0 attempts
0% avg
UBQ Credits
(a) Calculate the linear speed of the sphere when it reaches the bottom of the incline.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]h = L \sin(\theta)[/katex] | Calculate the vertical height [katex]h[/katex] fallen by the sphere using the length of the incline [katex]L[/katex] and the sine of the incline angle [katex]\theta[/katex]. |
| 2 | [katex]h = 7.0 \sin(35^\circ)[/katex] | Substitute [katex]L = 7.0 \, m[/katex] and [katex]\theta = 35^\circ[/katex]. |
| 3 | [katex]PE_{\text{top}} = KE_{\text{trans}} + KE_{\text{rot}}[/katex] | Use the conservation of mechanical energy, where potential energy at the top is equal to the sum of transnational kinetic energy and rotational kinetic energy at the bottom. |
| 4 | [katex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2[/katex] | Express the conservation of energy equation in terms of [katex]v[/katex] (linear velocity) and [katex]\omega[/katex] (angular velocity). |
| 5 | [katex]I = \frac{2}{5}MR^2[/katex] | Substitute the given moment of inertia for a solid sphere, where [katex]I = \frac{2}{5}MR^2[/katex]. |
| 6 | [katex]v = R\omega[/katex] | Relation between linear velocity and angular velocity for rolling without slipping. |
| 7 | [katex]\omega = \frac{v}{R}[/katex] | Rearrange the equation for [katex]\omega[/katex]. |
| 8 | [katex]mgh = \frac{1}{2}mv^2 + \frac{1}{2}(\frac{2}{5}MR^2)(\frac{v}{R})^2[/katex] | Substitute [katex]I[/katex] and [katex]\omega[/katex] in terms of [katex]v[/katex] and [katex]R[/katex]. |
| 9 | [katex]mgh = \frac{1}{2}mv^2 + \frac{1}{5}mv^2[/katex] | Simplify the equation by canceling [katex]M[/katex] and [katex]R[/katex]. |
| 10 | [katex]mgh = \frac{7}{10}mv^2[/katex] | Combine like terms. |
| 11 | [katex]v^2 = \frac{10}{7}gh[/katex] | Isolate [katex]v^2[/katex]. |
| 12 | [katex]v = \sqrt{\frac{10}{7}gh}[/katex] | Take the square root to find [katex]v[/katex]. |
| 13 | [katex]v = \sqrt{\frac{10}{7}(9.8)(7.0 \sin(35^\circ))}[/katex] | Substitute the values of [katex]g[/katex] and [katex]h[/katex]. |
| 14 | [katex]\boxed{v \approx 7.5 \, \text{m/s}}[/katex] | Final answer. |
(b) Determine the angular speed of the sphere at the bottom of the incline.
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | [katex]\omega = \frac{v}{R}[/katex] | Use the relation between linear and angular velocities for a rolling object. |
| 2 | [katex]\omega = \frac{7.5}{0.15}[/katex] | Substitute [katex]v = 7.5 \, \text{m/s}[/katex] and [katex]R = 0.15 \, m[/katex] (converted from cm). |
| 3 | [katex]\boxed{\omega \approx 50 \, \text{rad/s}}[/katex] | Final answer. |
(c) Does the linear speed depend on the radius or mass of the sphere?
| Step | Analysis | Conclusion |
|---|---|---|
| 1 | From the energy conservation equation, the mass [katex]m[/katex] cancelled out and the final expression for [katex]v[/katex] didn’t include the radius [katex]R[/katex]. | The linear speed does not depend on the mass or radius of the sphere as both factors were eliminated in deriving [katex]v[/katex]. |
(d) Does the angular speed depend on the radius or mass of the sphere?
| Step | Analysis | Conclusion |
|---|---|---|
| 1 | Angular speed [katex]\omega[/katex] was found from [katex]v[/katex] divided by [katex]R[/katex], but it did not involve mass [katex]m[/katex]. | Angular speed does depend on the radius and does not depend on the mass of the sphere. |
Just ask: "Help me solve this problem."
A disk of known radius and rotational inertia can rotate without friction in a horizontal plane around its fixed central axis. The disk has a cord of negligible mass wrapped around its edge. The disk is initially at rest, and the cord can be pulled to make the disk rotate. Which of the following procedures would best determine the relationship between applied torque and the resulting change in angular momentum of the disk?
The diagram above shows a top view of a child of mass \(M\) on a circular platform of mass \(2M\) that is rotating counterclockwise. Assume the platform rotates without friction. Which of the following describes an action by the child that will increase the angular speed of the platform-child system and gives the correct reason why?
A horizontal uniform rod of length L and mass M is pivoted at one end and is initially at rest. A small ball of mass M (same masses) is attached to the other end of the rod. The system is released from rest. What is the angular acceleration of the rod just immediately after the system is released?
The figure shows a person’s foot. In that figure, the Achilles tendon exerts a force of magnitude F = 720 N. What is the magnitude of the torque that this force produces about the ankle joint?
What is the net torque acting on the pivot supporting a 10-kilogram beam 2 meters long as shown above? Assume that the positive direction is clockwise.
By continuing you (1) agree to our Terms of Use and Terms of Sale 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 |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| 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] |
The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.
per month
Billed Monthly. Cancel Anytime.
Trial –> Phy Pro
We crafted the ultimate A.P Physics 1 course that simplifies everything so you can learn faster and score higher.
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.
10 Free Credits To Get You Started
By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data.
