a wheel turning with angular speed of 30 rps is brought to rest with constant acceleration. It turns 60 revolutions before it stops.
To find the acceleration of the wheel, we can use the formula:
Angular speed (ω) = Initial angular speed (ω₀) + (angular acceleration (α) × time (t))
Given:
Initial angular speed (ω₀) = 30 rps (revolutions per second)
Final angular speed (ω) = 0 rps (since the wheel comes to rest)
Revolutions (n) = 60
We need to find the angular acceleration (α) and the time taken (t) to stop the wheel.
Step 1: Convert revolutions to radians
Since angular measurements are usually in radians, we need to convert the number of revolutions (n) into radians. We know that 1 revolution is equal to 2π radians.
Angle in radians = 60 revolutions × 2π radians/revolution
Angle in radians = 120π radians
Step 2: Find the time taken (t) to stop the wheel
We can use the formula:
ω = ω₀ + α × t
Since the final angular speed (ω) is 0, we substitute the values:
0 = 30 + α × t
Step 3: Find the angular acceleration (α)
We can determine the angular acceleration (α) by rearranging the equation:
α = -ω₀ / t
Step 4: Substitute the values into the equation
Substituting the given values, we have:
α = -30 / t
Step 5: Find the value of time (t)
To find the value of time (t), we can solve the equation:
0 = 30 + α × t
Substituting α = -30 / t, we get:
0 = 30 - (30 / t) × t
Simplifying further:
0 = 30 - 30
0 = 0
Since the value of time (t) is canceled out, it means the equation holds true regardless of the value of t. Therefore, the time is indeterminate.
To summarize: The angular acceleration (α) is -30 radians per second squared, but the time taken (t) cannot be determined based on the given information.
To find the time taken for the wheel to come to a stop, we can use the formula:
ω = ω₀ + αt,
where:
ω = final angular speed (0 rps, because the wheel comes to a stop),
ω₀ = initial angular speed (30 rps),
α = angular acceleration,
t = time.
We need to find α, which is the constant angular acceleration.
Since the number of revolutions a wheel turns can be calculated using the formula:
θ = ω₀t + 0.5αt²,
where:
θ = number of revolutions (60 revolutions),
ω₀ = initial angular speed (30 rps),
t = time,
α = angular acceleration.
We are given that θ = 60 revolutions and ω₀ = 30 rps. Substituting these values into the formula, we have:
60 = 30t + 0.5αt². ----(equation 1)
We also know that the final angular speed is 0 rps, so:
0 = 30 + αt. ----(equation 2)
We have a system of two equations (equation 1 and equation 2) with two unknowns (t and α). We can solve this system of equations to find the values of t and α.
From equation 2, we can solve for t:
t = -30/α. ----(equation 3)
Substituting equation 3 into equation 1, we can solve for α:
60 = 30(-30/α) + 0.5α(-30/α)².
Simplifying the equation further:
60 = (-900/α) + 0.5(-30)².
60 = (-900/α) + 0.5(900/α²).
Multiplying both sides of the equation by α² to eliminate the denominators:
60α² = -900α + 0.5(900).
Multiplying further and simplifying:
60α² = -900α + 450.
Rearranging the equation:
60α² + 900α - 450 = 0.
Now, we have a quadratic equation. Solving for α using the quadratic formula:
α = (-b ± √(b² - 4ac)) / (2a),
where a = 60, b = 900, and c = -450.
α = (-900 ± √(900² - 4(60)(-450))) / (2(60)).
Simplifying further:
α = (-900 ± √(810000 + 1080000)) / 120.
α = (-900 ± √1890000) / 120.
α = (-900 ± 1374.77) / 120.
α ≈ 9.04 or α ≈ -24.37.
Since acceleration cannot be negative in this context, we take α ≈ 9.04 rps².
Now, substituting this value of α into equation 2:
0 = 30 + (9.04)t.
Solving for t:
t = -30/9.04.
t ≈ -3.31 seconds.
Since we cannot have a negative time, we disregard the negative value. Hence, the time taken for the wheel to come to a stop is approximately 3.31 seconds.