You drive down the road at 39 m/s (88 mi/h) in a car whose tires have a radius of 34 cm.
What is the period of rotation of the tires?
2pi*34cm/39= 5.4776 rev/ sec is this right or has to be sec/rev
Circumference = 2pi*r = 6.28 * 0.34m = 2.14m/rev.
Period = 2.14m/rev/(39m/s) =
NOTE: you cannot use cm AND meters. Use 0.34m instead of 34cm to make your units compatible.
To find the period of rotation of the tires, we need to determine the time it takes for one full revolution.
The formula to calculate the period of rotation is:
Period = 1 / frequency
Since the frequency is the number of revolutions per unit of time, in this case, it is measured in revolutions per second (rev/s).
In the given scenario, we have determined the frequency to be 5.4776 rev/s. Therefore, the period of rotation can be calculated as:
Period = 1 / 5.4776 rev/s
Hence, the period of rotation of the tires is approximately 0.1823 seconds per revolution or sec/rev.
To find the period of rotation of the tires, you need to know the angular speed of the tires, which is given in revolutions per second (rev/sec).
You have correctly calculated the angular speed as 5.4776 rev/sec using the formula (2*pi*r) / v, where r is the radius of the tires and v is the velocity of the car.
Now, to convert the angular speed to the period of rotation, you need to use the reciprocal of the angular speed.
1 rev/sec is equal to 1/rev/sec (or sec/rev). So, to convert from rev/sec to sec/rev, you can take the reciprocal of the angular speed.
Reciprocal of 5.4776 rev/sec is 1 / 5.4776 sec/rev, which is approximately 0.1823 sec/rev.
Therefore, the period of rotation of the tires is approximately 0.1823 seconds per revolution (sec/rev).