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| 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."
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 student is asked to design an experiment to determine the change in angular momentum of a disk that rotates about its center and the product of the average torque applied to the disk and the time interval in which the torque is exerted. A net force is applied tangentially to the surface of the disk. The rotational inertia of the disk about its center is [katex]I = MR^2[/katex]. Which two of the following quantities should the student measure to determine the change in angular momentum of the disk after 10 s? Select two answers.
A motorcycle has tires with a diameter of \( 44.0 \) \( \text{cm} \). Cruising down the highway, they are rotating at \( 1150 \) \( \text{rpm} \) (revolutions per minute).
A solid sphere is rotating about an axis through its center at a constant rotation rate. Another hollow sphere of the same mass and radius is rotating about its axis through the center at the same rotation rate. Which sphere has a greater rotational kinetic energy?
A grinding wheel is in the form of a uniform solid disk of radius \( 7.00 \) \( \text{cm} \) and mass \( 2.00 \) \( \text{kg} \). It starts from rest and accelerates uniformly under the action of the constant torque of \( 0.600 \) \( \text{N m} \) that the motor exerts on the wheel.
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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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