A rotating wheel requires 9.00 s to rotate 29.0 revolutions. Its angular velocity at the end of the 9.00 s interval is 98.0 rad/s. What is the constant angular acceleration of the wheel in radians/second squared? (Do not assume that the wheel starts at rest.)
displacement=Wi*t+1/2 alpha*t^2
where alpha is the angular acceleration in rev/sec
you know:
29=wi*9+1/2 alpha*81
and you know
wf=98/2PI rev/sec=wi+alpha*9
wi= 98/2PI-alpha*9
putting that in to the first equation
29=(98/2PI-alpha*9)*9 +1/2 alpha*81
solve for alpha. Now this alpha will be in rev/s^2, in radians/sec^2, divide rev by 2PI to get that.
To find the constant angular acceleration of the wheel, we can use the formula:
angular acceleration = (angular velocity - initial angular velocity) / time
Given data:
Initial angular velocity (ωi) = 0 (since the question does not mention the initial angular velocity, we assume it starts at rest)
Final angular velocity (ωf) = 98.0 rad/s
Time (t) = 9.00 s
Therefore, the formula becomes:
angular acceleration = (98.0 rad/s - 0) / 9.00 s
Simplifying the equation:
angular acceleration = 98.0 rad/s / 9.00 s
angular acceleration = 10.89 rad/s^2
Therefore, the constant angular acceleration of the wheel is 10.89 radians/second squared.
To find the constant angular acceleration of the wheel, we can use the formulas of angular velocity and angular acceleration.
The formula for angular velocity is:
ω = ω0 + αt
where ω is the final angular velocity, ω0 is the initial angular velocity, α is the angular acceleration, and t is the time interval.
In this problem, we are given that the final angular velocity ω is 98.0 rad/s and the time interval t is 9.00 s. We need to find the angular acceleration α.
We are also given that the wheel completes 29.0 revolutions in the 9.00 s interval. To convert this to radians, we multiply by 2π radians per revolution:
29.0 revolutions * 2π radians/revolution = 58π radians
We can use the kinematic equation for rotational motion:
θ = ω0t + 1/2 αt^2
where θ is the angular displacement.
In this problem, the angular displacement θ is 58π radians.
Substituting the known values into the equation:
58π radians = ω0 * 9.00 s + 1/2 * α * (9.00 s)^2
Simplifying, we have:
58π = 9.00ω0 + 40.5α
We also know that the final angular velocity ω is related to the initial angular velocity ω0 and the angular acceleration α:
ω = ω0 + αt
Substituting the given values:
98.0 rad/s = ω0 + α * 9.00 s
From this equation, we can solve for ω0:
ω0 = 98.0 rad/s - α * 9.00 s
Now, we can substitute this expression for ω0 into the first equation:
58π = 9.00(98.0 rad/s - α * 9.00 s) + 40.5α
Simplifying further:
58π = 882 rad/s - 81α + 40.5α
Combining like terms:
-81α + 40.5α = 58π - 882
-40.5α = 58π - 882
Divide both sides by -40.5:
α = (58π - 882) / -40.5
Calculating this value:
α ≈ -5.27 rad/s² (to three significant figures)
Therefore, the constant angular acceleration of the wheel is approximately -5.27 rad/s².