A hollow sphere of radius 0.30 m, with rotational inertia I = 0.022 kg m2 about

Question

A hollow sphere of radius 0.30 m, with rotational inertia I = 0.022 kg m2 about a line through its center of mass, rolls without slipping up a surface inclined at 25° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 50 J.
(a) How much of this initial kinetic energy is rotational?
J

(b) What is the speed of the center of mass of the sphere at the initial position?
m/s
Now, the sphere moves 1.0 m up the incline from its initial position.
(c) What is its total kinetic energy now?
J

(d) What is the speed of its center of mass now?

Explanation / Answer

total kinetic energy=I*w^2/2 + m*v*v/2 I of hollow sphere=m*r*r*2/3=.022 now since rolling without slipping v=w*r kinetic energy=m*w*w*r*r/2 =I*w*w*3/4 total kinetic energy=I*w*w*5/4=50 w=47.67 a)rotational=(1/2)/(5/4)=2/5 b) v=47.67*.30=14.301m/s c)total kinetic energy=50-m*g*1 we can find the m from I d)new total kinetic energy will be 5*I*w1*w1/4=50-m*g we can find the w1 from here new v=w1*r hope this will help u..

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