The angular position of a point on the rim of a rotating wheel is given by = 4.0t - 1.0t2 + t3, where is in radians and t is in seconds.
(a) What is the angular velocity at t = 2 s?
(b)What is the angular velocity at t = 4.0 s?
(c) What is the average angular acceleration for the time interval that begins at t = 2 s and ends at t = 4.0 s?
(d) What is the instantaneous angular acceleration at the beginning of this time interval?
(e)What is the instantaneous angular acceleration at the end of this time interval?
(a) differentiate the theta(t) equation you provided. This will yield the following:
d(theta)/dt = omega(t) = 4 - 2t + 3 t^2
Then plug in t = 2 fr your answer.
The Greek letter "omega" is usually used for angular velocity.
(b) Use the same omega(t) formula but use t=4.
(c) Subtract the omega(2) from omega (4) for the angular velocity change, and divinde by 2 s.
(d) The formula for instantaneous angular acceleration is
alpha (t) = d(omega)/dt = -2 + 6t. Insert t = 2 s
(e) Insert t=4 into the alpha(t) equation above
To solve these questions, we need to differentiate the given angular position equation with respect to time (t). Let's start with part (a).
(a) Differentiating the given equation θ(t) = 4.0t - 1.0t^2 + t^3 with respect to t gives:
dθ/dt = ω(t) = 4 - 2t + 3t^2
Now, we can plug in t = 2 s into the ω(t) equation:
ω(2) = 4 - 2(2) + 3(2)^2 = 4 - 4 + 12 = 12 rad/s
So, the angular velocity at t = 2 s is 12 rad/s.
Moving on to part (b):
Similarly, we can plug in t = 4 s into the ω(t) equation:
ω(4) = 4 - 2(4) + 3(4)^2 = 4 - 8 + 48 = 44 rad/s
Therefore, the angular velocity at t = 4 s is 44 rad/s.
For part (c):
To find the average angular acceleration between t = 2 s and t = 4 s, we need to find the change in angular velocity over this time interval.
Δω = ω(4) - ω(2) = 44 - 12 = 32 rad/s
The time interval is 4 s - 2 s = 2 s.
Average angular acceleration = Δω / Δt = 32 rad/s / 2 s = 16 rad/s^2
Hence, the average angular acceleration for the given time interval is 16 rad/s^2.
Moving on to part (d):
The formula for instantaneous angular acceleration is:
α(t) = d(ω)/dt = -2 + 6t
Substituting t = 2 s into the α(t) equation:
α(2) = -2 + 6(2) = -2 + 12 = 10 rad/s^2
Therefore, the instantaneous angular acceleration at the beginning of the time interval (t = 2 s) is 10 rad/s^2.
Lastly, for part (e):
Substituting t = 4 s into the α(t) equation:
α(4) = -2 + 6(4) = -2 + 24 = 22 rad/s^2
So, the instantaneous angular acceleration at the end of the time interval (t = 4 s) is 22 rad/s^2.