr

Question

A small round solid object beta equals 3/4 of mass m and radius r is released from rest at height h on a circular loop track of radius R. The objects rolls without slipping and the radius of the object is much less than that of the track r<<R. Assume air resistance is negligible. A) calculate the objects moment of inertia I. B) as with any rotating body, the objects moment of inertia I measures it's......? C) calculate the objects center of mass speed vcm at the bottom of the ramp( just before it begins the loop) given that it's released from rest at height h. D) calculate the minimum center of mass speed vcm the object can have at the top of the loop and still successfully complete the loop. E) calculate the minimum height h from which the object can be released and still successfully complete the loop. F) suppose that the object is released from rest at height h=2.50R. Find the magnitude of the net force the track exerts on the object in the x direction when it reaches point P, on the far right side of the circular loop. G) suppose that the object is released from rest at height h = 2.50R. Find the friction force the track exerts on the object in the y direction when it reaches point P, on the far right side of the circular loop. You may assume that the coefficient of static friction is known and has value, and that the object is just about to start slipping when it reaches point P. H) same thing as G but this time find the net torque on the object when it reaches point P.

Explanation / Answer

A)

I = 2/5*mr^2

B)

Resistance to rotation.

C)

Initial Potential energy = mgh

At bottom, all of it gets converted to KE. = KE_translation + KE_rotation

mgh = 1/2*mv^2 + 1/2*I*w^2

But w = v/r and I = 2/5 *mr^2

So, mgh = 1/2*mv^2 + 1/2 * 2/5 *mv^2

mgh = 7/10 *mv^2

v = sqrt [10*gh/7]

D)

At top, Centrifugal force > Weight

mv^2 / R > mg

v > sqrt (Rg)

E)

mgh = mg*(2R) + 7/10*mv^2

mgh = mg*2R + 7/10*m*(R*g)

h = 2R + 7/10*R

h = 2.7*R

F)

Fx = Centrifugal force = mv^2 / R

mg*(2.5*R) = mg*R + 7/10*mv^2

v = sqrt (15/7 *gR)

Fx = 15/7 *mg

G)

Let f = friction force.

Torque = f*r = I*ang.acceleration = (2/5*mr^2) *(a/r)

f = 2/5*ma

Also, ma = mg + f

Therefore, f = 2/5*(mg + f)

f = 2/3 *mg

H)

Torque = f*r = 2/3 *mgr

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