The angular velocity of a process control motor is (21−12t2) rad/s, where t is in seconds.
A)At what time does the motor reverse direction?
I got 6.5s
B) Through what angle does the motor turn between t =0 s and the instant at which it reverses direction?
Can you help with B
Sure, I'd be happy to help with part B. To find the angle through which the motor turns between t = 0s and the instant it reverses direction, we need to integrate the angular velocity with respect to time.
First, let's find the equation for the angular displacement. Integrate the angular velocity equation (21−12t^2) with respect to t:
∫(21−12t^2) dt = 21t - 4t^3/3 + C
Now, we can find the displacement between t = 0s and the time when the motor reverses direction (6.5s):
θ = 21(6.5) - 4(6.5)^3/3 - [21(0) - 4(0)^3/3]
θ ≈ 52.125 - 4(6.5)^3/3
Calculating this, we get:
θ ≈ 52.125 - 4(274.625)/3
θ ≈ 52.125 - 365.833
θ ≈ -313.708 radians
Hmm, it seems we have a negative angle. This implies that the motor has gone back in time! Maybe it's trying to help Marty McFly. But in reality, a negative angle doesn't make physical sense in this context, so we can say that the motor didn't actually reverse direction between t = 0s and 6.5s.
I hope this helped! If you have any other questions, feel free to ask.
To find the angle through which the motor turns between t = 0s and the instant at which it reverses direction, we can integrate the angular velocity equation.
The integral of the angular velocity function will give us the angle function.
∫(21 - 12t^2) dt
To integrate this, we will use the power rule of integration:
∫t^n dt = (t^(n+1))/(n+1) + C
Applying the power rule, the integral becomes:
∫(21 - 12t^2) dt = 21t - (4t^3)/3 + C
Now, we can find the angle by evaluating the integral at the endpoints of t = 0s and t = t_rev, where t_rev is the time at which the motor reverses direction.
Angle_turned = ∫(21 - 12t^2) dt from 0 to t_rev
Angle_turned = [21t - (4t^3)/3] from 0 to t_rev
Angle_turned = [21(t_rev) - (4(t_rev)^3)/3] - [21(0) - (4(0)^3)/3]
Angle_turned = 21(t_rev) - (4(t_rev)^3)/3
Therefore, the angle through which the motor turns between t = 0s and the instant it reverses direction is given by 21(t_rev) - (4(t_rev)^3)/3.
To find the angle through which the motor turns between t = 0 seconds and the instant at which it reverses direction, you need to integrate the angular velocity function with respect to time over the given interval.
Given that the angular velocity function is (21 - 12t^2) rad/s, we can integrate it to obtain the angle function.
To do this, we integrate (21 - 12t^2) with respect to t:
∫(21 - 12t^2) dt
The integral of 21 with respect to t is simply 21t, and the integral of -12t^2 with respect to t is -4t^3.
So, the integral becomes:
∫(21 - 12t^2) dt = 21t - 4t^3/3
Now, we need to evaluate the integral between the given limits of t = 0 s and the time at which the motor reverses direction.
Let's assume that the motor reverses direction at time t = x seconds. Then the angle through which the motor turns is given by:
Angle = ∫[from 0 to x] (21 - 12t^2) dt = [21t - 4t^3/3] (from 0 to x)
Plugging in the limits of integration, we get:
Angle = (21x - 4x^3/3) - (21(0) - 4(0)^3/3)
= (21x - 4x^3/3)
Therefore, the motor turns through an angle of (21x - 4x^3/3) radians between t = 0 seconds and the instant at which it reverses direction (t = x seconds).