The angle∅ swept by a particle moving round a circle at time t is given by ∅=t^2+3t-5. Find the angular velocity and acceleration at t=5s
To find the angular velocity and acceleration at t=5s, we need to find the derivative of the given function.
Given ∅ = t^2 + 3t - 5, we can find the angular velocity by taking the derivative with respect to t:
ω = d∅/dt = d/dt (t^2 + 3t - 5)
To find the acceleration, we take the second derivative of ∅ with respect to t:
α = d²∅/dt² = d/dt (d∅/dt) = d/dt (t^2 + 3t - 5)
First, let's find the angular velocity:
ω = d∅/dt = 2t + 3
Now, let's find the acceleration:
α = d²∅/dt² = d/dt (2t + 3) = 2
Therefore, the angular velocity at t=5s is:
ω = 2(5) + 3 = 13 rad/s
And the angular acceleration at t=5s is:
α = 2 rad/s²
To find the angular velocity and acceleration at t = 5s, we need to differentiate the angle function ∅(t) with respect to time.
Let's start by finding the angular velocity:
Angular velocity (ω) is the derivative of the angle function with respect to time (∅'(t)).
∅(t) = t^2 + 3t - 5
Differentiating ∅(t) with respect to time, we get:
∅'(t) = 2t + 3
Now we can substitute t = 5s into the derivative to find the angular velocity:
∅'(t) = 2(5) + 3
∅'(t) = 10 + 3
∅'(t) = 13
So, the angular velocity at t = 5s is 13 rad/s.
Next, let's find the angular acceleration:
Angular acceleration (α) is the derivative of the angular velocity with respect to time (∅''(t)).
∅'(t) = 2t + 3
Differentiating ∅'(t) with respect to time, we get:
∅''(t) = 2
The angular acceleration is a constant, which means it does not depend on time.
Therefore, the angular acceleration at t = 5s is 2 rad/s^2.