A uniform thin-walled hollow sphere with a radius of 0.120 m has a mass of 1.60

Question

A uniform thin-walled hollow sphere with a radius of 0.120 m has a mass of 1.60 kg. The sphere is rolling along a horizontal surface, without slipping, at a constant speed of 4.75 m/s.

a) What is the kinetic energy of the sphere due to its translational motion along the surface?

b) What is the rotational inertia of the sphere about an axis of rotation through its center of mass?

c) What is the rotational speed of the sphere about its center of mass?

d) What is the kinetic energy of the sphere due to its rotational motion about its

center of mass?

e) What is the total kinetic energy of the sphere as it rolls without slipping?

Explanation / Answer

Here,

radius ,r = 0.120 m

mass , m = 1.60 Kg

constant speed , v = 4.75 m/s

a) kinetic energy of sphere = 0.5 * m * v^2

kinetic energy of sphere = 0.5 * 0.120 * 4.75^2

kinetic energy of sphere = 1.354 J

b)

rotational inertia of sphere = 0.4 * m * r^2

rotational inertia of sphere = 0.4 * 1.60 * 0.120^2

rotational inertia of sphere = 9.216 *10^-3 Kg.m^2

c)
let the rotational speed of the sphere is w

as v = r * w

4.75 = 0.120 * w

w = 39.5 rad/s

the rotational speed of the sphere is 39.5 rad/s

d)

kinetic energy of sphere due to rotational motion = 0.5 * I * w^2

kinetic energy of sphere due to rotational motion = 0.5 * 9.216 *10^-3 * 39.5^2

kinetic energy of sphere due to rotational motion = 7.22 J

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