AP Physics

Unit 1 - Vectors and Kinematics

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Step Reasoning
Identify the target equation for angular acceleration and the components needed for each case.
\[ \alpha = \dfrac{\sum \tau}{I} \]
The question asks for the ratio of angular accelerations, which are determined by the net torque and the rotational inertia of the system about the pivot.
Calculate the rotational inertia for Case 1 and Case 2.
\[ I_1 = I_{rod} + I_{block1} = \dfrac{1}{3}ML^2 + M\left(\dfrac{L}{2}\right)^2 = \dfrac{1}{3}ML^2 + \dfrac{1}{4}ML^2 = \dfrac{7}{12}ML^2 \]
\[ I_2 = I_{rod} + I_{block2} = \dfrac{1}{3}ML^2 + M(L)^2 = \dfrac{4}{3}ML^2 \]
The system consists of a uniform rod and a point mass. The total inertia is the sum of the rod’s inertia about its end and the block’s inertia at its specific location.
Calculate the net torque for Case 1 and Case 2.
\[ \sum \tau_1 = (Mg)\left(\dfrac{L}{2}\right) + (Mg)\left(\dfrac{L}{2}\right) = MgL \]
\[ \sum \tau_2 = (Mg)\left(\dfrac{L}{2}\right) + (Mg)(L) = \dfrac{3}{2}MgL \]
The torque is provided by gravity acting at the center of mass of the rod and the location of the block. Since the rod is horizontal, the lever arms are the horizontal distances from the pivot.
Express the angular accelerations and find their ratio.
\[ \alpha_1 = \dfrac{MgL}{\frac{7}{12}ML^2} = \dfrac{12g}{7L} \]
\[ \alpha_2 = \dfrac{\frac{3}{2}MgL}{\frac{4}{3}ML^2} = \dfrac{3}{2} \cdot \dfrac{3}{4} \cdot \dfrac{g}{L} = \dfrac{9g}{8L} \]
\[ \dfrac{\alpha_2}{\alpha_1} = \dfrac{9/8}{12/7} = \dfrac{9}{8} \cdot \dfrac{7}{12} = \dfrac{21}{32} \]
Substituting the torque and inertia values into the rotational version of Newton’s second law allows for a direct comparison.

Why each choice is correct or incorrect:

(A) Uses the ratio of the inertias \(I_1 / I_2\) but ignores the fact that the torque increases when the block is moved further from the pivot.

(B) This is the correct answer.

(C) Calculates the ratio correctly but omits the torque contribution of the rod’s own weight, using \(\tau_1 = MgL/2\) and \(\tau_2 = MgL\).

(D) Ignores the rod’s mass in both the torque and inertia calculations, which simplifies the ratio to \(\tau_2 I_1 / \tau_1 I_2 = (L)(L/2)^2 / (L/2)(L)^2 = 1/2\).

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KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque 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 MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[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
VariableSI 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]
VariableDerived 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. 

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

  2. 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]

  3. 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]

  4. 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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