a car wheel turning at 0.5 is brought to rest by the brakes in exactly two revolutions. what is the angular acceleration of the wheel?
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To find the angular acceleration of the wheel, we can use the equation:
θ = ω_i * t + (1/2) * α * t^2
Where:
θ is the angular displacement (in radians),
ω_i is the initial angular velocity (in radians per second),
t is the time interval (in seconds), and
α is the angular acceleration (in radians per second squared).
Given that the initial angular velocity ω_i is 0.5 revolutions per second and the wheel is brought to rest after exactly two revolutions, we need to convert the initial angular velocity from revolutions per second to radians per second. Since 1 revolution is equal to 2π radians, we have:
ω_i = (0.5 revolutions/second) * (2π radians/revolution)
= π radians/second
Next, we need to find the angular displacement θ when the wheel stops. Since the wheel stops after exactly two revolutions, the angular displacement is:
θ = (2 revolutions) * (2π radians/revolution)
= 4π radians
Now, we can plug in the values we know into the equation:
4π = (π) * t + (1/2) * α * t^2
Simplifying the equation:
4 = t + (1/2) * α * t^2
Since we are given that the wheel stops in exactly two revolutions, we know that the time interval t is equal to the time taken for two revolutions. Therefore, t = 2 revolutions / ω_i = 2 seconds. Plugging this value into the equation:
4 = 2 + (1/2) * α * (2^2)
Simplifying further:
4 = 2 + 2α
Rearranging the equation to solve for α, the angular acceleration:
2α = 4 - 2
2α = 2
α = 1 radian/second^2
Therefore, the angular acceleration of the wheel is 1 radian per second squared.