A bug rides on a turntable which is rotating at 3 rad/s. The moto is turned off, and the bug covers 36 radians before coming to rest.
a. What is the angular acceleration of the bug?
b. How long does it take the bug to stop?
To find the angular acceleration of the bug, we can use the formula:
Angular acceleration (α) = Change in angular velocity (Δω) / Change in time (Δt)
However, since the bug covers 36 radians before coming to rest, it means that the initial angular velocity (ωi) is 3 rad/s, and the final angular velocity (ωf) is 0 rad/s.
a. Finding the angular acceleration:
Δω = ωf - ωi
= 0 - 3
= -3 rad/s
Since the bug is decelerating, the change in angular velocity is negative.
Now, to find the change in time (Δt), we can use the formula:
θ = ωi * Δt + (1/2) * α * Δt^2
Since the bug comes to rest and the final angular velocity is zero (ωf = 0), we have:
θ = ωi * Δt + (1/2) * α * Δt^2
36 = 3 * Δt + (1/2) * (-3) * Δt^2
Rearranging the equation, we have:
36 = 3Δt - (3/2)Δt^2
Multiplying everything by 2 to eliminate fractions, we get:
72 = 6Δt - 3Δt^2
Rearranging further, we have:
3Δt^2 - 6Δt + 72 = 0
Now we can solve this quadratic equation using the quadratic formula:
Δt = (-b ± √(b^2 - 4ac)) / 2a
Substituting the values a = 3, b = -6, and c = 72 into the quadratic formula, we get:
Δt = (-(-6) ± √((-6)^2 - 4 * 3 * 72)) / (2 * 3)
Δt = (6 ± √(36 - 864)) / 6
Δt = (6 ± √(-828)) / 6
Since the square root of a negative number is not real, it means that there are no real solutions to this equation. Therefore, the bug does not come to rest and there is no change in time (Δt).
b. Since the bug does not stop, the bug does not take any time to stop.
To solve this problem, we can use the equations of rotational motion. There are two key equations we can use:
1. Angular displacement (θ) = Initial angular velocity (ω) × Time (t) + 0.5 × Angular acceleration (α) × Time squared (t^2)
2. Final angular velocity (ω_f) = Initial angular velocity (ω) + Angular acceleration (α) × Time (t)
a. To find the angular acceleration, we need to rearrange Equation 1 and solve for α:
θ = ω * t + 0.5 * α * t^2
Rearranging, we get:
α = (θ - ω * t) / (0.5 * t^2)
Given values:
θ = 36 radians
ω = 3 rad/s
t is unknown
We can solve for t by using Equation 2:
At rest, the final angular velocity ω_f is 0 rad/s:
ω_f = ω + α * t
0 = 3 + α * t
Now we have two equations with two unknowns (α and t). We can substitute the value of α from the first equation into the second equation:
0 = 3 + [(θ - ω * t) / (0.5 * t^2)] * t
Simplifying:
0 = 3 + (2 * (θ - ω * t)) / t
Multiplying through by t:
0 = 3t + 2(θ - ω * t)
Expanding:
0 = 3t + 2θ - 2ω * t
Rearranging:
2θ = (2ω - 3) * t
Solving for t:
t = (2θ) / (2ω - 3)
Now we can substitute this value of t back into Equation 1 to find α:
α = (θ - ω * t) / (0.5 * t^2)
b. To find the time it takes for the bug to stop, we substitute the value of α back into Equation 2:
0 = ω + α * t
0 = 3 + α * t
Rearranging:
t = -3 / α
Now we can substitute the value of α we found in part a into this equation to find the time it takes for the bug to stop.
a)2 (alpha) 36 = 3^2
b) t = 3 /(alpha)