A hollow sphere of radius 0.130 m, with rotational inertia I = 0.0859 kg·m2 abou
ID: 1695804 • Letter: A
Question
A hollow sphere of radius 0.130 m, with rotational inertia I = 0.0859 kg·m2 about a line through its center of mass, rolls without slipping up a surface inclined at 17.9° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 130 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 1.20 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
SOL:
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hallow sphere radious r = 0.13m
rotational inertia I= 0.859 kg.m^2
spher's total kinetic energy = 130J
inclined plane angle = 17.9 deg
momentum of inertia I= 2/3 M r^2
0.859 = 2/3 M (0.13)^2
mass of the sphere M= 76.24 kg
the spher's total K.E = 1/2 MV^2 + 1/2 I W^2
=1/2 MV^2 + 1/2 (2/3Mr^2)(V^2/r^2) (since: w=v/r)
=5/6 MV^2
130 J =5/6 MV^2
130*6/5 =MV^2
V=sqrt( 156/76.24 ) m/s
=1.43 m/s
the kinetic energy 1/2 MV^2 =1/2 (76.24)*(1.43)^2 =77.95 J
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(a) rotational K.E =total energy - kinetic energy
=130J - 77.95 J
= 52 J
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(b)
at intial position its speed of centre of mass Vcm= 1.43 m/s
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(c) when a sphere moves a distance 1.2m from the incine plane.
the acceleration of the sphere a= g sin (17.9)
= 9.8*0.3
= 2.94 m/s^2
moved distance (s) = 1.2 m
by using the kinematic relation ,
V^2 -U^2 =2gs ( these velocities are at s=1.2 m height of the plane.)
V^2= 2(2.94)(1.2)
=7.056
V=sqrt(7.056) m/s
=2.656 m/s
its total K.E at this position =1/2 M V^2
=1/2 *76.24 *(2.656)^2 J
= 268.9 J
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(d) the speed of its C.M at this positon = 2.656 m/s
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