c) A penny sits 12 cm from the center of a record turning at 33 1/3 rpm. What is the tangential speed in m/s?
d) A mass spins with a tangential speed of 1.2 m/s on the end of a string which is 0.75m long. What is the period of the rotation in seconds?
These problems are not hard. You need to show your work, or evidence of thought about the problem.
Start (c) by converting rpm to radians per second
I was not taught how to do this
We went over this in my physics class yesturday, and I understand most of the concepts of this section of the book, but the examples we were given confuse me. I can find no connection between the numbers of tangential speed and RPM. :(
To find the tangential speed of a point on a rotating object, you need to know the distance of the point from the center of rotation and the rotation speed of the object. The tangential speed is the linear speed at which the object moves along a circular path.
For question c), you are given that the penny sits 12 cm (or 0.12 m) from the center of the record and the record is turning at 33 1/3 rpm (or 33.33 revolutions per minute). To find the tangential speed in m/s, you can use the following formula:
Tangential speed (v) = 2πr/T
Where:
- v is the tangential speed
- r is the distance from the center (0.12 m in this case)
- T is the period, which is the time taken for one complete revolution (expressed in seconds)
To find T, you can convert the rotation speed from rpm to seconds by dividing by 60:
33 1/3 rpm = (33 + 1/3) revolutions/minute = 33.33/60 revolutions/second
Now you can substitute the values into the formula:
v = 2π(0.12 m) / (33.33/60 s)
Simplifying:
v = 0.24π m / (33.33/60 s)
v ≈ 0.0360 m/s (rounded to four decimal places)
Therefore, the tangential speed of the penny is approximately 0.0360 m/s.
Moving on to question d), you are given the tangential speed of 1.2 m/s and the length of the string as 0.75 m. To find the period of rotation in seconds, you can use the following formula:
Tangential speed (v) = (2πr) / T
Rearranging the formula to solve for T:
T = (2πr) / v
Substituting the known values:
T = (2π * 0.75 m) / 1.2 m/s
Simplifying:
T = 4.71239 s (rounded to five decimal places)
Therefore, the period of rotation is approximately 4.71239 seconds.