| Derivation/Formula | Reasoning |
|---|---|
| \[\tau = I \alpha\] | Torque needed to spin a disk scales with its moment of inertia \( I \) and angular acceleration \( \alpha \). The disks are the same size, so \( I \) is the same in both cases. |
| \[\alpha = \frac{\Delta \omega}{\Delta t}\] | Angular acceleration is change in angular speed \( \Delta \omega \) over spin-up time \( \Delta t \). Both start from rest \( \omega_i = 0 \), and the times are about equal. |
| \[\frac{\tau_{75}}{\tau_{45}} = \frac{I \alpha_{75}}{I \alpha_{45}} = \frac{\alpha_{75}}{\alpha_{45}}\] | With the same \( I \), the torque ratio equals the ratio of angular accelerations. |
| \[\frac{\alpha_{75}}{\alpha_{45}} = \frac{\omega_{x,75}/\Delta t}{\omega_{x,45}/\Delta t} = \frac{\omega_{x,75}}{\omega_{x,45}}\] | Since \( \omega_i = 0 \), \( \Delta \omega = \omega_x \). Similar \( \Delta t \) cancels. |
| \[\frac{\omega_{x,75}}{\omega_{x,45}} = \frac{75}{45} = \frac{5}{3}\] | The conversion from RPM to rad/s cancels in the ratio, so the numeric RPM ratio suffices. Thus \( \tau_{75} \) must exceed \( \tau_{45} \) by a factor \( \tfrac{5}{3} \). |
| \[\boxed{\tau_{75} = \tfrac{5}{3}\,\tau_{45}}\] | Therefore the torque is larger in the second case; correct choice is (c). |
| \[\tau = I \alpha\] | Why (a) is incorrect: Having the same \( I \) does not force the same \( \tau \); different \( \alpha \) values imply different \( \tau \). |
| \[L = I \omega\] | Why (b) is incorrect: Angular momentum \( L \) is not conserved while an external driving torque is applied; the system is not isolated. |
| \[\tau = rF\] | Why (d) is incorrect: The lever arm \( r \) (disk radius) is the same for both, so the difference does not come from a larger \( r \); the larger \( \tau \) arises from the larger required \( \alpha \). |
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Which of the following situations will increase the moment of inertia of a solid cylinder \( I = \tfrac{1}{2} M R^{2} \) by the same amount?

The figure shows scale drawings of four objects, each of the same mass and uniform thickness, with the mass distributed uniformly. Which one has the greatest moment of inertia when rotated about an axis perpendicular to the plane of the drawing at point P?
An object is moving in a horizontal circle at a constant speed. Which of the following correctly describes the linear and angular velocities of the object between any point along the circular path?
The angular velocity of a rotating disk of radius \(20 \, \text{cm}\) increases from \(1 \, \text{rad/s}\) to \(3 \, \text{rad/s}\) in \(0.5 \, \text{s}\). What is the linear tangential acceleration of a point on the rim of the disk during this time interval?
To increase the moment of inertia of a body about an axis, you must
The driver of a car traveling at \( 30.0 \) \( \text{m/s} \) applies the brakes and undergoes a constant negative acceleration of \( 2.00 \) \( \text{m/s}^2 \). How many revolutions does each tire make before the car comes to a stop, assuming that the car does not skid and that the tires have radii of \( 0.300 \) \( \text{m} \)?
Consider a uniform hoop of radius \( R \) and mass \( M \) rolling without slipping. Which is larger, its translational kinetic energy or its rotational kinetic energy? Hint: The moment of inertia of a uniform hoop is \(I = M R^2\)
An \( 80 \, \text{kg} \) block is placed \( 2 \, \text{m} \) away from the endpoint of a horizontal steel beam of length \( 6.6 \, \text{m} \) and mass \( 1,450 \, \text{kg} \). The plank makes contact with a vertical wall on one end (assume it does not slip). The other end of the beam is attached to a massless cable that makes an angle of \( 30^\circ \) with the horizontal and ties into the vertical wall as well. Calculate the (1) tension force in the cable and (2) the total force the wall exerts on the beam.

A disk of radius \( R = 0.5 \) \( \text{cm} \) rests on a flat, horizontal surface such that frictional forces are considered to be negligible. Three forces of unknown magnitude are exerted on the edge of the disk, as shown in the figure. Which of the following lists the essential measuring devices that, when used together, are needed to determine the change in angular momentum of the disk after a known time of \( 5.0 \) \( \text{s} \)?
A turntable rotates through \( 6 \) \( \text{rad} \) in \( 3 \) \( \text{s} \) as it accelerates uniformly from rest. What is its angular acceleration in \( \text{rad/s}^2 \)?
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
Metric Prefixes
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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