# relationship between circumference and revolution

i would like to clarify two things by this problem:first what is relationship between circumference and revolution and also revolution and distant traveled by round object.let us consider following problem:

A tire on a car rotates at $500$ RPM (revolutions per minute) when the car is traveling at $50$km/hr (kilometers per hour). What is the circumference of the tire, in meters? there is also picture

as i know circumference represents as a total distance around outside,so for our case first of all let us convert from hour to minute,$50$ km is $50000 meter$,so per minute it is equal to $50000/60$ meter,but now in which revolution fact will help us?from this link

http://www.algebra.com/algebra/homework/Trigonometry-basics/Trigonometry-basics.faq.question.110618.html

it seems that distance in this case is equal circumference multiplied number of revolution,so in our case it would be $50000/60$ divided by number of revlution,but how could i clarify that distance in this case is exactly $50000/60$ and not other term?do you see i am confused in terms of relationship between revolution,circumference and distance

• In one revolution of the tire, the car travels forward one circumference. Think of taking the piece of road covered in one revolution, and wrapping it around the tire. Commented Jul 23, 2013 at 8:43
• Forget the numbers for now. How far does a wheel travel in one revolution? Commented Jul 23, 2013 at 8:44
• it is confusion for me,that is why i have asked it Commented Jul 23, 2013 at 9:14

For every $1 \text{ revolution}$, the tire will travel a distance equal to its circumference. Hence, we should look for a quantity whose units are $\text{metres}/\text{revolution}$. Using conversion factors, we obtain: $$\left(\dfrac{50 \text{ km}}{1 \text{ hr}}\right) \left(\dfrac{1000 \text{ m}}{1 \text{ km}}\right) \left(\dfrac{1\text{ hr}}{60 \text{ min}}\right) \left(\dfrac{1\text{ min}}{500 \text{ revolutions}}\right) = \dfrac{5 \text{ m}}{3 \text{ revolutions}} = \dfrac{1.\overline{6} \text{ metres}}{1\text{ revolution}}$$

So the circumference is: $$(1\text{ revolution}) \left(\dfrac{1.\overline{6} \text{ metres}}{1\text{ revolution}}\right) = 1.\overline{6} \text{ metres}$$

• could you make a little detailed answer Commented Jul 23, 2013 at 9:20
• You found that the tire travels a distance of $50000 \text{ m}$ for every $60 \text{ min}$. We also know that the tire completes $500 \text{ revolutions}$ for every $1 \text{ min}$, which means that the tire completes $30000 \text{ revolutions}$ for every $60 \text{ min}$. Hence, the tire travels $50000 \text{ m}$ for every $30000 \text{ revolutions}$. Dividing these two numbers yields $1.\overline{6} \text{ metres}$ per revolution. Commented Jul 23, 2013 at 10:42

You don't need to get confused...just remember the definition of circumference and you will solve this.In case of a circle ,let me tell you the definition in simplest terms:
"if u cut a circle anywhere at its border,and stretch the border line to be a straight line,the distance that you will get will be your circumference and that is equal to 2πr".
Now we know that circumference is equal to one complete revolution,so as per 500RPM it takes 1revolution in 3/25 seconds.
Now the speed is given as 50kmph.So calculate the distance travelled in 3/25sec which will be equal to 20/12mts=1.6mts

• This old question already had an accepted answer. You have not provided anything new Commented Oct 7, 2016 at 4:35