# Rotation, Angular and Linear Speed

Questions related to angular and linear speeds of rotating objects are presented. The solutions and answers are also provided.

### Question 1

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

Solution to Question 1:

• For every one rotation of the wheel, the bicycle moves a distance equal to the circumference of the wheel. The circumference C of the wheel is given by
C = 40 Pi cm
• The number of rotations N of the wheel is obtained by dividing the total distance traveled, 100 m = 10000 cm, by the circumference.
N = 10000 cm / 40 Pi cm = 80 rotations (rounded to the nearest unit)

### 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?

Solution to Question 2:

• The circumference C of the wheel is given by

C = 60 Pi cm

• For each rotation of the wheel, the car travel a distance equal to the circumference of the wheel. 100 rotations correspond a distance d traveled by the car where d is given by

d = 100 * 60 Pi cm = 18850 cm (rounded to the nearest cm)

### Question 3

The wheel of a machine rotates at the rate of 300 rpm (rotation per minute). If the diameter of the wheel is 80 cm, what are the angular (in radian per second) and linear speed (in cm per second) of a point on the wheel?

Solution to Question 3:

• Each rotation corresponds to 2 Pi radians. Hence 300 rotations per minute correspond to an angular speed a given by

a = 300 * 2 Pi radians / minute

• We now substitute 1 minute by 60 seconds above

a = 300 * 2 Pi radians / 60 seconds

= 31.41 rad/sec (rounded to 2 decimal places)

• The linear speed is obtained by noting that one rotation corresponds to the circumference of the wheel. Hence the linear speed s is given by

s = 300 * (80 Pi) cm / minute

= 300 * 80 Pi / 60 cm/sec

= 1257 cm/sec

### Question 4

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

Solution to Question 4:

• One rotation every 24 hours (or 24 *3600 seconds) gives an angular speed a equal to

a = 2 Pi / (24*3600) = 0.0000727 rad/sec

• We first convert the radius R in feet

R = 4000 * 5280 = 21,120,000 feet

• One rotation every 24 hours (or 24 *3600 seconds) gives a linear speed s equal to

s = 2 Pi R / (24*3600)

= 2 * Pi * 21,120,000 / 86,400

= 1,536 feet / sec