Circular Applications

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In this lesson we will explore a number of applications of circular motion, and where appropriate, examine the applications quantitatively.
We will look at the conical pendulum, banked curves, rollercoaster and some other amusement park rides

The conical pendulum and banked curves

The conical pendulum and banking both are more complex examples of centripetal motion. This video examines these situations. An understanding of the basics of circular motion, normal and friction is assumed.



​A summary on the principles of banked curves, useful for review
Interactive
This is a useful interactive that examines the conical pendulum 
Sample Problems
We are now ready to try some sample problems
Below is are sample problems with a video that explain how to solve them. It is suggested you try the problems beforehand, as this actually aids understanding, even if you are unsure if you are correct.
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Some more problems for you to try

1. A car (m = 1500 kg) is travelling around a circular corner that has a radius of 20 m. The coefficient of static friction between the road surface and the tyres is μ = 0.8.
  • a. What is the maximum frictional force the car can experience in the turn?
  • b. What is the maximum speed that the car can travel around the curve without slipping?
  • c. As it enters the turn, it encounters a wet section that reduces μ to 0.5. Explain what happens to the car.

a. Maximum Frictional Force:

Ff = μFN = μmg = 0.8 × 1500 kg × 9.8 m/s2 = 11,760 N

b. Maximum Speed:

Setting centripetal force equal to max friction: (m v2) / r = μmg → v = √(μgr)

v = √(0.8 × 9.8 × 20) = √(156.8) ≈ 12.52 m/s (approx. 45.1 km/h)

c. Effect of Wet Section (μ = 0.5):

The maximum available frictional force drops to Ff = 0.5 × 1500 × 9.8 = 7,350 N, reducing max safe speed to v = √(0.5 × 9.8 × 20) = 9.9 m/s. If the car is travelling faster than 9.9 m/s, the required centripetal force exceeds available friction, causing the car to slide outwards tangent to the curve.

2. You are in a roller coaster approaching a circular vertical loop with a diameter of 60 m. You forgot to buckle in. What minimum speed must the roller coaster have at the top of the loop to ensure you do not fall out? (HINT: Consider the condition where normal force at the top becomes zero, g = 9.8 m/s2)

Given: Diameter d = 60 m → Radius r = 30 m

Condition: At the minimum speed at the top, gravity alone provides the necessary centripetal force (FN = 0):

m g = (m v2) / r → v = √(g r)

Calculation: v = √(9.8 × 30) = √(294) ≈ 17.15 m/s (approx. 61.7 km/h)

Answer: Minimum speed at top = 17.15 m/s
3. A conical pendulum is rotating with a frequency of 1.5 Hz in a circle of radius 5 cm (0.05 m). What is the angle θ that the string makes with the vertical?

Given: Frequency f = 1.5 Hz → Angular velocity ω = 2 π f = 2 × π × 1.5 = 3 π rad/s ≈ 9.425 rad/s; Radius r = 0.05 m

Formula: For a conical pendulum, tan(θ) = (v2) / (g r) = (ω2 r) / g

Calculation: tan(θ) = (9.4252 × 0.05) / 9.8 = (88.826 × 0.05) / 9.8 = 4.4413 / 9.8 ≈ 0.4532

θ = arctan(0.4532) ≈ 24.38°

Answer: θ ≈ 24.4°
4. A 70-tonne (70,000 kg) Boeing 737 makes a banked turn as it approaches the airport. The radius of the turn is 2 km (2,000 m) and its speed is 300 km/h (83.33 m/s).
  • a. What will be its banking angle θ?
  • b. What is the total aerodynamic lift force (Normal equivalent) experienced by the aircraft during this turn?

Given: Mass m = 70,000 kg, Radius r = 2,000 m, Speed v = 300 / 3.6 = 83.33 m/s

a. Banking Angle:

tan(θ) = v2 / (g r) = (83.332) / (9.8 × 2000) = 6944.44 / 19600 ≈ 0.3543

θ = arctan(0.3543) ≈ 19.51°

b. Lift / Normal Force:

FL = (m g) / cos(θ) = (700,000 × 9.8) / cos(19.51°)

FL = 686,000 / 0.9426 ≈ 727,780 N (or 728 kN)

 

Rollercoasters and looped tracks


Roller coasters provide a great way to examine physics concepts: everything from forces, momentum, and energy - from both a linear perspective and rotational.
This covers these concepts and is here since loops are examined.
Many motion graph analysis video examine simplistic examples where velocity is constant or acceleration is constant.

​In this video I analyse a well known toy which has more complex motion, and I apply the same skills taught to analyse its graph. This video helps students consolidate their understanding of motion graphs, as well as review concepts such as friction, energy and forces.

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