A uniform spherical shell of mass M = 4.5 kg and radius R = 8.5 cm can rotate ab

Question

A uniform spherical shell of mass M = 4.5 kg and radius R = 8.5 cm can rotate about a vertical axis on frictionless bearings (See Figure). A massless cord passes around the equator of the shell, over a pulley of rotational inertia I = 3.0 x 10 ^-3 kg m ^2and radius r = 5.0 cm, and is attached to a small object of mass m = 0.60 kg. There is no friction on the pulley’s axle; the cord does not slip on the pulley.

What is the angular acceleration of the sphere?

Hint: You should attempt the problem by first writing Newton's second law for each of the three objects. Remember that since they are moving together they share the same linear acceleration.

Explanation / Answer

ince they are moving together they share the same linear acceleration

mass of spherical shell M = 4.5 kg and radius R = 8.5 cm = 0.085 m

Moment of inertia of sphere = 2/3 MR2 = 0.021675

pulley of rotational inertia I = 3.0 x 10 ^-3 kg m ^2 and radius r = 5.0 cm = 0.05 m

small object of mass m = 0.60 kg

We have to find the angular acceleration of sphere.

Let the angular acceleration of sphere be lpha
The acceleration of string will be same throughout.

Let the acceleration be a

a = lphaR = 0.085lpha
Newton's second law for mass m = 0.60 kg

mg - T1 = ma

0.6*9.8 - T1 = 0.6a

T1 = 5.88 - 0.6a

Newton's second law for pulley

torque = I(a/0.05) = (3.0 x 10 ^-3 )*(20a)

(T1 - T2 )(0.05) = (3.0 x 10 ^-3 )*(20a)

5.88 - 0.6a - T2 = 1.2

T2 = 4.68- 0.6a

Now

Newton's second law for spherical shell

T2 (0.085) = 0.021675 (lpha)

4.68- 0.6a = 0.021675 (lpha)/0.085

4.68- 0.6(lpha/0.085) = 0.021675 (lpha)/0.085

lpha = 0.6398 rad/sec

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