AP Physics Unit

Unit 3 - Circular Motion




A neighbor’s child wants to go to a carnival to experience the wild rides. The neighbor is worried about safety because one of the rides looks particularly dangerous. She knows that you have taken physics and so asks you for advice.

The ride in question has a 4 kg chair which hangs freely from a 10 m long chain attached to a pivot on the top of a tall tower. When the child enters the ride, the chain is hanging straight down. The child is then attached to the chair with a seat belt and shoulder harness. When the ride starts up, the chain rotates about the tower. Soon the chain reaches its maximum speed and remains rotating at that speed, which corresponds to one rotation about the tower every 3 seconds.

When you ask the operator, he says the ride is perfectly safe. He demonstrates this by sitting in the stationary chair. The chain creaks but holds, and he weighs 90 kg.

  1. (a) Has the operator shown that the ride is safe for a 25 kg child? (2 points)
  2. (b) Using equations, support your argument and show why it aligns with your response from part (a). (5 points)
Step Calculation/Analysis Description
1 T_{\text{max, stationary}} = m_{\text{operator}} \times g Maximum stationary tension with the operator’s weight. m_{\text{operator}} = 90 , \text{kg}, g = 9.81 , \text{m/s}^2.
2 v = \frac{2\pi \times L}{T} Rotation speed calculation. L = 10 , \text{m}, T = 3 , \text{s}.
3.1 T_{\text{horizontal}} = F_{\text{centripetal}} Horizontal component of tension equals centripetal force.
3.2 F_{\text{centripetal}} = \frac{m_{\text{child}} \times v^2}{r} Centripetal force formula. m_{\text{child}} = 25 , \text{kg}.
3.3 r = L \sin(\theta) Effective radius related to the angle .
3.4 F_{\text{centripetal}} = \frac{m_{\text{child}} \times v^2}{L \sin(\theta)} Substitute r in centripetal force formula.
3.5 T_{\text{vertical}} = m_{\text{child}} \times g Vertical component of tension equals child’s weight.
3.6 \tan(\theta) = \frac{T_{\text{horizontal}}}{T_{\text{vertical}}} Relation between tension components.
3.7 \tan(\theta) = \frac{m_{\text{child}} \times v^2 / (L \sin(\theta))}{m_{\text{child}} \times g} Substitute velocity and radius in the tangent formula.
3.8 cos\theta = \frac{9}{4\pi^2} Final, simplified equation
3.9 \theta = \approx 76.8 Approximate value of the angle to the vertical of any rider (angle does not depend of mass).
4 T_{\text{ride}}cos\theta = mg Solve for tension to get a total tension of ~ 1290 N for when the child sits in the seat. Remember the to add the mass of the seat to the child.
5 Safety Assessment Tension in the chain is greater while child is rotating than when operator is sitting in it stationary, indicating the chain might break when the child is rotating in it.


No it is not safe for the child. Based on the rotational speed, the child would be raised to an angle of ~77° from the initial vertical position. At this point the tension in the chain (~1290 N), while rotating, is considerable greater than the maximum force of tension (~920 N) the chain can with stand.

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No it is not safe for the child. Based on the rotational speed, the child would be raised to an angle of ~77° from the initial vertical position. At this point the tension in the chain (~1290 N), while rotating, is considerable greater than the maximum force of tension (~920 N) the chain can with stand.

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\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}}
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}}



Power of Ten




















(Base unit)


Deca- or Deka-


















  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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