A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0448 kg·m2 abou
ID: 1492584 • Letter: A
Question
A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0448 kg·m2 about a line through its center of mass, rolls without slipping up a surface inclined at 37.2° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 9.10 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 0.530 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
a) rotational KE = 0.5 I w^2
Translation KE = 0.5 mv^2 = 0.5mr^2 w^2
Rotational KE / translational KE = I/mr^2 = 2/3
40% is rotational
Rotational KE = 0.4*9.1 = 3.64 J
B)0.5 I w^2 = 3.64
0.5* 0.0448 w^2 = 3.64
w ^2= 3.64/( 0.5* 0.0448) = 162.5
w = 12.75
v = wr = 12.75*0.2= 2.55 m/s
C) i =2/3 mr^2= 0.0448
m = 3/2 * 0.0448 /r^2 = 1.5*0.0448 /(0.2*0.2) = 1.68 kg
New total KE = 9.1 J - mgh
= 9.1 - 1.68*9.8*0.530 sin 37.2 degree
=9.1 - 5.28
= 3.82 J
d) new translational KE = 60% of 3.82J =2.29 J
0.5 m v^2 = 2.29
v =sqrt ( 2*2.29/m) = sqrt( 2*2.29/1.68)
= 1.65 m/s
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