A wheel mounted on a stationary axle starts at rest and is given the following angular acceleration: 9-12t where t is the time after the wheel begins to rotate. Find the number of revolutions that the wheel turns before it stops ( and begins to turn in the opposite direction).
Why did the wheel refuse to continue rotating? Because it realized that it wasn't getting anywhere!
To find the number of revolutions that the wheel turns before it stops and begins to turn in the opposite direction, we need to determine the final time at which the angular acceleration is zero.
Given: Angular acceleration, α = 9 - 12t
To find the time when the angular acceleration is zero, we equate α to zero and solve for t:
0 = 9 - 12t
12t = 9
t = 9/12
t = 3/4
Now, to determine the number of revolutions, we need to find the change in the angle of rotation for the wheel from t = 0 to t = 3/4.
The angular displacement, Δθ, is given by the formula:
Δθ = ω₀t + 1/2αt²
where ω₀ is the initial angular velocity, which is zero.
Since ω₀ = 0, the equation simplifies to:
Δθ = 1/2αt²
Substituting the values:
Δθ = 1/2(9 - 12t)(3/4)²
Δθ = 1/2(9 - 12 × 3/4)(9/16)
Δθ = 1/2(9 - 9/4)(9/16)
Δθ = 1/2(36/4 - 9/4)(9/16)
Δθ = 1/2(27/4)(9/16)
Δθ = 27/8 × 9/16
Simplifying, we get:
Δθ = 243/128
Since one revolution is equal to 2π radians, we can find the number of revolutions by dividing Δθ by 2π:
Number of revolutions = (243/128) / (2π)
Calculating this value will give the number of revolutions the wheel turns before stopping and changing direction.
To find the number of revolutions that the wheel turns before it stops, we need to find the time at which the angular velocity becomes zero.
The angular acceleration is given as:
α = 9 - 12t
We know that angular acceleration is the derivative of angular velocity with respect to time:
α = dω/dt
Integrating both sides of the equation with respect to t will give us the equation for angular velocity:
∫α dt = ∫dω
∫(9 - 12t) dt = ∫dω
Integrating 9 gives us 9t, and integrating -12t gives us -6t^2:
9t - 6t^2 = ω + C
Where C is a constant of integration.
Now we can determine the value of C. Since the wheel starts from rest, the initial angular velocity (ω_initial) would be 0. So we substitute ω=0 and t=0 into the equation:
0 = 9(0) - 6(0)^2 + C
C = 0
Thus, the equation for angular velocity becomes:
9t - 6t^2 = ω
To find the time when the wheel stops, we need to solve for t when ω becomes zero:
9t - 6t^2 = 0
Factor out t:
t(9 - 6t) = 0
Setting each factor equal to zero gives us two possible solutions:
t = 0 (the initial time when the wheel starts rotating)
9 - 6t = 0
Solving the second equation for t gives us:
9 - 6t = 0
6t = 9
t = 1.5 seconds
So, the wheel stops at t = 1.5 seconds.
Now we can calculate the number of revolutions the wheel turns before it stops. The number of revolutions is related to the total angle turned by the wheel.
The angle (θ) turned by the wheel is given by the equation:
θ = ∫ω dt
Integrating the equation for ω:
θ = ∫(9t - 6t^2) dt
θ = 4.5t^2 - 2t^3/3 + K
Where K is a constant of integration.
Since the wheel starts from rest, the initial angle turned (θ_initial) would be 0. So we substitute θ=0 and t=0 into the equation:
0 = 4.5(0)^2 - 2(0)^3/3 + K
K = 0
Thus, the equation for the angle turned becomes:
θ = 4.5t^2 - 2t^3/3
To find the number of revolutions, we need to find the value of θ at t = 1.5 seconds (when the wheel stops):
θ = 4.5(1.5)^2 - 2(1.5)^3/3
θ = 9
Therefore, the wheel turns 9 revolutions before it stops and begins to turn in the opposite direction.
dv/dt = 9 - 12 t and v(0)= 0
v = 9 t - 6 t^2 + constant
when t = 0, v = 0 so constant = 0
v is radians/sec
v = 9 t - 6 t^2
at v = 0
0 = t(9-6t)
so v = 0 at t = 0 as we knew
and
v = 0 at t=9/6 = 3/2 = 1.5
so how far did it go?
x is radians
x = (9/2) t^2 - (6/3)t^3 + 0
x = (9/2)(9/4) - (6/3)(9/4)(3/2)
= 81/8 - 54/12
= (243 - 108)/24
=135/24 radians
check my arithmetic
divide by 2 pi to get revolutions