The angular position of a point on the rim of a rotating wheel is given by è = 8.0t - 3.0t2 + t3, where è is in radians and t is in seconds.
(a) What is the angular velocity at t = 5 s?
___________rad/s
(b)What is the angular velocity at t = 7.0 s?
_________rad/s
(c) What is the average angular acceleration for the time interval that begins at t = 5 s and ends at t = 7.0 s?
_____________rad/s2
(d) What is the instantaneous angular acceleration at the beginning of this time interval?
_____________rad/s2
(e)What is the instantaneous angular acceleration at the end of this time interval?
_____________ rad/s2
You cant do it, its impossible.
To find the answers to these questions, we need to differentiate the given equation for angular position with respect to time, t.
Given: θ = 8.0t - 3.0t^2 + t^3
(a) Angular velocity is the derivative of angular position with respect to time.
ω = dθ/dt
Differentiating the given equation with respect to t:
ω = d(8.0t - 3.0t^2 + t^3)/dt
= 8.0 - 6.0t + 3.0t^2
To find the value of angular velocity at t = 5s, substitute t = 5 into the equation:
ω = 8.0 - 6.0(5) + 3.0(5)^2
= 8.0 - 30.0 + 75.0
= 53.0 rad/s
Therefore, the angular velocity at t = 5s is 53.0 rad/s.
(b) Similarly, substitute t = 7 into the equation to find the angular velocity at t = 7s:
ω = 8.0 - 6.0(7) + 3.0(7)^2
= 8.0 - 42.0 + 147.0
= 113.0 rad/s
Therefore, the angular velocity at t = 7s is 113.0 rad/s.
(c) Average angular acceleration is the change in angular velocity divided by the change in time.
Average angular acceleration = (ω2 - ω1) / (t2 - t1)
Substituting the values:
Average angular acceleration = (113.0 - 53.0) / (7.0 - 5.0)
= 60.0 / 2.0
= 30.0 rad/s^2
Therefore, the average angular acceleration for the time interval from t = 5s to t = 7s is 30.0 rad/s^2.
(d) Instantaneous angular acceleration is the derivative of angular velocity with respect to time.
α = dω/dt
To find the instantaneous angular acceleration at the beginning of the time interval, substitute t = 5 into the equation for angular acceleration:
α = d(8.0 - 6.0t + 3.0t^2)/dt
= -6.0 + 6.0t
Substituting t = 5:
α = -6.0 + 6.0(5)
= -6.0 + 30.0
= 24.0 rad/s^2
Therefore, the instantaneous angular acceleration at the beginning of the time interval is 24.0 rad/s^2.
(e) To find the instantaneous angular acceleration at the end of the time interval, substitute t = 7 into the equation for angular acceleration:
α = -6.0 + 6.0t
Substituting t = 7:
α = -6.0 + 6.0(7)
= -6.0 + 42.0
= 36.0 rad/s^2
Therefore, the instantaneous angular acceleration at the end of the time interval is 36.0 rad/s^2.