A hollow sphere of radius 0.15 m, with rotational inertia I = 0.040 kg middot m^

Question

A hollow sphere of radius 0.15 m, with rotational inertia I = 0.040 kg middot m^2 about a line through its center of mass, rolls without slipping up a surface inclined at 30 degree to the horizontal. At a certain initial position, the sphere's total kinetic energy is 20 J. How much of this initial kinetic energy is rotational? What is the speed of the center of mass of the sphere at the initial position? When the sphere has moved 1.0 m up the incline from its initial position, what are its total kinetic energy; and the speed of its center of mass?

Explanation / Answer

radius of the sphere, r=0.15m

moment of inertia I=0.04 kg.m^2

angle of inclination, theta=30 degrees

total K.E=20J

a)

total K.E=1/2*m*v^2+1/2*I*w^2

20=1/2*m*v^2+1/2*(2/3*m*r^2)*w^2

20=1/2*m*v^2+1/3*m*r^2*(v^2/r^2)

20=1/2*m*v^2+1/3*m*v^2

20=5/6*m*v^2 ---(1)

==>

rotational K.E=1/3*m*v^2

=(1/3)*20*(6/5)

=8 J

b)

Total K.E=5/6*m*v^2

here

moment of inertia, I=2/3*m*r^2

0.04=2/3*m*(0.15)^2

==> mass, m=2.67 m

and

total K.E=5/6*m*v^2

5/6*m*v^2=20

5/6*2.67*v^2=20

==> v=2.99 m/sec

c)

using enegry relation,

P.E=m*g*h

P.E=m*g*l*sin(theta)

P.E=2.67*9.8*1*sin(30)

P.E=13.083 J

and

K.E=T.E-P.E

K.E=20-13.083

K.E=6.917 J

d)

by using energy relation,

5/6*m*v^2=K.E

5/6*2.67*v^2=6.917

===> v=1.76 m/sec

speed of the center of mass, v=1.76 m/sec

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