A hollow sphere of radius 0.130 m, with rotational inertia I = 0.0859 kg·m2 abou
ID: 1696151 • Letter: A
Question
A hollow sphere of radius 0.130 m, with rotational inertia I = 0.0859 kg·m2 about a line through its center of mass, rolls without slipping up a surface inclined at 17.9° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 130 J. (a) How much of this initial kinetic energy is rotational? (b) What is the speed of the center of mass of the sphere at the initial position? When the sphere has moved 1.20 m up the incline from its initial position, what are (c) its total kinetic energy and (d) the speed of its center of mass?Explanation / Answer
Given:
radius of hollow sphere = 0.13 m
Inertia = 0.0859 kg/m^2
angle = 17.9
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Inertia of hollow sphere = I = 2/5mr^2
0.0859= 2/5 *m*(0.13)^2
mass of the sphere m = 14 kg
Total kinetic energy = translatory kinetic energy+rotational kinetic energy
= 1/2mv^2+1/2I^2
= 1/2mv^2+1/2I (v^2/r^2)
130= 1/2(m+I/r^2)V^2
v = 3.69 m/s^2
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(a)
Rotational KInetic Energy = 1/2I^2
= 1/2*0.0859*(3.69^2/0.13^2)
= 34.58 J
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(b) velocity of the sphere at height 1.20 m
By conservation of energy
130 = 1/2(m+I/r^2)v^2+mglsin
v = 2.88 m/s^2
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(c)
Total kinetic energy = K.E.- mglsin
= 79.4 J
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(d)
speed of centre of mass = 2.88 m/s^2
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