a wheel of radius 50 centimeters accelerates uniformly from a frequency of 4 hertz to 6 hertz during 50 revolutions. what are its angular acceleration and a period of time for this frequency change? find the tangential and normal accelerations at the beginning and the end this period of time .
To find the angular acceleration and the period of time for the frequency change, we can use the formula:
Angular acceleration (α) = (ω₂ - ω₁) / t
where:
- α is the angular acceleration
- ω₂ is the final angular velocity (in radians per second)
- ω₁ is the initial angular velocity (in radians per second)
- t is the time taken for the frequency change
Since the problem mentions the frequency change in terms of hertz, we need to convert it to angular velocity in radians per second. The formula to convert frequency (f) to angular velocity (ω) is:
ω = 2πf
Now let's calculate the values:
Given:
- Radius of the wheel (r) = 50 centimeters = 0.5 meters
- Initial frequency (f₁) = 4 hertz
- Final frequency (f₂) = 6 hertz
- Number of revolutions (n) = 50
Step 1: Convert frequencies to angular velocities
ω₁ = 2πf₁ = 2π * 4 = 8π radians per second
ω₂ = 2πf₂ = 2π * 6 = 12π radians per second
Step 2: Calculate time (t) for the frequency change
Since we are given the number of revolutions, we need to find the time taken for those revolutions. The formula to find time (t) is:
t = n / f
where:
- n is the number of revolutions
- f is the frequency
t = 50 / 4 = 12.5 seconds
Step 3: Calculate the angular acceleration (α)
α = (ω₂ - ω₁) / t = (12π - 8π) / 12.5 = 4π / 12.5 = 0.32π radians per second squared
Therefore, the angular acceleration is 0.32π radians per second squared.
To find the tangential and normal accelerations at the beginning and the end of this period of time, we can use the formulas:
Tangential acceleration (at) = α * r
Normal acceleration (an) = ω² * r
Step 4: Calculate the tangential acceleration at the beginning and the end
at_beginning = α * r = 0.32π * 0.5 = 0.16π meters per second squared
at_end = α * r = 0.32π * 0.5 = 0.16π meters per second squared
Step 5: Calculate the normal acceleration at the beginning and the end
an_beginning = ω₁² * r = (8π)² * 0.5 = 32π² meters per second squared
an_end = ω₂² * r = (12π)² * 0.5 = 72π² meters per second squared
Therefore, the tangential acceleration at the beginning and the end of this period of time is 0.16π meters per second squared, and the normal acceleration at the beginning and the end is 32π² meters per second squared and 72π² meters per second squared, respectively.