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| Step | Derivation/Formula | Reasoning |
|---|---|---|
| A1 | [katex] T – Mg = \frac{Mv^2}{L} [/katex] | This is the force equilibrium equation at the bottom point. [katex] T [/katex] is the tension in the string, [katex] M [/katex] is the mass of the ball, [katex] g [/katex] is the acceleration due to gravity, [katex] v [/katex] is the velocity of the ball, and [katex] L [/katex] is the length of the string. |
| A2 | [katex] T = 3Mg [/katex] | The tension at the bottom is given to be three times the weight of the ball. |
| A3 | [katex] 3Mg – Mg = \frac{Mv^2}{L} [/katex] | Substituting the tension value into the equilibrium equation. |
| A4 | [katex] v = \sqrt{2gL} [/katex] | Simplifying the equation for [katex] v [/katex], observe [katex] M [/katex] cancels out. This is velocity at any given point around the circle. |
| A5 | [katex] F_{\text{centripetal}} = \frac{Mv^2}{L} [/katex] | At the top, we can use the velocity calculated in the previous step to find the centripetal force required to keep the ball moving in the circle. |
| A6 | [katex] 2mg [/katex] | Substitute in the equation for velocity (from step A4) so that the final equation is in terms of [katex] M \, g \, L [/katex] |
| B1 | [katex] v_{\text{top}} = \sqrt{v^2 – 4gL} [/katex] | Using conservation of energy. [katex] KE_{\text{bottom}} + PE_{\text{bottom}} = KE_{\text{top}} + PE_{\text{top}} [/katex]. The velocity at the top is found by noting the potential energy difference between top and bottom. Simplify by substituting [katex] v [/katex] from A4. |
| B2 | [katex] v_{\text{top}} = \sqrt{2gL – 4gL} [/katex] | [katex] = \sqrt{-2gL} \rightarrow [/katex] which is zero since [katex] 2gL > 4gL [/katex] |
| C1 | [katex] t = \sqrt{\frac{4L}{g}} [/katex] | Ball falls freely under gravity and has no initial vertical velocity, so [katex] \delta y = \frac{1}{2}gt^2 [/katex]; solving for [katex] t [/katex] gives the time to fall a distance [katex] L [/katex]. Note that the displacement from the top to the bottom is twice the radius of the circle or [katex] 2L [/katex]. |
| D1 | [katex] \Delta x = v_0t [/katex] | The horiztontal distance traveled by any projectile is the product of the horiztonal speed and the time in air. |
| D2 | [katex] \sqrt{2gL} \times \sqrt{\frac{4L}{g}} [/katex] | Substitute in velocity from Step A4 and time from step C1 |
| D3 | [katex] \sqrt{8}L [/katex] | Simplify |
(a) The net force on the ball at the top is [katex] 2Mg [/katex], downward.
(b) The velocity of the ball at the top is [katex] v = \sqrt{2gL} [/katex].
(c) The time it takes to reach the ground is [katex] \sqrt{\frac{4L}{g}} [/katex].
(d) The horizontal distance traveled is [katex] \sqrt{8}L [/katex]
Just ask: "Help me solve this problem."
A particle of mass m slides down a fixed, frictionless sphere of radius R, starting from rest at the top.
In terms of m, g, R, and O, determine each of the following for the particle while it is sliding on the sphere.
If a baseball pitch leaves the pitcher’s hand horizontally at a velocity of \( 150 \) \( \text{km/h} \), by what \( \% \) will the pull of gravity change the magnitude of the velocity when the ball reaches the batter, \( 18 \) \( \text{m} \) away? For this estimate, ignore air resistance and spin on the ball.
A communications satellite orbits the Earth at an altitude of 35,000 km above the Earth’s surface. Take the mass of Earth to be [katex]6 \times 10^{24} \text{ kg}[/katex] the the radius of Earth to be [katex]6.4 \times 10^6 \text{ m}[/katex]. What is the satellite’s velocity?
The Moon does not crash into the Earth because:
A projectile is fired with an initial speed of \( 36.6 \) \( \text{m/s} \) at an angle of \( 42.2^\circ \) above the horizontal on a long flat firing range.
(a) [katex] 2Mg [/katex], downward.
(b) [katex] v = \sqrt{2gL} [/katex].
(c) [katex] \sqrt{\frac{4L}{g}} [/katex].
(d) [katex] \sqrt{8}L [/katex]
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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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