A uniform spherical shell of mats M = 14.0 kg and radius r = 0.600 m can rotate

Question

A uniform spherical shell of mats M = 14.0 kg and radius r = 0.600 m can rotate about a vertical axis on frictionless bearings (see the figure). A massless cord passes around the equator of the shell, over a pulley of rotational inertia I = 0.120 kg-m^2 and radius r = 0.0850 m, and is attached to a small object of mass m = 4.50 kg. There Is no friction on the pulley's axie; the cord does not slip on the pulley. What is the speed of the object when it has fallen a distance 1.01 m after being relased from rest? Use energy considerations.

Explanation / Answer

here,

Initially, the only energy in the system is the potential energy of the hanging block.

PE = m * g * h

PE = (4.5 kg)*(9.8 m/s)*(1.01 m)

PE = 44.54 J

After falling a distance h, that energy is converted to

* kinetic energy of the block
* rotational energy of the pulley
* rotational energy of the sphere

Let V be the speed of the block at that time.

The kinetic energy of the block is:

KE = m * V^2 / 2

KE = (4.5 kg) * V^2 / 2

KE = (2.25 kg) * V^2

The rotational energy of the pulley is:

RE1 = I1 * w^2 / 2
where
w = V / r

so

RE1 = I1 * (V/r)^2 / 2

RE1 = I1 * V^2 / 2r^2

RE1 = (8.3) * V^2

The moment of inertia of a uniform spherical shell is:

2 * M * R^2 / 3

so

I2 = 2 * (14 kg) * (0.6 m)^2 / 3

I2 = 3.36 kg m^2

The rotational energy of the sphere is:

RE2 = I2 * w^2 / 2
where

w = V / R

so

RE2 = I2 * (V/R)^2 / 2

RE2 = I2 * V^2 / 2R^2

RE2 = (3.36 kgm^2) * V^2 / 2(0.6 m)^2

RE2 = (4.67 kg) * V^2

Applying conservation of energy, you get:

PE = KE + RE1 + RE2

(44.54 ) = (2.25)*V^2 + (8.3 )*V^2 + (4.67)*V^2

solving for V

V = 1.71 m/s

the speed of the object is 1.71 m/s

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