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Using the energy conservation principle: the kinetic energy at release will be converted to potential energy at the highest point. Calculate height [katex] h [/katex] above the wheel center.
Alternatively, you can use both rotational and linear kinematics to determine the max height as shown below.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex] r = \frac{D}{2} [/katex] | Calculate the radius [katex] r [/katex] of the wheel by dividing the diameter [katex] D [/katex] by 2. The given diameter [katex] D [/katex] is 24 cm, so [katex] r = 24 \, \text{cm} / 2 = 12 \, \text{cm} = 0.12 \, \text{m} [/katex]. |
| 2 | [katex] {\omega_f }^2 ={\omega_i }^2 + 2\alpha \Delta \theta [/katex] | Use the angular kinematic equation to find the final angular velocity [katex] \omega_f [/katex]. Assume initial angular velocity [katex] \omega_i [/katex] is zero since the wheel starts from rest and angular acceleration [katex] \alpha [/katex] is given. |
| 3 | [katex] \omega_f = \sqrt{2 \alpha \theta} [/katex] | Rearrange equation in step 2 and calculate the final angular velocity [katex] \omega_f [/katex]. |
| 4 | [katex] v = r \omega_f [/katex] | To find the height ( a linear distance) we need to find the linear speed [katex] v [/katex] of the ball at the point of release. Use the equation [katex] v = r \omega [/katex]. |
| 5 | [katex] v = 0.12 \times \sqrt{2 \times 470 \times \frac{3\pi}{2}} [/katex] | Substitute values and calculate [katex] v = 7.99 \, m/s [/katex]. |
| 6 | [katex] {v_f}^2 = {v_0}^2 + 2a\Delta x [/katex] | Linear kinematic equation to solve for the change in vertical height, given that the final speed is 0 at the max height. |
| 7 | [katex] \Delta x = h_{max} = \frac{-{v_0}^2}{2g} [/katex] | Solve for [katex] \Delta x [/katex] which is the maximum height. |
| 8 | [katex] h = \frac{-(0.12 \times \sqrt{2 \times 470 \times \frac{3\pi}{2}})^2}{2 \times -9.81} [/katex] | Calculate [katex] h [/katex] by substituting [katex] v [/katex] from step 7 and solve. |
| 9 | [katex] h_{max} = 3.25 \, m [/katex] | This is the final step where the height [katex] h [/katex] is computed with all values substituted and simplified. |
This table gives a detailed breakdown of calculating the height that a steel ball reaches above the center of a wheel, considering the kinematic equations for rotational and linear motions and energy conservation principles.
Phy can also check your working. Just snap a picture!
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[katex] h_{max} = 3.25 \, m [/katex]
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| Kinematics | Forces |
|---|---|
| [katex]\Delta x = v_i t + \frac{1}{2} at^2[/katex] | [katex]F = ma[/katex] |
| [katex]v = v_i + at[/katex] | [katex]F_g = \frac{G m_1m_2}{r^2}[/katex] |
| [katex]a = \frac{\Delta v}{\Delta t}[/katex] | [katex]f = \mu N[/katex] |
| [katex]R = \frac{v_i^2 \sin(2\theta)}{g}[/katex] |
| Circular Motion | Energy |
|---|---|
| [katex]F_c = \frac{mv^2}{r}[/katex] | [katex]KE = \frac{1}{2} mv^2[/katex] |
| [katex]a_c = \frac{v^2}{r}[/katex] | [katex]PE = mgh[/katex] |
| [katex]KE_i + PE_i = KE_f + PE_f[/katex] |
| Momentum | Torque and Rotations |
|---|---|
| [katex]p = m v[/katex] | [katex]\tau = r \cdot F \cdot \sin(\theta)[/katex] |
| [katex]J = \Delta p[/katex] | [katex]I = \sum mr^2[/katex] |
| [katex]p_i = p_f[/katex] | [katex]L = I \cdot \omega[/katex] |
| Simple Harmonic Motion |
|---|
| [katex]F = -k x[/katex] |
| [katex]T = 2\pi \sqrt{\frac{l}{g}}[/katex] |
| [katex]T = 2\pi \sqrt{\frac{m}{k}}[/katex] |
| 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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