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Step Derivation/Formula Reasoning
A1 T – Mg = \frac{Mv^2}{L} This is the force equilibrium equation at the bottom point. T is the tension in the string, M is the mass of the ball, g is the acceleration due to gravity, v is the velocity of the ball, and L is the length of the string.
A2 T = 3Mg The tension at the bottom is given to be three times the weight of the ball.
A3 3Mg – Mg = \frac{Mv^2}{L} Substituting the tension value into the equilibrium equation.
A4 v = \sqrt{2gL} Simplifying the equation for v , observe M cancels out. This is velocity at any given point around the circle.
A5 F_{\text{centripetal}} = \frac{Mv^2}{L} 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 2mg Substitute in the equation for velocity (from step A4) so that the final equation is in terms of M \, g \, L
B1 v_{\text{top}} = \sqrt{v^2 – 4gL} Using conservation of energy. KE_{\text{bottom}} + PE_{\text{bottom}} = KE_{\text{top}} + PE_{\text{top}} . The velocity at the top is found by noting the potential energy difference between top and bottom. Simplify by substituting v from A4.
B2 v_{\text{top}} = \sqrt{2gL – 4gL} = \sqrt{-2gL} \rightarrow which is zero since 2gL > 4gL
C1 t = \sqrt{\frac{4L}{g}} Ball falls freely under gravity and has no initial vertical velocity, so \delta y = \frac{1}{2}gt^2 ; solving for t gives the time to fall a distance L . Note that the displacement from the top to the bottom is twice the radius of the circle or 2L .
D1 \Delta x = v_0t The horiztontal distance traveled by any projectile is the product of the horiztonal speed and the time in air.
D2   \sqrt{2gL} \times \sqrt{\frac{4L}{g}} Substitute in velocity from Step A4 and time from step C1
D3 \sqrt{8}L Simplify

(a) The net force on the ball at the top is 2Mg , downward.
(b) The velocity of the ball at the top is v = \sqrt{2gL} .
(c) The time it takes to reach the ground is \sqrt{\frac{4L}{g}} .
(d) The horizontal distance traveled is \sqrt{8}L

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(a) 2Mg , downward.
(b) v = \sqrt{2gL} .
(c) \sqrt{\frac{4L}{g}} .
(d) \sqrt{8}L

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KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_g = \frac{G m_1m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu N
R = \frac{v_i^2 \sin(2\theta)}{g}
Circular MotionEnergy
F_c = \frac{mv^2}{r}KE = \frac{1}{2} mv^2
a_c = \frac{v^2}{r}PE = mgh
KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion.
kSpring constant, in \text{N/m}
M_E = 5.972 \times 10^{24} , \text{kg} Mass of the Earth
M_M = 7.348 \times 10^{22} , \text{kg} Mass of the Moon
M_M = 1.989 \times 10^{30} , \text{kg} Mass of the Sun
VariableSI Unit
s (Displacement)\text{meters (m)}
v (Velocity)\text{meters per second (m/s)}
a (Acceleration)\text{meters per second squared (m/s}^2\text{)}
t (Time)\text{seconds (s)}
m (Mass)\text{kilograms (kg)}
VariableDerived SI Unit
F (Force)\text{newtons (N)}
E, PE, KE (Energy, Potential Energy, Kinetic Energy)\text{joules (J)}
P (Power)\text{watts (W)}
p (Momentum)\text{kilogram meters per second (kgm/s)}
\tau (Torque)\text{newton meters (Nm)}
I (Moment of Inertia)\text{kilogram meter squared (kgm}^2\text{)}
f (Frequency)\text{hertz (Hz)}

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters.

1. Start with the given measurement: \text{5 km}

2. Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}

3. Perform the multiplication: \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}

4. Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

10^{-12}

Nano-

n

10^{-9}

Micro-

µ

10^{-6}

Milli-

m

10^{-3}

Centi-

c

10^{-2}

Deci-

d

10^{-1}

(Base unit)

10^{0}

Deca- or Deka-

da

10^{1}

Hecto-

h

10^{2}

Kilo-

k

10^{3}

Mega-

M

10^{6}

Giga-

G

10^{9}

Tera-

T

10^{12}

1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
4. Bookmark questions that you can’t solve so you can come back to them later.
5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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