Essential Physics

Section 2 review

Rotational inertia is at the heart of the rotation of rigid bodies. The moment of inertia of an object represents its rotational inertia about a particular axis. Moment of inertia depends not just on mass but also on the distribution of that mass. Even if a solid sphere and a hollow sphere have the same mass, they have different moments of inertia because their mass is distributed differently. Rolling objects possess not just kinetic energy but also rotational energy (sometimes called the kinetic energy of rotation). When rolling downhill, objects will be accelerated differently, depending on their moment of inertia. Read the text aloud

Read the text aloud

rotational energy, precession

E_{r} = \frac{1}{2} I ω^{2}

Review problems and questions

In bowling, usually the bowling ball will initially slide down the lane without spinning. Partway down the lane, however, the ball begins to roll. What happens to the velocity of the ball when it begins to roll?

What is the moment of inertia for each of the following? (Mass of the Earth: 6.0×10²⁴ kg. Mass of the Moon: 7.3×10²² kg. Radius of the Earth: 6.4×10⁶ m. Radius of the Moon: 1.7×10⁶ m. Radius of the Earth’s orbit around the Sun: 1.5×10²⁴ m. Radius of the Moon’s orbit around the Earth: 3.9×10⁸ m.)
1. Moon’s orbit around the Earth
2. Earth’s orbit around the Sun
3. rotation of the Moon about its axis
4. rotation of the Earth about its axis

Answer:

1.1×10⁴⁰ kg m²
1.4×10⁷³ kg m²
8.4×10³⁴ kg m²
9.8×10³⁷ kg m²

Solution:

$I_{o r b i t, M o o n} = m r^{2} = (7.3 \times 10^{22} kg) {(3.9 \times 10^{8} m)}^{2} = 1.1 \times 10^{40} {kg m}^{2}$
$I_{o r b i t, E a r t h} = m r^{2} = (6.0 \times 10^{24} kg) {(1.5 \times 10^{24} m)}^{2} = 1.4 \times 10^{73} {kg m}^{2}$
The moment of inertia of a rotating, solid sphere is ⅖mr², so $I_{r o t, M o o n} = \frac{2}{5} m r^{2} = \frac{2}{5} (7.3 \times 10^{22} kg) {(1.7 \times 10^{6} m)}^{2} = 8.4 \times 10^{34} {kg m}^{2}$
$I_{r o t, E a r t h} = \frac{2}{5} m r^{2} = \frac{2}{5} (6.0 \times 10^{24} kg) {(6.4 \times 10^{6} m)}^{2} = 9.8 \times 10^{37} {kg m}^{2}$

A beginning bowler pushed a 6 kg bowling ball so that it rolled down the lane without slipping at 2 m/s. How much work did the bowler do on the ball to start it rolling?

Answer: The bowler did 16.8 J of work to roll the bowling ball.
Asked: work W done by the bowler to roll the bowling ball
Given: mass of the ball, m = 6 kg; (linear) velocity of the ball, v = 2 m/s
Relationships: rolling motion equation v = ωr; kinetic energy E_k = ½mv²;
rotational (kinetic) energy E_k = ½mv²; moment of inertia of a solid sphere, I = ⅖mr²
Solve: From the work–energy theorem, the work done is the change in the kinetic energy of the ball. In this case, the kinetic energy of the rolling ball is the sum of its (linear) kinetic energy and rotational (kinetic) energy:

W = Δ E_{k} + Δ E_{r} = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}

Rewrite the equation of rolling motion as ω = v/r, then substitute for this and the moment of inertia I = ⅖mr² to get

\begin{array}{l} W & = & \frac{1}{2} m v^{2} + \frac{1}{2} (\frac{2}{5} m r^{2}) {(\frac{v}{r})}^{2} = \frac{1}{2} m v^{2} + \frac{1}{5} m v^{2} = \frac{7}{10} m v^{2} \\ = & \frac{7}{10} (6 kg) {(2 m/s)}^{2} = 16.8 J \end{array}

Will a hoop or a solid disk be accelerated more when rolling down the same inclined plane? (Assume that the two objects have the same radius.)

Two designs for a wheel and axle system

You are considering two different designs for the wheel that forms part of a wheel-and-axle simple machine. Although both designs have the same mass, one puts most of the mass near the rim while the other distributes the mass evenly throughout the wheel. What are the advantages or disadvantages of each design if the input force will be applied to the axle?


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