The angular position of a point on the rim of a rotating wheel is given by Theta = 4.0t-3t^2+t^3, where theta is in radians and t is in seconds. At t=0, what are (a.) the point's angular position and (b.) its angular velocity? (c.) What is its angular velocity at t=4.0s? What are the instantaneous angular accelerations at (d.) the beginning and (e.) the end of this time interval?
w=dTheta/dtime. I assume you know calculus.
d) alpha=dw/dtime, at t=0 and at t=tfinal
So far i got angular velocities at t=2s and t=4s
W2 = 3(2)^2-6(2)+4 = 4 rad/s
W4 = 3(4)^2-6(4)+4 = 28 rad/s
For average angular acceleration i got
W2-W1 / T2-T1 = 28-4/4-2 = 12 rad/s^2
But I don't know how to figure out part d and e. Thanks for the help.
To find the answers to these questions, we need to differentiate the given equation to obtain expressions for angular position (θ), angular velocity (ω), and angular acceleration (α) in terms of time (t).
(a.) The angular position (θ) at any given time (t) is given by the equation θ = 4.0t - 3t^2 + t^3. To find the angular position at t = 0, simply substitute t = 0 into the equation:
θ(at t=0) = 4.0(0) - 3(0)^2 + (0)^3
θ(at t=0) = 0 - 0 + 0
θ(at t=0) = 0 radians
Therefore, at t = 0, the point's angular position is 0 radians.
(b.) The angular velocity (ω) is the derivative of the angular position (θ) with respect to time (t). Differentiate the equation θ = 4.0t - 3t^2 + t^3 to find ω:
ω = dθ/dt = d/dt(4.0t - 3t^2 + t^3)
Using the power rule of differentiation, we get:
ω = 4.0 - 6t + 3t^2
To find the angular velocity (ω) at t = 0, substitute t = 0 into the equation:
ω(at t=0) = 4.0 - 6(0) + 3(0)^2
ω(at t=0) = 4.0 - 0 + 0
ω(at t=0) = 4.0 rad/s
Therefore, at t = 0, the point's angular velocity is 4.0 rad/s.
(c.) To find the angular velocity (ω) at t = 4.0 s, substitute t = 4.0 into the equation for angular velocity:
ω(at t=4.0) = 4.0 - 6(4.0) + 3(4.0)^2
ω(at t=4.0) = 4.0 - 24.0 + 48.0
ω(at t=4.0) = 28.0 rad/s
Therefore, at t = 4.0 s, the point's angular velocity is 28.0 rad/s.
(d.) The instantaneous angular acceleration (α) is the derivative of the angular velocity (ω) with respect to time (t). Differentiate the equation ω = 4.0 - 6t + 3t^2 to find α:
α = dω/dt = d/dt(4.0 - 6t + 3t^2)
Using the power rule of differentiation, we get:
α = -6 + 6t
To find the instantaneous angular acceleration (α) at the beginning of the time interval (t = 0), substitute t = 0 into the equation:
α(at t=0) = -6 + 6(0)
α(at t=0) = -6 rad/s^2
Therefore, at the beginning of the time interval, the instantaneous angular acceleration is -6 rad/s^2.
(e.) To find the instantaneous angular acceleration (α) at the end of the time interval (t = 4.0 s), substitute t = 4.0 into the equation for angular acceleration:
α(at t=4.0) = -6 + 6(4.0)
α(at t=4.0) = -6 + 24.0
α(at t=4.0) = 18.0 rad/s^2
Therefore, at the end of the time interval, the instantaneous angular acceleration is 18.0 rad/s^2.