a wheel of radius R rotates making f revolutions per second. the quantity f is know as the frecuency of the motion. show tha 1/f is the time to complete one revolution, called the period T of the motion. show that the centripetal acceleration of a point at the rim of the wheel is a=4(3.14)^2Rf^2
To show that 1/f is the time to complete one revolution, we can start by converting the frequency f from revolutions per second to seconds per revolution. Since frequency is the number of revolutions per second, the time taken for one revolution is the inverse of the frequency.
Let's denote the time taken for one revolution as T. We have:
f revolutions / 1 second = 1 revolution / T seconds
Cross multiplying, we get:
f * T = 1
Dividing both sides by f, we have:
T = 1/f
Hence, we have shown that 1/f is the time to complete one revolution, which is referred to as the period T of the motion.
Now let's derive the formula for the centripetal acceleration of a point at the rim of the wheel. The centripetal acceleration is given by the equation:
a = (v^2) / r
where v is the linear velocity and r is the radius.
In this case, the point at the rim of the wheel travels in a circular path, completing one revolution in the time T. The linear velocity v of a point at the rim of the wheel is equal to the circumference of the wheel divided by the time period T.
Circumference of the wheel = 2 * π * R (where R is the radius)
Linear velocity v = (2 * π * R) / T
Substituting this value of v into the centripetal acceleration equation, we get:
a = ((2 * π * R) / T)^2 / R
Simplifying further, we have:
a = (4 * (π^2) * (R^2)) / (T^2) [since (ab)^2 = a^2 * b^2]
Since we have already shown that T = 1/f, we can substitute this value in:
a = (4 * (π^2) * (R^2)) / ((1/f)^2)
Simplifying further:
a = (4 * (π^2) * (R^2)) * (f^2)
Hence, we have shown that the centripetal acceleration of a point at the rim of the wheel is equal to a = 4 * (π^2) * R * f^2.