| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \( v_{\text{critical}} = \sqrt{gR} \) | The critical speed \( v_{\text{critical}} \) is the minimum speed at the top of the loop which allows the roller coaster to remain in contact with the track and complete the loop without falling off. It is found by equating the gravitational force to the required centripetal force for circular motion at the top of the loop. |
| 2 | \( v = 2v_{\text{critical}} = 2\sqrt{gR} \) | It is given that the roller coaster crosses at twice the critical speed. Therefore, the actual speed \( v \) is twice the critical speed. |
| 3 | \( F_{\text{net}} = m \frac{v^2}{R} \) | The net force needed for circular motion at the top of the loop is given by the centripetal force formula where \( m \) is the mass of the roller coaster, \( v \) is the velocity, and \( R \) is the radius of the loop. |
| 4 | \( F_{\text{net}} = m \frac{(2\sqrt{gR})^2}{R} = 4mg \) | Substitute \( v = 2\sqrt{gR} \) into the centripetal force formula and simplify. |
| 5 | \( F_{\text{net}} = F_{\text{n}} + mg = 4mg \) | The net force \( F_{\text{net}} \) at the top of the loop is the sum of the normal force \( F_{\text{n}} \) exerted by the track on the roller coaster and the gravitational force \( mg \). |
| 6 | \( F_{\text{n}} + mg = 4mg \) | Setting the net inward force (which includes the normal and gravitational forces) equal to the required centripetal force for motion at this speed. |
| 7 | \( F_{\text{n}} = 4mg \,-\, mg = 3mg \) | Solving for \( F_{\text{n}} \) by subtracting \( mg \) from each side. |
| 8 | \( \frac{F_{\text{n}}}{F_{\text{g}}} = \frac{3mg}{mg} = 3 \) | The ratio of the normal force to the gravitational force is \( \frac{F_{\text{n}}}{F_{\text{g}}} \). Substituting the values from the previous step. |
| 9 | \( \frac{F_{\text{n}}}{F_{\text{g}}} = 3 \) | The final answer, indicating the ratio is 3, matching option (b). |
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A loop-de-loop roller coaster has a radius of \( 30 \) \( \text{m} \). Determine the apparent weight a \( 500 \) \( \text{N} \) person will feel at the bottom of the loop while traveling at a speed of \( 25 \) \( \text{m/s} \) and at the top of the loop while traveling at a speed of \( 20 \) \( \text{m/s} \).
A car of mass \( M \) moves around a circularly banked curve on a freeway off-ramp. The off-ramp has a radius of curvature \( R \) and is raised to an angle \( \theta \) from the horizontal. The road is slick, and friction is negligible.
Two satellites of equal mass, \( S_1 \) and \( S_2 \), orbit the Earth. \( S_1 \) is orbiting at a distance \( r \) from the Earth’s center at speed \( v \). \( S_2 \) orbits at a distance \( 2r \) from the Earth’s centre at speed \( \dfrac{v}{\sqrt{2}} \). The ratio of the centripetal force on \( S_1 \) to the centripetal force on \( S_2 \) is
While driving fast around a sharp right turn, you find yourself pressing against the car door. What is happening?
An object moves at constant speed in a circular path of radius \( r \) at a rate of \( 1 \) revolution per second. What is its acceleration in terms of \(r\)?
A car is safely negotiating an unbanked circular turn at a speed of 21 m/s. The road is dry, and the maximum static frictional force acts on the tires. Suddenly a wet patch in the road decreases the maximum static frictional force to one-third its dry road value. If the car is to continue safely around the curve, to what speed must the driver slow the car?
A rock is whirled on the end of a string in a horizontal circle of radius \(R\) with a constant period \(T\). If the radius of the circle is reduced to \(R/3\), while the period remains \(T\), what happens to the centripetal acceleration (\(a_c\)) of the rock?
Suppose you are a passenger traveling in car along a road that bends to the left. Why will you feel like you are being thrown against the door. What causes this force?

A particle of mass \(m\) slides down a fixed, frictionless sphere of radius \(R\), starting from rest at the top.
In terms of \(m\), \(g\), \(R\), and \(\theta\), determine each of the following for the particle while it is sliding on the sphere.
The International Space Station has a mass of \(4.2 \times 10^{5} \, \text{kg}\) and orbits Earth at a distance of \(4.0 \times 10^{2} \, \text{km}\) above the surface. Earth has a radius of \(6.37 \times 10^{6} \, \text{m}\) and a mass of \(5.97 \times 10^{24} \, \text{kg}\). Calculate the following:
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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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