A hollow sphere of radius 0.15 m, with rotational inertia I 0.040 kg m2 about a line through its center of mass, rolls without slipping up a surface inclined at 30 to the horizontal. At a certain initial position, the sphere’s total kinetic energy is 20 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 1.0 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:

Let v be the velocity of the c.m of sphere at the initial position .

Total kinetic energy = linear + rotational kinetic energy

= 1/2 mv² + 1/2 I ω² ; m is mass , I is moment of inertia and ω be angular velocity of sphere .

Total kinetic energy = 1/2 mv² + 1/2 x 1/2 m r² ω² ; (   I = 1/2 mr² , r is radius of the sphere)

Total kinetic energy = 1/2 mv² + 1/2 x 1/2 m r² ω²

= 1/2 mv² + 1/4 m v² (v = ω r , for perfect rolling )

= 3 /4 mv² = 20 J

mv² = 80 / 3

1/4 m v ² = 20 / 3

rotational kinetic energy = 20/ 3   J

b )mv² = 80 / 3

.04 v² = 80/3

v² = 666.67

v = 25.82 m /s

c )

When the sphere has moved 1.0 m up the incline from its initial position

height achieved h = 1 / sin30

= 2 m

potential energy attained by sphere = mgh

= .04 x 9.8 x 2

= .784 J

Total kinetic energy at the final position

= 20 - .784

= 19.216 J

d )

So total energy = 3 / 4 m v²

3 / 4 m v² = 19.216

3/4 x .04 v² = 19.26

v² = 642

v = 25.34 m /s

This is velocity at the final position.

0.55x745.7j and a calculator

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