#1
An electric motor rotating a workshop grinding wheel at a rate of 231 rev/min is switched
off. Assume constant angular deceleration of
magnitude 1.87 rad/s^2
.
Through how many revolutions does the
wheel turn before it finally comes to rest?
Answer in units of rev.
#2
At t = 0, a wheel rotating about a fixed axis at
a constant angular deceleration of 0.24 rad/s
2
has an angular velocity of 2.2 rad/s and an
angular position of 8.1 rad.
What is the angular position of the wheel
after 2 s?
Answer in units of rad.
#1 To find out how many revolutions the wheel turns before it comes to rest, we first need to determine the time it takes for the wheel to stop. We can use the formula:
Angular deceleration (α) = (Change in angular velocity (Δω)) / (Time taken (t))
Rearranging the formula, we can solve for t:
t = Δω / α
Given that the angular deceleration (α) is 1.87 rad/s^2 and the initial angular velocity (ω) is 231 rev/min, we need to convert the initial angular velocity to rad/s. There are 2π radians in one revolution, and 60 seconds in one minute. So,
Initial angular velocity (ω) = (231 rev/min)(2π rad/1 rev)(1 min/60 s) = 24.17 rad/s
Now we can substitute the values into our formula:
t = (0 - 24.17 rad/s) / (-1.87 rad/s^2) = 12.94 s
Next, we need to find the total number of revolutions the wheel makes during this time period. We can use the formula:
Number of revolutions (N) = (Angular velocity (ω) * Time taken (t)) / (2π radians)
Substituting the values we know:
N = (24.17 rad/s)(12.94 s) / (2π rad) = 49.09 rev
Therefore, the wheel turns approximately 49.09 revolutions before it comes to rest.
#2 To find the angular position of the wheel after 2 seconds, we can use the formula:
Angular position (θ) = Initial angular position (θ0) + (Initial angular velocity (ω0) * Time taken (t)) + (0.5 * Angular acceleration (α) * (Time taken (t))^2)
Given that the initial angular position (θ0) is 8.1 rad, the initial angular velocity (ω0) is 2.2 rad/s, the time taken (t) is 2 s, and the angular acceleration (α) is -0.24 rad/s^2 (negative deceleration):
θ = 8.1 rad + (2.2 rad/s * 2 s) + (0.5 * -0.24 rad/s^2 * (2 s)^2)
= 8.1 rad + 4.4 rad - 0.48 rad
= 12.02 rad
Therefore, the angular position of the wheel after 2 seconds is approximately 12.02 radians.
#1 To find the number of revolutions, we need to calculate the time it takes for the wheel to come to rest first.
Final angular velocity, ωf = 0 rad/s
Initial angular velocity, ωi = 231 rev/min * 2π rad/rev * 1 min/60 s = 24.18 rad/s
Angular acceleration, α = -1.87 rad/s^2 (negative because it's deceleration)
Using the equation ωf = ωi + αt, we can solve for the time, t, it takes for the wheel to come to rest:
0 = 24.18 rad/s + (-1.87 rad/s^2) * t
Rearranging the equation, we have:
1.87 t = 24.18
t = 24.18 rad/s / 1.87 rad/s^2
t = 12.94 s
Now, we can find the number of revolutions:
Number of revolutions = ωi * t / (2π) = 24.18 rad/s * 12.94 s / (2π) = 245.34 rev
Therefore, the wheel turns approximately 245.34 revolutions before it finally comes to rest.
#2 To find the angular position of the wheel after 2 seconds, we can use the equation:
θ = θi + ωi * t + 0.5 * α * t^2
Where:
θ = Final angular position
θi = Initial angular position = 8.1 rad
ωi = Initial angular velocity = 2.2 rad/s
t = Time = 2 s
α = Angular acceleration = -0.24 rad/s^2 (negative because it's deceleration)
Plugging in the values:
θ = 8.1 rad + (2.2 rad/s) * (2 s) + 0.5 * (-0.24 rad/s^2) * (2 s)^2
θ = 8.1 rad + 4.4 rad + 0.5 * (-0.24 rad/s^2) * 4 s^2
θ = 8.1 rad + 4.4 rad - 0.48 rad
θ = 12.02 rad
Therefore, the angular position of the wheel after 2 seconds is approximately 12.02 rad.