| Step | Derivation or Formula | Reasoning |
|---|---|---|
| 1 | \[ mgh = \frac{1}{2}mv_x^2 + \frac{1}{2}I\left(\frac{v_x}{R}\right)^2 \] | Apply energy conservation for rolling without slipping. The gravitational potential energy \(mgh\) converts to translational kinetic energy \(\frac{1}{2}mv_x^2\) and rotational kinetic energy \(\frac{1}{2}I\omega^2\) with the no-slip condition \(\omega=\frac{v_x}{R}\). |
| 2 | \[ mgh = \frac{1}{2}mv_x^2\left(1+\frac{I}{mR^2}\right) \] | Substitute \(\omega=\frac{v_x}{R}\) and factor out \(\frac{1}{2}mv_x^2\) to combine the terms into a single expression. |
| 3 | \[ v_x^2 = \frac{2gh}{1+\frac{I}{mR^2}} \] | Solve for the translational speed \(v_x\) in terms of the moment of inertia factor \(\frac{I}{mR^2}\). |
| 4 | \[ \text{For a solid sphere: } \frac{I}{mR^2}=\frac{2}{5} \quad\Rightarrow\quad v_{x,\text{sphere}}^2=\frac{2gh}{1+\frac{2}{5}}=\frac{10gh}{7} \] | Substitute the moment of inertia for a solid sphere \(I=\frac{2}{5}mR^2\), leading to a higher translational speed. |
| 5 | \[ \text{For a solid cylinder: } \frac{I}{mR^2}=\frac{1}{2} \quad\Rightarrow\quad v_{x,\text{cylinder}}^2=\frac{2gh}{1+\frac{1}{2}}=\frac{4gh}{3} \] | Substitute the moment of inertia for a solid cylinder \(I=\frac{1}{2}mR^2\). Its greater inertia compared to the sphere reduces its translational speed. |
| 6 | \[ \text{For a hollow pipe: } \frac{I}{mR^2}=1 \quad\Rightarrow\quad v_{x,\text{pipe}}^2=\frac{2gh}{1+1}=gh \] | Substitute the moment of inertia for a hollow pipe \(I=mR^2\). The high inertia value further reduces the translational speed. |
| 7 | \[ \frac{10gh}{7} > \frac{4gh}{3} > gh \] | Compare the expressions for \(v_x^2\): The solid sphere has the largest value, so it reaches the bottom first. The cylinder and pipe are slower because more energy goes into rotation. |
| 8 | \[ \boxed{(a)\;\text{Sphere}} \] | The sphere reaches the bottom first. The cylinder and pipe are slower due to their higher moments of inertia; hence, they devote a larger portion of the energy to rotation. |
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Two uniform disks have the same mass but different radii: disk \( 1 \) has a radius \( R \) and disk \( 2 \) has a radius \( 2R \). What is the ratio of the moment of inertia of the first disk to the second disk?
Three masses are attached to a \( 1.5 \, \text{m} \) long massless bar. Mass 1 is \( 2 \, \text{kg} \) and is attached to the far left side of the bar. Mass 2 is \( 4 \, \text{kg} \) and is attached to the far right side of the bar. Mass 3 is \( 4 \, \text{kg} \) and is attached to the middle of the bar. At what distance from the far left side of the bar can a string be attached to hold the bar up horizontally?
A centrifuge in a medical laboratory is rotating at an angular speed of \( 3600 \) \( \text{rev/min} \). When switched off, it rotates \( 50.0 \) times before coming to rest. Find the constant angular deceleration of the centrifuge.
A \( 50 \, \text{kg} \) person is sitting on a seesaw \( 1.2 \, \text{m} \) from the balance point. On the other side, a \( 70 \, \text{kg} \) person is balanced. How far from the balance point is the second person sitting?
Five forces act on a rod that is free to pivot at point \( P \), as shown in the figure. Which of these forces is producing a counter-clockwise torque about point \( P \)?
Two spheres of equal size and equal mass are rotated with an equal amount of torque. One of the spheres is solid with its mass evenly distributed throughout its volume, and the other is hollow with all of its mass concentrated at the edges. Which sphere would rotate faster if the same amount of torque is applied for the same period of time for both?
A uniform solid sphere of mass M and radius R is placed on a frictionless horizontal surface. A massless string is wrapped around the sphere and is pulled with a force F. The string makes an angle of θ with the horizontal. What is the minimum value of the coefficient of static friction between the sphere and the surface required for the sphere to start rolling without slipping?
Two disks, A and B, each experience a net external torque that varies over an interval of \( 5 \) \( \text{s} \). Disk B has a rotational inertia that is twice that of Disk A. The graph shown represents the angular momentum of the two disks as functions of time between \( t = 0 \) \( \text{s} \) and \( t = 5 \) \( \text{s} \). The average magnitudes of the net torques exerted on disks A and B from \( t = 0 \) \( \text{s} \) to \( t = 5 \) \( \text{s} \) are \( \tau_A \) and \( \tau_B \), respectively. Which of the following expressions correctly relates the magnitudes of the average torques?
A discus is held at the end of an arm that starts at rest. The average angular acceleration of \(54 \, \text{rad/s}^2 \) lasts for 0.25 s. The path is circular and has radius 1.1 m.
Note: A discuss is a heavy, flattened circular object for throwing.
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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] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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