| Derivation or Formula | Reasoning |
|---|---|
| \[m = 25\,\text{g} = 0.025\,\text{kg},\quad d = 24\,\text{cm} = 0.24\,\text{m},\quad r = \frac{d}{2} = 0.12\,\text{m}\] | Convert to SI units and compute the wheel radius \(r\) from the given diameter \(d\). |
| \[\alpha = 470\,\text{rad/s}^2,\quad \Delta\theta = \frac{3}{4}(2\pi) = \frac{3\pi}{2}\] | The angular acceleration is given, and \(\tfrac{3}{4}\) of a revolution corresponds to angular displacement \(\Delta\theta\). |
| \[\tau = I\alpha\] | Torque \(\tau\) relates to angular acceleration \(\alpha\) through rotational dynamics. |
| \[I = mr^2\] | The wheel is negligible mass, so the rotating inertia is dominated by the ball modeled as a point mass at radius \(r\). |
| \[W = \tau\Delta\theta\] | Work done by a constant torque over an angular displacement is \(W = \tau\Delta\theta\). |
| \[W = (I\alpha)\Delta\theta = (mr^2\alpha)\Delta\theta\] | Substitute \(\tau = I\alpha\) and \(I = mr^2\) into the work expression, as indicated in the hint. |
| \[mr^2\alpha\Delta\theta = mgh\] | Use conservation of energy from the very start (rest) to the maximum height: the work put in becomes gravitational potential energy \(mgh\). |
| \[r^2\alpha\Delta\theta = gh\] | Cancel \(m\) from both sides, showing the height does not depend on the ball’s mass. |
| \[h = \frac{r^2\alpha\Delta\theta}{g}\] | Solve algebraically for the maximum height above the starting reference implied by the energy equation. |
| \[h = \frac{(0.12)^2(470)\left(\frac{3\pi}{2}\right)}{9.8}\] | Substitute \(r = 0.12\,\text{m}\), \(\alpha = 470\,\text{rad/s}^2\), \(\Delta\theta = \frac{3\pi}{2}\), and \(g = 9.8\,\text{m/s}^2\). |
| \[h = \frac{(0.0144)(470)\left(\frac{3\pi}{2}\right)}{9.8} = \frac{6.768\left(\frac{3\pi}{2}\right)}{9.8}\] | Compute \(r^2\) and multiply by \(\alpha\) to simplify the numerator step-by-step using algebra. |
| \[h = \frac{6.768(1.5\pi)}{9.8} = \frac{10.152\pi}{9.8}\] | Rewrite \(\frac{3\pi}{2} = 1.5\pi\) and multiply constants. |
| \[\boxed{h \approx 3.25\,\text{m above the center}}\] | The maximum height reached above the center of the wheel. |
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The launching mechanism of a toy gun consists of a spring with an unknown spring constant, \( k \). When the spring is compressed \( 0.120 \, \text{m} \) vertically, a \( 35.0 \, \text{g} \) projectile is able to be fired to a maximum height of \( 25 \, \text{m} \) above the position of the projectile when the spring is compressed. Assume that the barrel of the gun is frictionless.
An object weighing 120 N is set on a rigid beam of negligible mass at a distance of 3 m from a pivot, as shown above. A vertical force is to be applied to the other end of the beam a distance of 4 m from the pivot to keep the beam at rest and horizontal. What is the magnitude F of the force required?
The moment of inertia of a solid cylinder about its axis is given by \( I = \frac{1}{2}mR^2 \). If this cylinder rolls without slipping, the ratio of its rotational kinetic energy to its translational kinetic energy is
A spring with spring constant \( k = 2.3 \) \( \text{N/m} \) is attached to an object of mass \( 10 \) \( \text{kg} \). If the object is hung from the ceiling by this spring, how much would the spring be stretched?
A disk, a hoop, and a solid sphere are released at the same time at the top of an inclined plane. They are all uniform and roll without slipping. In what order do they reach the bottom?
\( \text{Solid sphere: } I = \frac{2}{5}mR^2, \quad \text{Solid disk: } I = \frac{1}{2}mR^2, \quad \text{Hoop: } I = mR^2 \)
A 0.72-m-diameter solid sphere can be rotated about an axis through its center by a torque of 10.8 N·m which accelerates it uniformly from rest through a total of 160 revolutions in 15.0 s. What is the mass of the sphere?
A \(100 \, \text{kg}\) person is riding a \(10 \, \text{kg}\) bicycle up a \(25^\circ\) hill. The hill is long and the coefficient of static friction is \(0.9\). The person rides \(10 \, \text{m}\) up the hill then takes a rest at the top. If she then starts from rest from the top of the hill and rolls down a distance of \(7 \, \text{m}\) before squeezing hard on the brakes locking the wheels, how much work is done by friction to bring the bicycle to a full stop, knowing that the coefficient of kinetic friction is \(0.65\)?
A snowboarder starts from rest and slides down a \(32^\circ\) incline that’s \(75 \, \text{m}\) long.
A massless rigid rod of length \(3d\) is pivoted at a fixed point \(W\), and two forces each of magnitude \(F\) are applied vertically upward as shown above. A third vertical force of magnitude \(F\) may be applied, either upward or downward, at one of the labeled points. With the proper choice of direction at each point, the rod can be in equilibrium if the third force of magnitude \(F\) is applied at point?
\(3.25\,\text{m}\)
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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] | |
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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