A hollow sphere of radius 0.50 m, with rotational inertia I = 0.028 kg m2 about

Question

A hollow sphere of radius 0.50 m, with rotational inertia I = 0.028 kg m2 about a line through its center of mass, rolls without slipping up a surface inclined at 10° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 85 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?

Now, the sphere moves 1.0 m up the incline from its initial position.
(c) What is its total kinetic energy now?


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

Explanation / Answer

(b)At some height h       Total kinetic energy   = potential energy + Rotational kinetic energy             95 J   = ( 1/2) Mv2+ ( 1/2) I2        Here    m = mass of the hollow sphere    v = r    95 J   =   ( 1/2) Mv2    + (1/2)  (2/3) Mr2 (v /r)2 95 J =   ( 1/2) Mv2    (   1+  (2/3) )    .............(1) moment of inetia of the hallow sphere I = (2/3) Mr2    = 0.036 kg.m2 radius of the sphere   r = 0.5 m   Then M =  ( 3/2)(0.036 kg.m2 ) ( 1/(0.5 m)2 )   = 0.216 kg           plug this values in equation (1) , we get         velocity   v   = 22.97 m/s   (a) Initial rotational kinetic energy   = (1/2)  (2/3) Mr2 (v /r)2     =   (1/3)Mv 2                    K (rotational) = 37.988 J (c) If the height speed up to the 1.0 m    Initial energy   Mgh =   95 J ==>         initial height h = (95 J) /( (0.216 kg) ( 9.8 m/s2 )   =   44.87 m    If   1.0 m height is increased h = 45.87 m (c) Then total energy = Mgh =   (  ( (0.216 kg) ( 9.8 m/s2 ) )( 45.87 m ) = 97.09 J   (d) From equation (1)   97.09 J =   ( 1/2) Mv2    (   1+  (2/3) )   solving for new position velocity    v =   23.22 m/s                    K (rotational) = 37.988 J (c) If the height speed up to the 1.0 m    Initial energy   Mgh =   95 J ==>         initial height h = (95 J) /( (0.216 kg) ( 9.8 m/s2 )   =   44.87 m    If   1.0 m height is increased h = 45.87 m (c) Then total energy = Mgh =   (  ( (0.216 kg) ( 9.8 m/s2 ) )( 45.87 m ) = 97.09 J   (d) From equation (1)   97.09 J =   ( 1/2) Mv2    (   1+  (2/3) )   solving for new position velocity    v =   23.22 m/s 97.09 J =   ( 1/2) Mv2    (   1+  (2/3) )   solving for new position velocity    v =   23.22 m/s

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