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Step Derivation/Formula Reasoning
1 \tau = FR The torque (\tau) exerted on the cylinder is due to the force F applied at a radius R. The formula for torque is the force times the perpendicular distance (radius in this case) from the axis of rotation.
2 \tau = I\alpha Newton’s second law for rotation states that the torque is equal to the moment of inertia (I) times the angular acceleration (\alpha).
3 I = \frac{1}{2}MR^2 The moment of inertia for a solid cylinder about its axis is given by this formula, where M is the mass and R is the radius of the cylinder.
4 FR = \frac{1}{2}MR^2 \alpha Substitute the moment of inertia of the cylinder into the torque equation.
5 \alpha = \frac{2F}{MR} Solve for the angular acceleration (\alpha) by isolating \alpha on one side of the equation.
6 \omega^2 = \omega_0^2 + 2\alpha \theta Use the kinematic equation for rotational motion to relate the angular displacement (\theta) to the final angular velocity (\omega). Here, \omega_0 (initial angular velocity) is zero as the cylinder starts from rest.
7 \omega^2 = 2\alpha \theta Substitute \omega_0 = 0 into the equation because the cylinder starts from rest.
8 \omega = \sqrt{2\alpha \theta} = \sqrt{\frac{4F\theta}{MR}} Substitute the value of \alpha from Step 5 into the equation to find \omega.
9 K = \frac{1}{2}I\omega^2 The total kinetic energy (K) of the rotating cylinder is given by the formula for rotational kinetic energy, where I is the moment of inertia and \omega is the angular velocity.
10 K = \frac{1}{2} \times \frac{1}{2}MR^2 \times \left(\frac{4F\theta}{MR}\right) Substitute the expressions for I and \omega into the kinetic energy formula.
11 K = \frac{F\theta R}{2} Simplify the equation to get the final expression for the kinetic energy.
12 K = \frac{F\theta R}{2} Conclude with the neat, simplified expression for the kinetic energy of the cylinder after it has rotated through an angle \theta.

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K = \frac{F\theta R}{2}

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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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