Rotation, Angular and Linear Speed

Here we present questions related to the angular and linear speeds of rotating objects. Each solution is explained step by step so learners can follow the reasoning.

Questions with Answers

Question 1

A bicycle traveled a distance of 100 meters. The diameter of the wheel is 40 cm. Find the number of rotations of the wheel.

• Each rotation of the wheel moves the bicycle a distance equal to the circumference of the wheel.
  Circumference of the wheel:
  \[ C = \pi \times \text{diameter} = \pi \times 40 \text{ cm} = 40 \pi \text{ cm} \]
• The total distance traveled by the bicycle is 100 meters, which is 10000 cm. The number of rotations is obtained by dividing the distance by the circumference:
  \[ N = \frac{10000}{40 \pi} \approx 80 \text{ rotations} \]

Question 2

The wheel of a car made 100 rotations. What distance has the car traveled if the diameter of the wheel is 60 cm?

• The circumference of the wheel is:
  \[ C = \pi \times 60 \text{ cm} = 60 \pi \text{ cm} \]
• The total distance \(d\) traveled is:
  \[ d = \text{number of rotations} \times C = 100 \times 60 \pi \text{ cm} \approx 18850 \text{ cm} \]
• This shows that for each rotation, the wheel moves the car by its circumference, and multiplying by the number of rotations gives the total distance.

Question 3

The wheel of a machine rotates at 300 rpm (rotations per minute). If the diameter of the wheel is 80 cm, find the angular speed (in rad/s) and the linear speed (in cm/s) of a point on the wheel.

• Each rotation corresponds to \(2 \pi\) radians. Therefore, 300 rotations per minute correspond to an angular speed:
  \[ \omega = 300 \times 2\pi \text{ radians per minute} \]
• Convert minutes to seconds (1 minute = 60 seconds):
  \[ \omega = \frac{300 \times 2\pi}{60} = 10 \pi \approx 31.42 \text{ rad/s} \]
• The linear speed \(v\) is the distance traveled per unit time by a point on the edge of the wheel. One rotation corresponds to the circumference:
  \[ v = \frac{300 \times 80 \pi}{60} \approx 1257 \text{ cm/s} \]
• Here we see that linear speed and angular speed are related by \(v = r \omega\), where \(r = 40 \text{ cm}\) is the radius of the wheel.

Question 4

The Earth rotates about its axis once every 24 hours. The radius of the equator is approximately 4000 miles. Find the angular speed (rad/s) and linear speed (ft/s) of a point on the equator.

• One rotation every 24 hours (or 24 × 3600 seconds) gives an angular speed:
  \[ \omega = \frac{2\pi}{24 \times 3600} \approx 7.27 \times 10^{-5} \text{ rad/s} \]
• Convert the radius to feet:
  \[ R = 4000 \times 5280 = 21,120,000 \text{ ft} \]
• The linear speed is the circumference divided by the rotation time:
  \[ v = \frac{2 \pi R}{24 \times 3600} \approx 1,536 \text{ ft/s} \]
• This calculation shows how the Earth's rotation speed at the equator can be derived from its radius and rotation period.

More References

Trigonometry Problems with Answers