Return to the Purplemath home page

 The Purplemath Forums
Helping students gain understanding
and self-confidence in algebra

powered by FreeFind


Return to the Lessons Index  | Do the Lessons in Order  |  Get "Purplemath on CD" for offline use  |  Print-friendly page

Sectors, Areas, and Arcs: Worked Examples (page 2 of 2)

  • A wheel has a diameter of 100 cm. If the wheel is supporting a craft moving at 45 kilometers per hour, then what is the angular velocity of the wheel, to the nearest whole number of revolutions per minute?
  • The "angular velocity" is the number of times the wheel revolves in some named time period. So this question is asking me to find the number of times the wheel twirls around in one minute. To do this, I'll need to find the distance covered (per minute) when moving at 45 kph. Then I'll need to find the circumference of the wheel, and divide the total per-minute distance by this "once around" distance. The number of circumferences which fit inside the total distance is the number of times the wheel revolves in that time period.

    First, I'll find the distance, using what I've learned about converting units:

      conversion from kilometers-per-hour to centimeters-per-minute

    So the distance covered in one minute is 75,000 centimeters. The diameter of the wheel is 100 cm, so the radius is 50 cm, and the circumference is 100π cm. How many of these circumferences (wheel revolutions) fit inside the 75,000 cm?

      (75,000 cm)/(100pi cm) = 750 / pi = 238.7324... (approximately)

    The angular velocity is ω = 239 rpm

The abbreviation "rpm" for "revolutions per minute" is standard, so you can safely use this notation. The character "ω" is the Greek lower-case letter "omega", and is often the variable used for angular velocity.

Note: This speed isn't as fast as it might appear: it's just under four revolutions per second. You can do that on your bike without breaking a sweat.




  • A bicycle wheel has a diameter of 78 cm. If the wheel revolves at a rate of 120 revolutions per minute, what is the linear velocity of the bike, in kilometers per hour? Round your answer to one decimal place.
  • The linear velocity will be the distance, stretched out in the straight line, that a point on the wheel moves during a defined period of time. They've given me the number of times the wheel revolves each minute. A fixed point on the tire (say, a pebble in the tire's tread) moves the length of the circumference for each revolution. Unrolling this distance onto the ground, the bike will move along the ground the same distance, one circumference, for each revolution.

    So this question is asking me to find the circumference length, and then use this to find the total distance covered per minute. Since the diameter is 78 cm, then the circumference is C = 78π cm. This means that the bike moves 78π cm forward for each revolution of the tire. There are 120 such revolutions per minute, so:

      (78pi cm/rev)×(120 rev/min) = 9,360pi cm/min

    Now I need to convert this from centimeters-per-minute to kilometers-per-hour:

      9360pi cm/min = 17.64318434 kph (approximately)

    The bike is moving at about 17.6 kph.

This is about eleven miles an hour.

  • Assume that the Earth's orbit is circular, with a radius of 93,000,000 miles, and let "one year" equal 365.25 days. Under these conditions, find the linear velocity of the Earth in miles per second. Round to the nearest whole number of miles.
  • The circumference of the circle with r = 93,000,000 mi will be the linear distance that the Earth covers in one year.

      C = 2π(93,000,000 mi)/year = 186,000,000π mi/yr

    This is the distance covered, in miles, in one year. There are twenty-four hours in a day, sixty minutes in an hour, and sixty seconds in a minute, so the total number of seconds for that year is:

      (365.25 days/yr)(24 hr/day)(60 min/hr)(60 sec/min) = 31,557,600 sec/yr

    Then the linear velocity, being the total linear distance divided by the total time and expressed as a unit rate, is:

      (186,000,000π mi/yr)/(31,557,600 sec/yr) = 18.51649788... mi/sec

    The linear velocity of the Earth is about 19 miles per second.

  • A train is travelling at the rate of 10 mph on a curve of radius 3000 ft. Through what angle will the train turn in one minute? Round to the nearest whole number of degrees.
  • "A curve of radius 3000 ft" means that, if you tried to fit a circle snugly inside the curve, the best fit would be a circle with a radius of r = 3000 feet. In other words, I can use circle facts to answer this question. Copyright © Elizabeth Stapel 2010-2011 All Rights Reserved

    Since the radius of the curve is in feet and since I need to find the angle for one minute, I'll start by converting the miles-per-hour speed to feet-per-second:

      (10 mi/hr)(5280 ft/mi)(1 hr / 60 min) = 880 ft/min

    The amount of the curved track that the train covers is also a portion of the circumference of the circle. So the 880 feet is the arc length, and now I need to find the subtended angle:

      s = (theta)r, 880 = 3000(theta), (theta) = 880/3000 = 0.29333...

    But this value is in radians, and I need my answer to be in degrees, so I need to convert:

      (0.293333... radians)(180°/π radians) = 16.80676199...°

    The train turns through an angle of about 17°

  • On a certain vehicle, one windshield wiper is 60 cm long, and is afixed to a swing arm which is 72 cm long from pivot point to tip. If the swing arm turns through 105°, what area of the windshield, to the nearest square centimeter, is swept by the wiper blade?
  • They gave me the radius of a circle and a subtended angle, and want me to find the area. So I'll be needing to use the sector-area formula. However, since the wiper blade itself does not go all the way down to the pivot point for the swing arm, so I'll need to subtract out a portion of the sector to find the area that is actually covered by the blade.

    Since they gave me the angle in degrees, I'll need to be careful to adjust the formulas accordingly.

      A = (105*/2)(pi/180*)(72 cm)^2 = 1512pi cm^2

    This is the total area swept by the swing arm. The wiper blade only covers the outer 60 cm of the length of the swing arm, so the inner 72 – 60 = 12 cm is not covered by the blade.

      A = (105*/2)(pi/180*)(12 cm)^2 = 42pi cm^2

    I need to subtract this area:

      1512pi cm^2 - 42pi cm^2 = 1470pi cm^2 = 4618.141201 cm^2 (approx)

    The blade sweeps about 4618 cm2 of the windshield.

<< Previous  Top  |  1 | 2  |  Return to Index

Cite this article as:

Stapel, Elizabeth. "Sectors, Areas, and Arcs: Worked Examples." Purplemath. Available from Accessed


  Linking to this site
  Printing pages
  School licensing

Reviews of
Internet Sites:
   Free Help
   Et Cetera

The "Homework

Study Skills Survey

Tutoring from Purplemath
Find a local math tutor

This lesson may be printed out for your personal use.

Content copyright protected by Copyscape website plagiarism search

  Copyright © 2010-2012  Elizabeth Stapel   |   About   |   Terms of Use


 Feedback   |   Error?