| 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]
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A group of astronauts in a spaceship are attempting to land on Mars. As they approach the planet, they begin to plan their descent to the surface.
A \(5.0 \, \text{g}\) coin is placed \(15 \, \text{cm}\) from the center of a turntable. The coin has coefficients of static and kinetic friction of \(\mu_s = 0.80\) and \(\mu_k = 0.50\). The turntable slowly speeds up to \(60 \, \text{rpm}\). Does the coin slide off the turntable?
A ball is attached to the end of a string. It is swung in a vertical circle of radius \( 0.33 \) \( \text{m} \). What is the minimum velocity that the ball must have to make it around the circle?
A baseball rolls off a 0.70 m high desk and strikes the floor 0.25 m away from the base of the desk. How fast was the ball rolling?
Two balls are thrown off a building with the same speed, one straight up and one at a 45° angle. Which statement is true if air resistance can be ignored?
Consider a neutron star with a mass equal to the sun, a radius of 10 km, and a rotation period of 1.0 s. What is the speed of a point on the equator of the star?
A projectile has the least speed at what point in its path?
A circus cannon fires an acrobat into the air at an angle of \( 45^\circ \) above the horizontal, and the acrobat reaches a maximum height \( y \) above her original launch height. The cannon is now aimed so that it fires straight up, at an identical speed, into the air at an angle of \( 90^\circ \) to the horizontal. In terms of \( y \), what is the acrobat’s new maximum height?
A ball of mass \(m\) is released from rest at a distance \(h\) above a frictionless plane inclined at an angle of \(45^\circ\) to the horizontal as shown above. The ball bounces horizontally off the plane at point \(P_1\) with the same speed with which it struck the plane and strikes the plane again at point \(P_2\). In terms of \(g\) and \(h\), determine each of the following quantities:
A compressed spring mounted on a disk can project a small ball. When the disk is not rotating, as shown in the top view above, the ball moves radially outward. The disk then rotates in a counterclockwise direction as seen from above, and the ball is projected outward at the instant the disk is in the position shown above. Which of the following best shows the subsequent path of the ball relative to the ground?
(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] |
Metric Prefixes
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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