If the displacement of an object moving in a circular motion in terms of time is given by 6t^2 + 8
what would be the expression for its angular velocity?
Angular velocity, ω = dθ/dt
In circular motion, the displacement of the object is related to the angle through the formula s = rθ, where r is the radius of the circle.
Differentiating both sides of this equation with respect to time, we get:
ds/dt = r(dθ/dt)
But ds/dt is simply the velocity of the object, v. So we have:
v = rω
where ω is the angular velocity.
Now, given that the displacement in terms of time is 6t^2 + 8, we can write:
s = rθ = 6t^2 + 8
Taking the derivative with respect to time, we get:
ds/dt = 12t
Substituting this into the equation v = rω, we have:
v = rω = 12tr
But we don't know the value of r. However, we do know that the displacement is given by 6t^2 + 8, which means that the object is moving in a circle of radius r = (6t^2 + 8)/θ. Substituting this into the equation above, we get:
v = rω = 12t(6t^2 + 8)/θ
Therefore, the expression for the angular velocity, ω, is:
ω = 12t(6t^2 + 8)/rθ
or
ω = 12t(6t^2 + 8)/(6t^2 + 8)θ
which simplifies to:
ω = 12t/θ