# Math Help - Spin Balancing Tires

1. ## Spin Balancing Tires

A spin balancer rotates the wheel of a car at 480 revolutions per minute. If the diamteer of the wheel is 26 inches, what road speed is being tested? Express your answer in miles per hour. At how many revolutions per minute should the balancer be set to test a road speed of 80 miles per hour?

2. Originally Posted by magentarita
A spin balancer rotates the wheel of a car at 480 revolutions per minute. If the diamteer of the wheel is 26 inches, what road speed is being tested? Express your answer in miles per hour.

If an object moves along a circular path of radius r units, then its linear velocity, v, is given by:

$v=r\frac{\theta}{t}$

where $\frac{\theta}{t}$ represents the angular velocity in radians per unit of time.

480 revolutions = $480 \cdot 2\pi$ radians per minute or $960 \pi$ radians per minute.

$r=13in$

Here it is in one fell swoop:

$13in \times \frac{1ft}{12in} \times \frac{1mi}{5280ft} \times \frac{480rev}{1min} \times \frac{2\pi}{1rev} \times \frac{60min}{1h}\approx37.13mph$

And here it is all broken out:

$v=13in \cdot \frac{440 \cdot 2\pi \ \ rad}{1 \ \ min}=39207.0763 \ \ in/min$

This translates to $39207.0763 \times 60min = 2352424.579\ \ in/hr$.

$1\ \ mi = 5280 \ \ ft = 63360 \ \ in$

$2352424.579 \ \ in/hr \div 63360\ \ in/mile\approx 37.13\ \ mph$
Originally Posted by magentarita
At how many revolutions per minute should the balancer be set to test a road speed of 80 miles per hour?
Find the circumference of the wheel

$C = \pi D=26\pi in$

$\frac{80mi}{1hr}\times\frac{5280ft}{1mi}\times\fra c{12in}{1ft}\times\frac{1rev}{26\pi in}\times\frac{1hr}{60min}\approx1034.26$ rev/min

The miles, inches, feet, and hours cancel leaving you with revolutions over minutes

3. ## ok....

Originally Posted by masters
If an object moves along a circular path of radius r units, then its linear velocity, v, is given by:

$v=r\frac{\theta}{t}$

where $\frac{\theta}{t}$ represents the angular velocity in radians per unit of time.

480 revolutions = $480 \cdot 2\pi$ radians per minute or $960 \pi$ radians per minute.

$r=13in$

Here it is in one fell swoop:

$13in \times \frac{1ft}{12in} \times \frac{1mi}{5280ft} \times \frac{480rev}{1min} \times \frac{2\pi}{1rev} \times \frac{60min}{1h}\approx37.13mph$

And here it is all broken out:

$v=13in \cdot \frac{440 \cdot 2\pi \ \ rad}{1 \ \ min}=39207.0763 \ \ in/min$

This translates to $39207.0763 \times 60min = 2352424.579\ \ in/hr$.

$1\ \ mi = 5280 \ \ ft = 63360 \ \ in$

$2352424.579 \ \ in/hr \div 63360\ \ in/mile\approx 37.13\ \ mph$

Find the circumference of the wheel

$C = \pi D=26\pi in$

$\frac{80mi}{1hr}\times\frac{5280ft}{1mi}\times\fra c{12in}{1ft}\times\frac{1rev}{26\pi in}\times\frac{1hr}{60min}\approx1034.26$ rev/min

The miles, inches, feet, and hours cancel leaving you with revolutions over minutes
I knew this question had some work to it.