A hollow sphere of radius 0.170 m, with rotational inertia I = 0.0250 kg·m2 abou

Question

A hollow sphere of radius 0.170 m, with rotational inertia I = 0.0250 kg·m2 about a line through its center of mass, rolls without slipping up a surface inclined at 11.4° to the horizontal. At a certain initial position, the sphere's total kinetic energy is 18.0 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 0.930 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

given that

r = 0.170 m

Ir = 0.0250 kg*m^2

theta = 11.4 degree

total kinetic energy = KE = 18 J

we can say that

total KE = translational KE + rotational KE
KE = KEt + KEr

KEt = 1/2 * m*v^2

KEr = 1/2 * I *w^2                        ( I for hollo sphere = 2/3 m*r^2 )

so,

18 = 1/2*m*v^2 + 1/2 * (2/3)*m*r^2*w^2

w = v / r

18 = 1/2* m*v^2 + 1/3 m*v^2 = 5/6*m*v^2

18 = 5/6* m*v^2

KEr = 1/3 * m*v^2 = 1/3 * 18 * 6/5

KEr = 7.2 J

part(b)

l = 2/3* m*r^2

m = 3*I / 2*r^2 = 3 * 0.0250 / 2*(0.170)^2

m = 1.29 kg

KE = 5/6* m*v^2 = 18

v^2 = 18*6 / (5*1.29)

v = sqrt (16.74) = 4.09 m /s^2

part(c)

decrease in total KE at the height h = increase in potential energy.

h = 0.930 * sin (11.4 deg)

PE = m*g*h = 1.29 * 9.8 * sin (11.4)

PE = 2.32 J

total KE at height h = 18 - 2.32 = 15.68 J

part(d)

at height h  

let , speed of center of mass is v1

total KE at height h = 5/6 * m* v1^2

5/6 * m* v1^2 = 15 .68

v1^2 = 6*15.68 / 5*1.29

v1 = sqrt ( 14.58 )

v1 = 3.81 m/s

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