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# Part (a): Expression for the radius of the hoop
The solution involves converting the initial kinetic energy into gravitational potential energy at the maximum height [katex] h [/katex].
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]v = R\omega[/katex] | The velocity [katex] v [/katex] of the hoop at the bottom is related to the angular velocity [katex] \omega [/katex] and the radius [katex] R [/katex] of the hoop by the no-slip condition. |
| 2 | [katex]KE_{\text{bottom}} = \frac{1}{2}m v^2 + \frac{1}{2} I \omega^2[/katex] | Calculate the total kinetic energy at the bottom considering both translational ([katex] \frac{1}{2}m v^2 [/katex]) and rotational ([katex] \frac{1}{2} I \omega^2 [/katex]) kinetic energy. |
| 3 | [katex]I = m R^2[/katex] (for a hoop) | The moment of inertia [katex] I [/katex] of a hoop about its center is [katex] m R^2 [/katex]. |
| 4 | [katex]KE_{\text{bottom}} = \frac{1}{2}m (R\omega)^2 + \frac{1}{2} m R^2 \omega^2 = m R^2 \omega^2[/katex] | Substitute [katex] I [/katex] and [katex] v [/katex] into the kinetic energy expression and simplify. |
| 5 | [katex]PE_{\text{top}} = mgh[/katex] | Calculate the potential energy at the maximum height [katex] h [/katex] using the mass [katex] m [/katex] and gravitational acceleration [katex] g [/katex]. |
| 6 | [katex]KE_{\text{bottom}} = PE_{\text{top}}[/katex] | Apply conservation of mechanical energy, assuming no energy loss to friction or air resistance. |
| 7 | [katex]m R^2 \omega^2 = mgh[/katex] | Set the expressions for kinetic and potential energy equal and simplify. |
| 8 | [katex]R = \sqrt{\frac{gh}{\omega^2}}[/katex] | Solve for [katex] R [/katex]. This equation provides the radius in terms of the given quantities and constants. |
# Part (b): Direction of friction while the hoop rolls up the ramp
| Step | Explanation |
|---|---|
| Reasoning | The direction of friction must oppose the tendency of slipping. Whether the hoop slides up or down the ramp, it will always try to slide down the ramp. Thus, static friction acts in the direction of motion, up the ramp, to prevent the hoop from sliding back. |
# Part (c): Direction of friction while the hoop rolls down the ramp
| Step | Explanation |
|---|---|
| Reasoning | As explained in part (b), regardless of the direction the hoop travels on the ramp, friction will continue to point up the ramp. As the hoop rotates down the ramp, it want to slip down the ramp, but is countered by static friction that points up the ramp. |
Just ask: "Help me solve this problem."
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| Time \( (\text{s}) \) | Net Torque \( (\text{N} \cdot \text{m}) \) |
|---|---|
| 0 | 0 |
| 2 | 1.5 |
| 4 | 3.0 |
| 6 | 4.5 |
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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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