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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \(\tau = r \cdot F\) | Calculate the torque (\(\tau\)) applied to the sphere. Torque is the product of the force applied (F) and the distance (r) from the axis of rotation. |
| 2 | \(\tau = 0.30 \, \text{m} \cdot 2.0 \, \text{N}\) | Substitute the given values for \(r\) and \(F\) into the torque formula. |
| 3 | \(\tau = 0.60 \, \text{N} \cdot \text{m}\) | Calculate the torque. |
| 4 | \(\tau = I \cdot \alpha\) | Relate torque (\(\tau\)) to angular acceleration (\(\alpha\)) using the moment of inertia (\(I\)). |
| 5 | \(0.60 \, \text{N} \cdot \text{m} = 0.06 \, \text{kg} \cdot \text{m}^2 \cdot \alpha\) | Substitute the known values of torque and moment of inertia into the formula to solve for \(\alpha\). |
| 6 | \(\alpha = 10 \, \text{rad/s}^2\) | Solve for the angular acceleration by dividing the torque by the moment of inertia. |
| 7 | \(\omega_f = \omega_i + \alpha \cdot t\) | Use the angular kinematics equation that relates initial angular velocity (\(\omega_i\)), angular acceleration (\(\alpha\)), and time (\(t\)) to final angular velocity (\(\omega_f\)). Since we are bringing the sphere to rest, \(\omega_f = 0\). |
| 8 | \(0 = 20 \, \text{rad/s} + (-10 \, \text{rad/s}^2) \cdot t\) | Substitute \(\omega_i = 20 \, \text{rad/s}\) and \(\alpha = -10 \, \text{rad/s}^2\) (negative because the sphere is slowing down). |
| 9 | \(-20 \, \text{rad/s} = -10 \, \text{rad/s}^2 \cdot t\) | Rearrange the equation to solve for \(t\). |
| 10 | \(t = 2 \, \text{s}\) | Solve for \(t\) by dividing both sides by \(-10 \, \text{rad/s}^2\). |
| 11 | \(\boxed{2 \, \text{s}}\) | The time required to stop the sphere is 2 seconds, which matches option (b). |
Just ask: "Help me solve this problem."
Consider a solid uniform sphere of radius \(R\) and mass \(M\) rolling without slipping. Which form of its kinetic energy is larger, translational or rotational?
A solid ball of mass \( M \) and radius \( R \) has rotational inertia \( \frac{2}{5} M R^{2} \) about its center. It rolls without slipping along a level surface at speed \( v \) just before it begins rolling up an inclined plane. Which of the following expressions correctly represents the maximum vertical height the solid ball can ascend to when it rolls up the incline without slipping?
An isolated spherical star of radius \( R_o \), rotates about an axis that passes through its center with an angular velocity of \( \omega_o \). Gravitational forces within the star cause the star’s radius to collapse and decrease to a value \( r_o < R_o \), but the mass of the star remains constant. A graph of the star’s angular velocity as a function of time as it collapses is shown. Which of the following predictions is correct about the angular momentum \( L \) of the star immediately after the collapse?
The system in the Figure is in equilibrium. A concrete block of mass 225 kg hangs from the end of a uniform strut whose mass is 45.0 kg.
A friend is balancing a fork on one finger. Which of the following are correct explanations of how he accomplishes this? Select two answers.
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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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