The angular velocity of a wheel is given by w(t) = (2.00 rad/s^2)t + (1.00 rad/s^4)t^3.
A. What is the angular displacement of the wheel from time t=0.00s to time t=T?
B.What is the angular acceleration of the wheel as a function of time?
ω(t)=t³+2t rad/s
note that ω(0)=0 rad/s.
(A)
Angular displacement
=∫ω(t)dt [0≤t≤T]
=[t^4/4+t^2] from 0 to T
=T^4/4 + T²
(B)
The angular acceleration is obtained by differentiation of ω(t) with respect to t.
To find the angular displacement of the wheel from time t=0.00s to time t=T, we need to integrate the angular velocity function with respect to time over the interval [0, T].
A. Angular displacement is given by the integral of angular velocity with respect to time:
θ(t) = ∫[w(t)] dt
Given that w(t) = (2.00 rad/s^2)t + (1.00 rad/s^4)t^3, we can integrate this function to find the angular displacement:
θ(t) = ∫[(2.00 rad/s^2)t + (1.00 rad/s^4)t^3] dt
To integrate each term separately:
θ(t) = ∫(2.00 rad/s^2)t dt + ∫(1.00 rad/s^4)t^3 dt
Taking the antiderivative:
θ(t) = (1.00 rad/s^2)(t^2/2) + (1.00 rad/s^4)(t^4/4)
Now, evaluate the expression at t=T and t=0:
θ(T) = (1.00 rad/s^2)(T^2/2) + (1.00 rad/s^4)(T^4/4)
Therefore, the angular displacement of the wheel from time t=0.00s to time t=T is given by the expression:
θ(T) = (1.00 rad/s^2)(T^2/2) + (1.00 rad/s^4)(T^4/4)
B. The angular acceleration of the wheel can be found by taking the derivative of the angular velocity function with respect to time.
Given: w(t) = (2.00 rad/s^2)t + (1.00 rad/s^4)t^3
To find the angular acceleration, differentiate w(t) with respect to t:
a(t) = dw(t)/dt
Differentiating each term separately:
a(t) = d/dt[(2.00 rad/s^2)t + (1.00 rad/s^4)t^3]
a(t) = (2.00 rad/s^2) + (3.00 rad/s^4)t^2
Therefore, the angular acceleration of the wheel as a function of time is given by:
a(t) = (2.00 rad/s^2) + (3.00 rad/s^4)t^2