New Tool FRQ Atlas - Find, Solve, and Grade Any FRQ In Seconds.

AP Physics

Unit 3 - Circular Motion

FRQ
Mathematical
Intermediate

Pro Tier

Unlimited Grading Credits, Explanations, and AI Assist

0 attempts

0% avg

Explanation 0
0

(a) Tension in the Rope While the Swing is Held at Rest

Step Derivation/Formula Reasoning
1 \[ T \cos(\theta) = W \] At equilibrium, the vertical component of the tension must balance the weight of the child.
2 \[ T = \frac{W}{\cos(\theta)} \] Solve for the tension \( T \) in the rope by dividing both sides by \(\cos(\theta)\).
3 \[\boxed{T = \frac{W}{\cos(\theta)}}\] Final expression for the tension when the swing is held at an angle \(\theta\).

(b) Horizontal Force Exerted by the Adult

Step Derivation/Formula Reasoning
1 \[ F = T \sin(\theta) \] The horizontal component of the tension is equal to the horizontal force exerted by the adult.
2 \[ F = \frac{W}{\cos(\theta)} \sin(\theta) \] Substitute the expression for \( T \) from part (a).
3 \[ F = W \tan(\theta)\] Simplify using the trigonometric identity \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\).
4 \[\boxed{F = W \tan(\theta)}\] Final expression for the horizontal force exerted by the adult.

(c) Tension in the Rope at the Lowest Point

Step Derivation/Formula Reasoning
1 \[ h = L – L \cos(\theta) \] Calculate the vertical height the swing descends from the initial position.
2 \[ PE_{\text{initial}} = W h \] Potential energy at the held position is equal to the weight times the height.
3 \[ KE_{\text{lowest}} = \frac{1}{2} mv^2 \] Kinetic energy at the lowest point is expressed in terms of mass and velocity.
4 \[ W h = \frac{1}{2} mv^2 \] By conservation of energy, convert initial potential energy to kinetic energy at the lowest point.
5 \[ v = \sqrt{2gh} \] Solve for velocity using the relationship between potential energy, kinetic energy, and height.
6 \[ T – W = \frac{mv^2}{L} \] Net force at the lowest point, which provides centripetal force, is the difference between tension and weight.
7 \[ T = W + \frac{mv^2}{L} \] Rearrange to solve for \( T \).
8 \[ T = W + 2W(1 – \cos(\theta)) \] Substitute \( v = \sqrt{2gL(1-\cos(\theta))} \) from step 5 and \( m = \frac{W}{g} \).
9 \[\boxed{T = W(3 – 2\cos(\theta))}\] Final expression for the tension in the rope at the lowest point of the swing.

Need Help? Ask Phy To Explain

Just ask: "Help me solve this problem."

Just Drag and Drop!
Quick Actions ?
×

Topics in this question

We'll help clarify entire units in one hour or less — guaranteed.

NEW AI Quiz Builder

Be the first to use our new Quiz platform to create and grade quizzes from scratch. Join the waitlist and we'll email you for early access.

Go Pro to remove ads + unlimited access to our AI learning tools.
  1. \(\boxed{T = \frac{W}{\cos(\theta)}}\)
  2. \(\boxed{F = W \tan(\theta)}\)
  3. \(\boxed{T = W(3 – 2\cos(\theta))}\)

Nerd Notes

Discover the world's best Physics resources

Continue with

By continuing you (1) agree to our Terms of Use and Terms of Sale and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.

Error Report

Sign in before submitting feedback.

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]

Sign In to View Your Questions

Share This Question

Enjoying UBQ? Share the 🔗 with friends!

Link Copied!

Phy Pro

One price to unlock most advanced version of Phy across all our tools.

$11.99

per month

Billed Monthly. Cancel Anytime.

Physics is Hard, But It Does NOT Have to Be

We crafted THE Ultimate A.P Physics 1 Program so you can learn faster and score higher.

Trusted by 10k+ Students

📚 Predict Your AP Physics Exam Score

Try our free calculator to see what you need to get a 5 on the 2026 AP Physics 1 exam.

Feeling uneasy about your next physics test? We'll boost your grade in 3 lessons or less—guaranteed

We use cookies to improve your experience. By continuing to browse on Nerd Notes, you accept the use of cookies as outlined in our privacy policy.