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| Step | Derivation or Formula | Reasoning |
|---|---|---|
| 1 | \[ T\cos\theta = mg \] | This equation comes from vertical force equilibrium; the vertical component of the tension \(T\) must balance the gravitational force \(mg\). |
| 2 | \[ T = \frac{mg}{\cos\theta} \] | We solve the vertical equilibrium equation for \(T\) by dividing both sides by \(\cos\theta\). |
| 3 | \[ \sin\theta = \frac{R}{L} \quad \text{and} \quad \cos\theta = \sqrt{1-\frac{R^2}{L^2}} \] | Since the ball moves in a horizontal circle, the radius is given by \(R = L\sin\theta\). We then use the Pythagorean identity \(\cos\theta = \sqrt{1-\sin^2\theta}\) to express \(\cos\theta\) in terms of \(R\) and \(L\). |
| 4 | \[ \boxed{T = \frac{mg}{\sqrt{1-\frac{R^2}{L^2}}}} \] | Substitute the expression for \(\cos\theta\) into the formula for \(T\) to obtain the tension in terms of \(m\), \(g\), \(R\), and \(L\). |
| Step | Derivation or Formula | Reasoning |
|---|---|---|
| 1 | \[ T\sin\theta = m\frac{v^2}{R} \] | This equation represents the horizontal force balance. The horizontal component of the tension \(T\) provides the centripetal force \(m\frac{v^2}{R}\) required for circular motion. |
| 2 | \[ v^2 = \frac{T R \sin\theta}{m} \] | We solve the horizontal force equation for \(v^2\) by isolating it on one side. |
| 3 | \[ v^2 = \frac{(mg/\cos\theta) R \sin\theta}{m} = gR\tan\theta \] | Substitute \(T = \frac{mg}{\cos\theta}\) from the vertical balance and simplify; notice that \(m\) cancels out. |
| 4 | \[ v = \sqrt{gR\tan\theta} \] | Take the square root to find the speed \(v\) of the ball. |
| 5 | \[ P = \frac{2\pi R}{v} \] | The period \(P\) is found by dividing the circumference of the circle \(2\pi R\) by the speed \(v\) of the ball. |
| 6 | \[ P = \frac{2\pi R}{\sqrt{gR\tan\theta}} = 2\pi \sqrt{\frac{R}{gR\tan\theta}} = 2\pi \sqrt{\frac{1}{g\tan\theta}} \] | Simplify the expression by canceling \(R\) under the square root. |
| 7 | \[ \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{R/L}{\sqrt{1-\frac{R^2}{L^2}}} = \frac{R}{\sqrt{L^2-R^2}} \] | Express \(\tan\theta\) in terms of \(R\) and \(L\) using \(\sin\theta = \frac{R}{L}\) and \(\cos\theta = \sqrt{1-\frac{R^2}{L^2}}\). |
| 8 | \[ P = 2\pi \sqrt{\frac{1}{g\left(\frac{R}{\sqrt{L^2-R^2}}\right)}} = 2\pi \sqrt{\frac{\sqrt{L^2-R^2}}{gR}} \] | Substitute the expression for \(\tan\theta\) into the period formula. |
| 9 | \[ P = \frac{2\pi R}{\sqrt{gR\tan\theta}} = 2\pi\sqrt{\frac{L\cos\theta}{g}} \] | An alternative derivation: using \(R = L\sin\theta\), one can show that \(P = 2\pi\sqrt{\frac{L\cos\theta}{g}}\). Since \(\cos\theta = \sqrt{1-\frac{R^2}{L^2}}\), this gives the same result. |
| 10 | \[ \boxed{P = 2\pi \sqrt{\frac{L\sqrt{1-\frac{R^2}{L^2}}}{g}}} \] | This is the final expression for the period \(P\) expressed in terms of \(L\), \(R\), and \(g\). |
Just ask: "Help me solve this problem."
Two wires are tied to the \(500 \, \text{g}\) sphere as shown above. The sphere revolves in a horizontal circle at a constant speed of \(7.2 \, \text{m/s}\). What is the tension in the upper wire? What is the tension in the lower wire?
A car with speed \( v \) and an identical car with speed \( 3v \) both travel the same circular section of an unbanked (flat) road. If the frictional force required to keep the faster car on the road without skidding is \( F \), then the frictional force required to keep the slower car on the road without skidding is
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
A child whirls a ball in a vertical circle. Assuming the speed of the ball is constant (an approximation), when would the tension in the cord connected to the ball be greatest?
An adult exerts a horizontal force on a swing that is suspended by a rope of length \( L \), holding it at an angle \( \theta \) with the vertical. The child in the swing has a weight \( W \) and dimensions that are negligible. In terms of \( W \) and \( \theta \), determine:
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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] |
General Metric Conversion Chart
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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