1. ## [SOLVED] linear/angular speed?

If a rocket car with 20 inch radius tires brakes the sound barrier (a speed in excess of 760 miles/hour), what is the angular speed of the wheels in revolutions/second?

ok i am totally lost.....all i know is one revolution would probably be 40pi inches because that the circumfrence of the circle...and i thought i probably have to use the formula v=rw (v=linear speed, r=radius and w=angular speed)

so i started with

760mi/hr = 20w but im confused with the units??? i dont know if i even started it off right ???

thank you(:

2. ## Linear/angular speed

Hello desperate_on_sunday_night
Originally Posted by desperate_on_sunday_night
If a rocket car with 20 inch radius tires brakes the sound barrier (a speed in excess of 760 miles/hour), what is the angular speed of the wheels in revolutions/second?

ok i am totally lost.....all i know is one revolution would probably be 40pi inches because that the circumfrence of the circle...and i thought i probably have to use the formula v=rw (v=linear speed, r=radius and w=angular speed)

so i started with

760mi/hr = 20w but im confused with the units??? i dont know if i even started it off right ???

thank you(:
Use $v = r\omega$ if $\omega$ is measured in radians. Here, stick to your first idea, and work with the circumference of the tire, which is $40\pi$ inches.

So, in 1 revolution, the car moves forward $40\pi$ inches $= 125.68$ inches.

And we know that in one hour, the car would move forward $760$ miles. So in one second it moves forward $760 \div 3600$ miles, or $760 \times 5280 \times 12 \div 3600$ inches, which $= 13376$ inches.

So, how many revolutions do we need to make this many inches? Answer: $13376 \div 125.68 = 106.5$. So there's your answer: just over $106$ revolutions per second.