Question

A hollow sphere of radius $0.15 m$ with rotational inertia $I = 0.048 kg m^{2}$ 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?

Accepted Answer

Given,

Radius of hollow sphere, $R = 0.15 m$ ,

Moment of inertia of hollow sphere, $I = 0.048 kg m^{2}$

Angle made by the incline plane with the horizontal, $θ = 30^{0}$

Total Kinetic energy of sphere is, $K E_{total} = 20 J$

The moment of inertia of hollow sphere is given by,

$I = \frac{2}{3} M R^{2} . . . (1)$

Now the mass of sphere is given by,

$\Rightarrow M = \frac{3 I}{2 R^{2}} = \frac{3 \times 0.048}{2 \times {(0.15)}^{2}} = 3.2 kg$

Rotational Kinetic energy of hollow sphere is given by,

$K E_{r o t} = \frac{1}{2} I ω^{2} = \frac{1}{3} M R^{2} ω^{2}$ .

Furthermore, it rolls without slipping, $v_{com} = R ω$

(a) Fraction of kinetic initial kinetic energy is rotational is given by,

$\frac{K E_{rot}}{K E_{total}} = \frac{K E_{rot}}{K E_{com} + K E_{rot}}$

$\Rightarrow \frac{K E_{rot}}{K E_{total}} = \frac{\frac{1}{3} M R^{2} ω^{2}}{\frac{1}{2} M R^{2} ω^{2} + \frac{1}{3} M R^{2} ω^{2}}$

$\Rightarrow \frac{K E_{rot}}{K E_{total}} = 0.4$

Thus, $40 %$ of the total kinetic energy is rotational, or

The rotational kinetic energy,

$\Rightarrow K E_{rot} = 0.4 \times 20 = 8.0 J$

(b) From $K_{rot} = \frac{1}{3} M R^{2} ω^{2} = 8.0 J$

Angular velocity is given by,

$ω = \sqrt{\frac{3 K E_{rot}}{M R^{2}}}$

Now, the velocity of center of mass is given by,

$v_{com} = R ω = \sqrt{\frac{3 K E_{rot}}{M}}$

$\Rightarrow v_{com} = \sqrt{\frac{3 \times 8.0}{3.2}}$

$\Rightarrow v_{com} = 2.74 m s^{- 1}$

(c) We note that the inclined distance of $1.0 m$ corresponds to a height,

$h = 1.0 \sin 30 ° = 0.50 m$ .

Using the principle of conservation of mechanical energy is given by,

$∆ K E = - ∆ U$

$\Rightarrow K E_{f} = K E_{i} + U_{i} - U_{f}$

$\Rightarrow K E_{f} = 20 + 0 - M g h (∵ U_{i} = 0)$

$\Rightarrow K E_{f} = 20 - 3.2 \times 9.8 \times 0.5$

$\Rightarrow K E_{f} = 4.32 J$

(d) We found in part (a) that $40 %$ of this must be rotational, so

$K E_{rot} = \frac{1}{3} M R^{2} ω_{f}^{2} = \frac{1}{3} M v_{f}^{2} = (0.40) K E_{f}$

$\Rightarrow {(v_{com})}_{f} = \sqrt{\frac{3 \times 0.40 \times 4.32}{2.7}}$

$\Rightarrow {(v_{com})}_{f} = 1.39 m s^{- 1}$

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