The angular displacement in radians is given by: θ=8√(t^2+2)
Find the angular velocity and angular acceleration at t = 0.25 s.
To find the angular velocity and angular acceleration at t = 0.25 s, we need to differentiate the given angular displacement function, θ, with respect to time, t.
The angular velocity, ω, is the derivative of angular displacement with respect to time:
ω = dθ/dt
To find the angular acceleration, α, we differentiate the angular velocity function, ω, with respect to time:
α = dω/dt
Let's start by finding the expression for ω by taking the derivative of θ with respect to t:
θ = 8√(t^2+2)
To take the derivative, we can rewrite θ as:
θ = 8(t^2 + 2)^(1/2)
Now, let's differentiate θ with respect to t using the chain rule. The chain rule states that if we have a composite function f(g(t)), then the derivative of f(g(t)) with respect to t is given by:
d(f(g(t)))/dt = f'(g(t)) * g'(t)
In our case, f(u) = 8√u and g(t) = t^2+2, so we have:
θ = f(g(t))
f'(u) = 8/(2√u) = 4/√u
g'(t) = 2t
By applying the chain rule, we get:
dθ/dt = (4/√(t^2+2)) * 2t
Simplifying further:
dθ/dt = 8t / √(t^2+2)
Now that we have the expression for ω, let's substitute t = 0.25 s into it to find the angular velocity at t = 0.25 s:
ω = 8(0.25) / √(0.25^2+2)
= 2 / √(0.0625+2)
= 2 / √(2.0625)
= 2 / 1.4385
≈ 1.39 rad/s
Next, let's find the expression for α by taking the derivative of ω with respect to t:
ω = 8t / √(t^2+2)
To find α, we need to differentiate ω with respect to t. Using the quotient rule for differentiation, we have:
α = d(ω) / dt = (d(8t)/dt * √(t^2+2) - 8t * d(√(t^2+2))/dt) / (t^2+2)
Taking the derivatives, we get:
α = (8 * √(t^2+2) - 8t * (1/(2√(t^2+2))) * 2t) / (t^2+2)
= (8 * √(t^2+2) - 8t^2 / √(t^2+2)) / (t^2+2)
= (8 - 8t^2 / √(t^2+2)) / (t^2+2)
Now, let's substitute t = 0.25 s into α to find the angular acceleration at t = 0.25 s:
α = (8 - 8(0.25)^2 / √(0.25^2+2)) / (0.25^2+2)
= (8 - 8(0.0625) / √(0.0625+2)) / (0.0625+2)
= (8 - 8(0.0625) / √(2.0625)) / (0.0625+2)
= (8 - 8(0.0625) / 1.4385) / (0.0625+2)
≈ (7.95 / 1.4385) / 2.0625
≈ 5.5 rad/s^2
Therefore, at t = 0.25 s, the angular velocity is approximately 1.39 rad/s and the angular acceleration is approximately 5.5 rad/s^2.