A hollow sphere of radius 0.35 m, with rotational inertia I = 0.096 kg m2 about

Question

A hollow sphere of radius 0.35 m, with rotational inertia I = 0.096 kg m2 about a line through its center of mass, rolls without slipping up a surface inclined at 13° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 90 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?
___ m/s

Explanation / Answer

rotational kinetic energy= .5* .096*^2

=.5*.096*v^2/.35^2

=.39 v^2

2*m*.35^2/3= .096

or,m=1.176 kg

translation kinetic energy= .5*1.176*v^2=.588 v^2

so rotational kinetic energy= 90/(.39+.588)*.39=35.9 J

.39*v^2= 35.9

or,v=9.6 m/s

total kinetic energy after climbing 1 m= 90- 1.176*9.8*sin(13)= 87.4 J

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