Quantitative Analysis of Uniform Circular Motion

Expert reviewed • 22 November 2024 • 9 minute read

What is Uniform Circular Motion?

Uniform circular motion is a fundamental concept in physics that describes the motion of an object traveling in a circular path at a constant speed.

What are the key Equations in Uniform Circular Motion?

To understand and analyse uniform circular motion quantitatively, we need to familiarise ourselves with several important equations:

  • Centripetal Acceleration: ac=v2ra_c = \frac{v^2}{r}
  • Tangential Velocity: v=2πrTv = \frac{2\pi r}{T}
  • Centripetal Force: Fc=mv2rF_c = \frac{mv^2}{r}
  • Angular Velocity: ω=Δθt\omega = \frac{\Delta \theta}{t}

What is Tangential Velocity?

Tangential velocity (v)(v) is the linear speed of an object moving in a circular path. It is defined as the distance traveled along the circumference of the circle divided by the time taken for one complete revolution (period, T):

v=2πrTv = \frac{2\pi r}{T}

Where:

  • rr is the radius of the circular path
  • TT is the period (time for one complete revolution)

Practice Question 1

A satellite orbits Earth in a circular path with a radius of 6,700 km. If it completes one orbit in 90 minutes, what is its tangential velocity?

r = 6,700, 000mr = 6,700, 000 m\\ T = 90 × 60 = 5, 400sT = 90 × 60 \\= 5, 400 s v=2πrT=2π×6,700,0005,400≈7,800 m/sv = \frac{2\pi r}{T} \\= \frac{2\pi \times 6,700,000}{5,400} \\\approx 7,800 \text{ m/s}

What is Angular Velocity?

Angular velocity ω\omega represents the rate of change of angular position. It can be defined in two ways:

  • ω=2πT\omega = \frac{2\pi}{T}
  • ω=Δθt\omega = \frac{\Delta \theta}{t}

Where:

  • ΔθΔθ is the change in angular position
  • tt is the time taken (not necessarily the period)

The SI unit for angular velocity is radians per second (rad/s).

An important relationship between tangential velocity and angular velocity is:

v = ωrv = ωr

Practice Question 2

A wheel with a radius of 0.3 m rotates at an angular velocity of 10 rad/s. Calculate its tangential velocity.

v = ωr = 10 × 0.3 = 3m/sv = ωr \\= 10 × 0.3 \\= 3 m/s

What is Centripetal Acceleration?

Centripetal acceleration (ac)(a_c) is the acceleration directed towards the centre of the circular path. It is given by:

ac=v2ra_c = \frac{v^2}{r}

We can also express centripetal acceleration in terms of angular velocity:

ac = ω2ra_c = ω^2r

Practice Question 3

A car travels around a circular track with a radius of 50 m at a constant speed of 20 m/s. Calculate its centripetal acceleration.

ac=v2r=20250=8 m/s2a_c = \frac{v^2}{r} \\= \frac{20^2}{50} \\= 8 \text{ m/s}^2

What is Centripetal Force?

Centripetal force (Fc)(F_c) is the force required to keep an object in circular motion. It is given by:

Fc=mv2rF_c = \frac{mv^2}{r}

Where mm is the mass of the object.

We can also express centripetal force in terms of angular velocity:

Fc = mω2rF_c = mω^2r

Practice Question 4

A 0.5 kg object is attached to a string and swung in a horizontal circle with a radius of 1 m. If the object makes 2 revolutions per second, calculate the centripetal force.

First, let’s calculate the angular velocity:

ω=2πT=2π×2=4π rad/s\omega = \frac{2\pi}{T} \\= 2\pi \times 2 \\= 4\pi \text{ rad/s}

Now we can use the equation Fc = mω2rF_c = mω^2r:

Fc = 0.5 × (4π)2 × 1 ≈ 78.96NF_c = 0.5 × (4π)^2 × 1 \\≈ 78.96 N

Problem-Solving Strategies for Circular Motion

When approaching problems involving uniform circular motion, consider the following steps:

  • Identify the given quantities and the quantity you need to find.
  • Choose the appropriate equation(s) that relate the known and unknown quantities.
  • Substitute the known values into the equation(s).
  • Solve for the unknown quantity.
  • Check that your answer has the correct units and makes physical sense.

Practice Question 5

A satellite orbits Earth at an altitude of 200 km above the surface. The radius of Earth is 6,370 km. If the satellite’s orbital period is 90 minutes, calculate:

  • The satellite’s tangential velocity
  • Its centripetal acceleration
  • The centripetal force acting on the satellite if its mass is 1,000 kg
  • First, calculate the radius of the orbit:
r = 6, 370+ 200 = 6, 570km = 6, 570, 000mr = 6, 370+ 200 \\= 6, 570km \\= 6, 570, 000 m

Now, calculate the tangential velocity:

v=2πrT=2π×6,570,00090×60≈7,644 m/sv = \frac{2\pi r}{T} \\= \frac{2\pi \times 6,570,000}{90 \times 60} \\\approx 7,644 \text{ m/s}
  • Calculate the centripetal acceleration:
ac=v2r=7,64426,570,000≈8.89 m/s2a_c = \frac{v^2}{r} \\= \frac{7,644^2}{6,570,000} \\\approx 8.89 \text{ m/s}^2
  • Calculate the centripetal force:
Fc=mv2r=1,000×7,64426,570,000≈8,893 NF_c = \frac{mv^2}{r} \\= \frac{1,000 \times 7,644^2}{6,570,000} \\\approx 8,893 \text{ N}

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