AP Physics Unit

Unit 3 - Circular Motion

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GQ

Two wires are tied to the 500 g sphere shown below. The sphere revolves in a horizontal circle at a constant speed of 7.2 m/s. What is the tension in the upper wire?  What is the tension in the lower wire?

Upper wire: 22 N; Lower wire: 12 N

0

Before solving the question, we can find the radius of the ball using Pythagorean theorem to get .866 m. We can also use the trig to solve for the angle each rope makes with the horizontal (30° for both ropes).

Sum of Forces in the Horizontal Direction:

Step Formula Derivation Reasoning
1 \cos(30) = \frac{\sqrt{3}}{2} Cosine of 30^\circ.
2 F_{\text{centripetal}} = \frac{mv^2}{r} Centripetal force for circular motion.
3 T_1 \cos(\theta) + T_2 \cos(\theta) = \frac{mv^2}{r} Sum of horizontal components of tension equals centripetal force.
4 T_1 \frac{\sqrt{3}}{2} + T_2 \frac{\sqrt{3}}{2} = \frac{(0.5)(7.2)^2}{0.866} Substitute values for m, v, r, and \cos(\theta).
5 \frac{\sqrt{3}}{2}(T_1 + T_2) = 29.93 Calculate centripetal force and factor out \frac{\sqrt{3}}{2}.

Sum of Forces in the Vertical Direction:

Step Formula Derivation Reasoning
1 \sin(30) = \frac{1}{2} Sine of 30^\circ.
2 w = mg Weight of the sphere.
3 T_2 \sin(\theta) + mg – T_1 \sin(\theta) = 0 Vertical forces must balance: upward tensions and downward weight.
4 T_2 \frac{1}{2} + (0.5)(9.8) – T_1 \frac{1}{2} = 0 Substitute values for m, g, and \sin(\theta).
5 \frac{1}{2}(T_2 – T_1) + 4.9 = 0 Factor out \frac{1}{2} and calculate weight.

Solving for Tensions:

Step Formula Derivation Reasoning
1 Solve equations Use the system of equations to solve for T_1 and T_2.
2 T_1 \approx 22.18 \text{ N} Numerical solution for T_1.
3 T_2 \approx 12.38 \text{ N} Numerical solution for T_2.

Final Tensions:

  • Upper wire: \boxed{T_1 \approx 22.18 \text{ N}}
  • Lower wire: \boxed{T_2 \approx 12.38 \text{ N}}

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Upper wire: 22 N; Lower wire: 12 N

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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)}
\omega (Angular Velocity)\text{radians per second (rad/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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