| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[ mg\bigl(h_{\min}+r\bigr)=KE_{\text{trans}}+KE_{\text{rot}}+mg\bigl(2R-r\bigr) \] | Apply conservation of mechanical energy between the release point (centre of mass height \(h_{\min}+r\)) and the top of the loop (centre of mass height \(2R-r\)). |
| 2 | \[ KE_{\text{trans}}=\tfrac12 m v^2 \] | Translational kinetic energy of the marble’s centre of mass at the loop top. |
| 3 | \[ KE_{\text{rot}}=\tfrac12 I\omega^2 \] | Rotational kinetic energy about the centre of mass. |
| 4 | \[ I=\tfrac25 m r^2,\qquad \omega=\frac{v}{r} \] | For a solid sphere that rolls without slipping. |
| 5 | \[ KE_{\text{rot}} = \frac12\!\left(\frac25 m r^2\right)\!\left(\frac{v}{r}\right)^2 = \tfrac15 m v^2 \] | Substitute \(I\) and \(\omega\) into the rotational term. |
| 6 | \[ KE_{\text{total}} = \tfrac12 mv^2 + \tfrac15 mv^2 = \tfrac{7}{10} m v^2 \] | Combine translational and rotational kinetic energies. |
| 7 | \[ v^2 = g\,(R-r) \] | At the top, the normal force is zero for the marginal case. Centripetal requirement: \(mg=mv^2/(R-r)\). |
| 8 | \[ mg(h_{\min}+r)=mg(2R-r)+\tfrac{7}{10}m g (R-r) \] | Insert \(v^2\) from Step 7 into the energy equation from Step 1. |
| 9 | \[ h_{\min}+r = 2R – r + \tfrac{7}{10}(R-r) \] | Divide by \(g\) and cancel \(m\). |
| 10 | \[ h_{\min}=\tfrac{27}{10}R-\tfrac{27}{10}r = \tfrac{27}{10}(R-r) \] | Collect like terms to isolate \(h_{\min}\). |
Therefore, the minimum track height required is \[ h_{\min}=\tfrac{27}{10}\bigl(R-r\bigr). \]
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A communications satellite orbits the Earth at an altitude of \(35{,}000 \, \text{km}\) above the Earth’s surface. Take the mass of Earth to be \(6 \times 10^{24} \, \text{kg}\) and the radius of Earth to be \(6.4 \times 10^6 \, \text{m}\). What is the satellite’s velocity?
What is the net torque acting on the pivot supporting a \(10 \, \text{kilogram}\) beam \(2 \, \text{meters}\) long as shown above? Assume that the positive direction is clockwise.
A girl throws a stone from a bridge. Consider the following ways she might throw the stone. The speed of the stone as it leaves her hand is the same in each case.
Case A: Thrown straight up.
Case B: Thrown straight down.
Case C: Thrown out at an angle of 45° above horizontal.
Case D: Thrown straight out horizontally.
In which case will the speed of the stone be greatest when it hits the water below if there is no significant air resistance, assuming equal initial speeds?
A 0.2 kg object is attached to a horizontal spring undergoes SHM with the total energy of 0.4 J. The kinetic energy as a function of position presented by the graph.
A uniform meter stick has a mass of \( 45.0 \) \( \text{g} \) placed at the \( 20 \) \( \text{cm} \) mark. If a pivot is placed at the \( 42.5 \) \( \text{cm} \) mark and the meter stick remains horizontal in static equilibrium, what is the mass of the meter stick?
Three forces of equal magnitude are applied to a \(3 \, \text{m} \times 2 \, \text{m}\) rectangle. Force \(F_1\) and \(F_2\) act at \(45^\circ\) angles to the vertical as shown, while \(F_3\) acts horizontally.
In short:
\(F_1\): applied at \((0, -2)\), direction SW \(45^\circ\)
\(F_2\): applied at \((2, -2)\), direction NW \(45^\circ\)
\(F_3\): applied at \((3, -1)\), direction east
Points of rotation: \(A = (0, 0)\), \(B = (0, -1)\), \(C = (1, -1)\)
A disk, a hoop, and a solid sphere are released at the same time at the top of an inclined plane. They are all uniform and roll without slipping. In what order do they reach the bottom?
\( \text{Solid sphere: } I = \frac{2}{5}mR^2, \quad \text{Solid disk: } I = \frac{1}{2}mR^2, \quad \text{Hoop: } I = mR^2 \)
An object is projected vertically upward from ground level. It rises to a maximum height \( H \). If air resistance is negligible, which of the following must be true for the object when it is at a height \( H/2 \) ?
The elliptical orbit of a comet is shown above. Positions 1 and 2 are, respectively, the farthest and nearest positions to the Sun, and at position 1 the distance from the comet to the Sun is 10 times that at position 2. What is the ratio \(v_1\)/\(v_2\) of the speed of the comet at position 1 to the speed at position 2?
The maximum acceleration a pilot can withstand without blacking out is about \( 7.0 \) \( g \). In an endurance test for a jet plane’s pilot, what is the maximum speed he can tolerate if he is spun in a horizontal circle of diameter \( 85 \) \( \text{m} \)?
\(\frac{27}{10}(R-r)\)
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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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