0 attempts
0% avg
UBQ Credits
Step | Derivation/Formula | Reasoning |
---|---|---|
1 | \[ K_{\text{trans}} = \frac{1}{2} M_{\text{tot}} v^2 \] | All three wagons have the same total mass and are accelerated to the same speed, so the translational kinetic energy is identical for each. |
2 | \[ K_{\text{rot}} = \frac{1}{2} I \omega^2 \quad \text{with} \quad \omega = \frac{v}{R} \] | This is the rotational kinetic energy for a wheel rolling without slipping. |
3 | \[ I_{\text{disk}} = \frac{1}{2} M_{w} R^2 \quad \Rightarrow \quad K_{\text{rot,disk}} = \frac{1}{2}\left(\frac{1}{2} M_{w} R^2\right)\left(\frac{v}{R}\right)^2 = \frac{1}{4} M_{w} v^2 \] | For a solid disk wheel (used in Wagons A and B), the moment of inertia is substituted to find its rotational kinetic energy. |
4 | \[ I_{\text{hoop}} = M_{w} R^2 \quad \Rightarrow \quad K_{\text{rot,hoop}} = \frac{1}{2}\left(M_{w} R^2\right)\left(\frac{v}{R}\right)^2 = \frac{1}{2} M_{w} v^2 \] | For a hollow hoop wheel (used in Wagon C), the moment of inertia is larger, leading to greater rotational energy. |
5 | \[ K_{\text{total, wheel}} = K_{\text{trans, wheel}} + K_{\text{rot}} = \frac{1}{2} M_{w} v^2 + K_{\text{rot}} \] | Each wheel has translational kinetic energy (as it moves with the wagon) plus its rotational kinetic energy. |
6 | For a disk: \[ K_{\text{total, disk}} = \frac{1}{2} M_{w} v^2 + \frac{1}{4} M_{w} v^2 = \frac{3}{4} M_{w} v^2 \] |
This is the effective energy per disk wheel. If the wheel’s mass were a simple point mass, it would only have \(\frac{1}{2}M_{w}v^2\); thus, the extra energy due to rotation is \(\Delta K_{\text{disk}} = \frac{3}{4} M_{w} v^2 – \frac{1}{2} M_{w} v^2 = \frac{1}{4} M_{w} v^2\). |
7 | For a hoop: \[ K_{\text{total, hoop}} = \frac{1}{2} M_{w} v^2 + \frac{1}{2} M_{w} v^2 = M_{w} v^2 \] |
Similarly, the extra energy due to rotation for a hoop is \(\Delta K_{\text{hoop}} = M_{w} v^2 – \frac{1}{2} M_{w} v^2 = \frac{1}{2} M_{w} v^2\), which is larger than that for a disk of the same mass. |
8 | Wagon A (disk, \(M_{w}=0.5\,\text{kg}\)): \[ \Delta K_{A} = 4\left(\frac{1}{4}\times 0.5\, v^2\right) = 4(0.125\, v^2) = 0.5\, v^2 \]Wagon B (disk, \(M_{w}=0.2\,\text{kg}\)): \[ \Delta K_{B} = 4\left(\frac{1}{4}\times 0.2\, v^2\right) = 4(0.05\, v^2) = 0.2\, v^2 \]Wagon C (hoop, \(M_{w}=0.2\,\text{kg}\)): \[ \Delta K_{C} = 4\left(\frac{1}{2}\times 0.2\, v^2\right) = 4(0.1\, v^2) = 0.4\, v^2 \] |
Each wagon has four wheels. The extra energy due to wheel rotation is calculated by multiplying the additional energy per wheel by four. |
9 | With \(v = 10\,\text{m/s}\) (\(v^2 = 100\)): \[ \Delta K_{A} = 0.5 \times 100 = 50\, \text{J} \] \[ \Delta K_{B} = 0.2 \times 100 = 20\, \text{J} \] \[ \Delta K_{C} = 0.4 \times 100 = 40\, \text{J} \] |
Substitute the final speed into each expression to obtain the numerical extra energy that must be supplied to accelerate the wheels. |
10 | \[\boxed{\text{Wagon A}}\] | Since the translational energy is the same for all wagons, the wagon with the greatest extra wheel energy (50 J) requires the most energy input. Therefore, Wagon A requires the most energy. |
Just ask: "Help me solve this problem."
An object is moving in a horizontal circle at a constant speed. Which of the following correctly describes the linear and angular velocities of the object between any point along the circular path?
A discus is held at the end of an arm that starts at rest. The average angular acceleration of [katex]54 \, \text{rad/s}^2 [/katex] lasts for 0.25 s. The path is circular and has radius 1.1 m.
Note: A discuss is a heavy, flattened circular object for throwing.
An object moves at a constant speed of [katex] 9.0 \frac{m}{s} [/katex] in a circular path of radius of 1.5 m. What is the angular acceleration of the object?
In both cases, a massless rod is supported by fulcrum, and a 200-kg hanging mass is suspended from the left end of the rod by a cable. A downward force F keeps the rod in rest. The rod in Case A is 50 cm long, and the rod in case B is 40 cm long (each rod is marked at 10-cm intervals). The magnitude of each vertical force F exerted on the rod will be
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?
\( \text{Solid sphere: } I = \frac{2}{5}mR^2, \quad \text{Solid disk: } I = \frac{1}{2}mR^2, \quad \text{Hoop: } I = mR^2 \)
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] | 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
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.
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 AI accurately solves all questions
🔥 Get up to 30% off Elite Physics Tutoring
🧠 NEW! Learn Physics From Scratch Self Paced Course
🎯 Need exam style practice questions?