0 attempts
0% avg
UBQ Credits
Step | Derivation/Formula | Reasoning |
---|---|---|
1 | \[ \alpha = \frac{\tau}{I} = \frac{3.00}{5.00} = 0.600\ \text{rad/s}^2 \] | We use \(\tau = I\alpha\) to determine the angular acceleration \(\alpha\) given the torque \(\tau\) and moment of inertia \(I\). |
2 | \[ v_x = v_i + \alpha t = 0 + 0.600 \times 8.0 = 4.8\ \text{rad/s} \] | Since the wheel starts from rest (\(v_i=0\)), we calculate the final angular speed \(v_x\) after \(8.0\) s. |
3 | \[ KE = \frac{1}{2} I v_x^2 = \frac{1}{2} \times 5.00 \times (4.8)^2 = 57.6\ \text{J} \] | This step applies the rotational kinetic energy formula \(KE=\frac{1}{2}I v_x^2\) to compute the energy at the end of \(8.0\) s. |
4 | \[ \boxed{57.6\ \text{J}} \] | The final value represents the wheel’s rotational kinetic energy, corresponding to option (d). |
Just ask: "Help me solve this problem."
A sphere starts from rest and rolls down an incline of height \( H = 1.0 \) \( \text{m} \) at an angle of \( 25^\circ \) with the horizontal, as shown above. The radius of the sphere \( R = 15 \) \( \text{cm} \), and its mass \( m = 1.0 \) \( \text{kg} \). The moment of inertia for a sphere is \( \frac{2}{5}mR^2 \). What is the speed of the sphere when it reaches the bottom of the plane?
Two uniform solid balls, one of radius \( R \) and mass \( M \), the other of radius \( 2R \) and mass \( 8M \), roll down a high incline. They start together from rest at the top of the incline. Which one will reach the bottom of the incline first?
A uniform solid sphere of mass M and radius R is placed on a frictionless horizontal surface. A massless string is wrapped around the sphere and is pulled with a force F. The string makes an angle of θ with the horizontal. What is the minimum value of the coefficient of static friction between the sphere and the surface required for the sphere to start rolling without slipping?
A wheel of radius R and negligible mass is mounted on a horizontal frictionless axle so that the wheel is in a vertical plane. Three small objects having masses [katex]m[/katex], [katex]M[/katex], and [katex]2M[/katex], respectively, are mounted on the rim of the wheel, as shown above. If the system is in static equilibrium, what is the value of [katex]m[/katex] in terms of [katex]M[/katex] ?
Suppose a solid uniform sphere of mass M and radius R rolls without slipping down an inclined plane starting from rest. The angular velocity of the sphere at the bottom of the incline depends on
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 |
---|---|
\(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
\(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
\(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
\(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
\(v^2 = v_f^2 \,-\, 2a \Delta x\) |
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] | 0.000000000001 |
Nano- | n | [katex]10^{-9}[/katex] | 0.000000001 |
Micro- | µ | [katex]10^{-6}[/katex] | 0.000001 |
Milli- | m | [katex]10^{-3}[/katex] | 0.001 |
Centi- | c | [katex]10^{-2}[/katex] | 0.01 |
Deci- | d | [katex]10^{-1}[/katex] | 0.1 |
(Base unit) | – | [katex]10^{0}[/katex] | 1 |
Deca- or Deka- | da | [katex]10^{1}[/katex] | 10 |
Hecto- | h | [katex]10^{2}[/katex] | 100 |
Kilo- | k | [katex]10^{3}[/katex] | 1,000 |
Mega- | M | [katex]10^{6}[/katex] | 1,000,000 |
Giga- | G | [katex]10^{9}[/katex] | 1,000,000,000 |
Tera- | T | [katex]10^{12}[/katex] | 1,000,000,000,000 |
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.
A quick explanation
Credits are used to grade your FRQs and GQs. Pro users get unlimited credits.
Submitting counts as 1 attempt.
Viewing answers or explanations count as a failed attempts.
Phy gives partial credit if needed
MCQs and GQs are are 1 point each. FRQs will state points for each part.
Phy customizes problem explanations based on what you struggle with. Just hit the explanation button to see.
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.
NEW! PHY instantly solves any question
🔥 Elite Members get up to 30% off Physics Tutoring
🧠 Learning Physics this summer? Try our free course.
🎯 Need exam style practice questions? We’ve got over 2000.