The angle∅ swept by a particle moving round a circle at time t is given by ∅=t^3+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 first and second derivative of the angle with respect to time (∅ = t^3 + 3t - 5).
First derivative:
∅' = 3t^2 + 3
Second derivative:
∅" = 6t
Now, we can substitute t = 5s into the derivatives to find the angular velocity and acceleration at t = 5s.
Angular velocity at t = 5s:
∅'(5) = 3(5)^2 + 3 = 3(25) + 3 = 75 + 3 = 78 rad/s
Therefore, the angular velocity at t = 5s is 78 rad/s.
Angular acceleration at t = 5s:
∅"(5) = 6(5) = 30 rad/s^2
Therefore, the angular acceleration at t = 5s is 30 rad/s^2.
To find the angular velocity and acceleration at t=5s, we need to differentiate the expression for ∅ with respect to time (t).
Given ∅=t^3+3t-5, we can find the angular velocity by differentiating ∅ with respect to t:
d∅/dt = d/dt (t^3+3t-5)
To do this, we apply the power rule of differentiation:
d/dt (t^n) = n*t^(n-1)
Therefore, using the power rule, we differentiate each term:
d∅/dt = 3*t^2 + 3
Now, we can substitute t=5s into the expression to find the angular velocity at t=5s:
ω = d∅/dt
ω = 3*t^2 + 3
ω(5) = 3*(5^2) + 3
= 75 + 3
= 78 rad/s
So, the angular velocity at t=5s is 78 rad/s.
To find the angular acceleration, we differentiate the expression for angular velocity (ω) with respect to time:
dω/dt = d/dt (3*t^2 + 3)
Using the power rule of differentiation:
dω/dt = 3*(2*t)
= 6t
Again, we substitute t=5s into the expression to find the angular acceleration at t=5s:
α = dω/dt
α = 6*t
α(5) = 6*(5)
= 30 rad/s^2
Therefore, the angular acceleration at t=5s is 30 rad/s^2.