2018 A/L Physics Past Paper · Question 35 (English Medium)

Solid sphere, disc and hollow sphere — rotational kinetic energy

Figure shows vertical cross-sections of three uniform bodies $A, B$ and $C$ of identical masses. $A$ is a solid sphere of radius $r$ . $C$ is a hollow sphere of radius $r$ and having thin walls. The spheres can be rotated about vertical axes passing through respective centres. $B$ is a disc of radius $r$ which can be rotated about an axis normal to the plane of the disc passing through its centre. All figures are drawn to same scale. If $KE_A, KE_B$ and $KE_C$ respectively are the rotational kinetic energies to be provided for the bodies $A, B$ and $C$ to attain equal angular speeds, which of the following expressions is true?

(1) $KE_A < KE_B < KE_C$
(2) $KE_C < KE_A < KE_B$
(3) $KE_C < KE_B < KE_A$
(4) $KE_A < KE_C < KE_B$
(5) $KE_A = KE_B = KE_C$

Answer and explanation

Answer: (1) $KE_A < KE_B < KE_C$

What the question asks

Three bodies have the same mass $m$ and the same radius $r$ .

We spin each one up to the same angular speed $\omega$ .

We want: which one needs the most energy?

Relevant theory

The kinetic energy possessed by a rotating body is called rotational kinetic energy.

"I" depends on the position of the axis of rotation and how the mass of the body is distributed about it.

Moment of inertia of a uniform circular disc or solid cylinder of radius R, mass M about an axis normal to the plane passing through the centre is given by

Moment of Inertia of a uniform solid sphere about an axis through the centre is given by

Required equations

$KE = \tfrac{1}{2}I\omega^2$

$I = \sum m_i r_i^2$

$KE \propto I$

Reading the options

Each option is an order of $KE_A$ , $KE_B$ , $KE_C$ .

First find $I$ for all three. Then compare.

Reasoning and calculation

All three have the same $m$ , $r$ and $\omega$ .

So only $I$ decides the energy.

What is different? Where the mass sits.

Solid sphere: most of the mass is near the axis.

Disc: the mass is spread evenly over a flat plate.

Hollow sphere: all the mass is on the outside, far from the axis.

A — solid sphere

$KE_A = \tfrac12 \cdot \tfrac25 mr^2\omega^2 = 0.2\,mr^2\omega^2$

For a solid sphere about its centre:

$I_A = \tfrac{2}{5}mr^2$

B — disc

$KE_B = \tfrac12 \cdot \tfrac12 mr^2\omega^2 = 0.25\,mr^2\omega^2$

For a disc about the axis through its centre:

$I_B = \tfrac{1}{2}mr^2$

C — thin hollow sphere

$KE_C = \tfrac12 \cdot \tfrac23 mr^2\omega^2 \approx 0.33\,mr^2\omega^2$

All the mass is at the radius $r$ .

Add up $I$ about three axes at right angles: $I_x + I_y + I_z = 2mr^2$ .

The sphere looks the same every way, so the three are equal:

$I_C = \tfrac{2}{3}mr^2$

Now compare.

$\tfrac25 < \tfrac12$ , so $KE_A < KE_B$ .

$\tfrac12 < \tfrac23$ , so $KE_B < KE_C$ .

KE_A

KE_B

KE_C

$I_A : I_B : I_C = 0.40 : 0.50 : 0.67$

At the same $\omega$ , more $I$ means more energy.

$KE_A < KE_B < KE_C$

Checks

$0.5 * (2/3) - 0.5 * (1/2)$

Does it make sense?

Mass far from the axis moves faster. So it needs more energy.

The hollow sphere has all its mass at the edge. It needs the most.

The solid sphere has its mass near the middle. It needs the least.

Conclusion

$KE_A < KE_B < KE_C$

$I_A = \tfrac25 mr^2 < I_B = \tfrac12 mr^2 < I_C = \tfrac23 mr^2$

So $KE_A < KE_B < KE_C$ .

The answer is option (1).

Explore

All three bodies have the same mass and radius; only the way the mass is spread differs.

We spin all three up to the same angular speed.

Mass far from the axis moves faster, so it needs more energy.

The hollow sphere has all its mass at the edge, so $KE_A < KE_B < KE_C$ .

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