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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.
Just ask: "Help me solve this problem."
| Speed | \( 10 \, \mathrm{m/s} \) | \( 20 \, \mathrm{m/s} \) | \( 30 \, \mathrm{m/s} \) |
| Braking Distance | \( 6.1 \, \mathrm{m} \) | \( 23.9 \, \mathrm{m} \) | \( 53.5 \, \mathrm{m} \) |
A car of mass \( 1500 \, \mathrm{kg} \) is traveling at one of the speeds listed when the brakes are first applied. Using the data above, what is the magnitude of the average braking force required to stop the car?
A ball of radius \( r \) rolls on the inside of a circular track of radius \( R \). If the ball starts from rest at the left vertical edge of the track, what will be its speed when it reaches the lowest point of the track, rolling without slipping?
A man with mass \( m \) is standing on a rotating platform in a science museum. The platform can be approximated as a uniform disk of radius \( R \) that rotates without friction at a constant angular velocity \( \omega \).
Two students are discussing what the man should do if he wishes to change the angular velocity of the platform.
Student A says that the man should run towards the center of the platform, because this will decrease the moment of inertia of the man-platform system. Since \( L \propto I \), the angular momentum will decrease proportionately and the platform will slow down.
Student B says that since the platform is rotating counterclockwise, the man should run in a clockwise direction to slow the platform down. His feet will exert a frictional torque on the platform, which will cause an angular acceleration of the man-platform system.
Explain what is correct and incorrect about each students statement if anything.
Pulleys \( X \) and \( Y \) are each attached to a block by a string that wraps around the pulley. Both blocks are released and have the same linear acceleration \( a \). As the blocks fall, the pulleys rotate about their centers. Pulley \( Y \) has a larger radius than Pulley \( X \). How does the angular acceleration \( \alpha_X \) of Pulley \( X \) compare to the angular acceleration \( \alpha_Y \) of Pulley \( Y \)?
[katex] h_{max} = 3.25 \, m [/katex]
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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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