0 attempts
0% avg
UBQ Credits
| Derivation or Formula | Reasoning |
|---|---|
| \[ Mg\sin\theta – f = Ma \] | This is Newton’s second law for translation along the incline, where \(Mg\sin\theta\) is the component of gravity down the plane and \(f\) is the friction force. |
| \[ fR = I\alpha \quad \text{with} \quad a = \alpha R \] | The friction force provides the torque needed for rolling without slipping, and the rolling condition relates linear acceleration \(a\) to angular acceleration \(\alpha\). |
| \[ f = \frac{Ia}{R^2} \] | Rearranging the torque equation yields an expression for \(f\) in terms of \(a\). |
| \[ f = \frac{2}{5}Ma \] | Substitute the moment of inertia for a uniform solid sphere \(I = \frac{2}{5}MR^2\) into the previous expression. |
| \[ Mg\sin\theta – \frac{2}{5}Ma = Ma \] | Insert \(f = \frac{2}{5}Ma\) into Newton’s second law and set up the equation for \(a\). |
| \[ Mg\sin\theta = \left(1+\frac{2}{5}\right)Ma = \frac{7}{5}Ma \] | Simplify the equation to combine like terms and isolate \(a\). |
| \[ a = \frac{5}{7}g\sin\theta \] | Solve for the linear acceleration \(a\) of the sphere along the incline. |
| \[ f = \frac{2}{5}M\left(\frac{5}{7}g\sin\theta\right) = \frac{2}{7}Mg\sin\theta \] | Determine the friction force required for rolling by substituting \(a\) back into \(f = \frac{2}{5}Ma\). |
| \[ \frac{2}{7}Mg\sin\theta = \mu Mg\cos\theta \] | For the sphere to roll without slipping, the required friction force must be available, i.e., it must equal the maximum static friction \(\mu Mg\cos\theta\). |
| \[ \mu = \frac{2}{7}\tan\theta \] | Solve for the minimum coefficient of friction \(\mu\) needed to prevent slipping. |
| \[ \boxed{\mu = \frac{2}{7}\tan\theta} \] | This boxed expression is the final answer for part (a). |
| Derivation or Formula | Reasoning |
|---|---|
| \[ Mgh = \frac{1}{2}Mv_x^2 + \frac{1}{2}I\left(\frac{v_x}{R}\right)^2 \quad \Rightarrow \quad v_x = \sqrt{\frac{10}{7}gh} \] | For a rolling sphere, the gravitational potential energy converts into both translational and rotational kinetic energy. |
| \[ Mgh = \frac{1}{2}Mv_x^2 \quad \Rightarrow \quad v_x = \sqrt{2gh} \] | Without friction (\(\mu = 0\)), the sphere slides without rotating, so all potential energy becomes translational kinetic energy. |
| \[ \sqrt{2gh} > \sqrt{\frac{10}{7}gh} \] | Since \(2 > \frac{10}{7}\), the translational speed of a sliding sphere is higher than that of a rolling sphere, where some energy goes into rotation. |
| \(\text{Speed is greater when } \mu = 0\) | Thus, with zero friction the sphere attains a higher speed at the bottom because no energy is diverted to rotational motion. |
Just ask: "Help me solve this problem."
A horizontal uniform meter stick of mass 0.2 kg is supported at its midpoint by a pivot point. A mass of 0.1 kg is attached to the left end of the meter stick, and another mass of 0.15 kg is attached to the right end of the meter stick. The meter stick is free to rotate in the horizontal plane around the pivot point. What is the tension in the string supporting the left end of the meter stick?
Four identical lead balls with large mass are connected by rigid but very light rods in the square configuration shown in the preceding figure. The balls are rotated about the three labeled axes. Which of the following correctly ranks the rotational inertia \(I\) of the balls about each axis?
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 high-speed flywheel in a motor is spinning at \( 500 \) \( \text{rpm} \) when a power failure suddenly occurs. The flywheel has a mass of \( 40 \) \( \text{kg} \) and a diameter of \( 75 \) \( \text{cm} \). The power is off for \( 30 \) \( \text{s} \) and during this time the flywheel slows due to friction in its axle bearings. During this time the flywheel makes \( 200 \) complete revolutions.
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?
\(\mu = \frac{2}{7}\tan\theta\)\n\(\text{Greater}\)
By continuing you (1) agree to our Terms of Sale and Terms of Use and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.
| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) | Â |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| Â | \(W = Fd \cos\theta\) |
| 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.Â
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] |
The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
Try our free calculator to see what you need to get a 5 on the upcoming AP Physics 1 exam.
Understand you mistakes quicker.
Phy automatically provides feedback so you can improve your responses.Â
10 Free Credits To Get You StartedÂ
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.Â
