Rotational motion/angular acceleration

Question

A barrel ride at an amusement park

is turning 2.30 rad/s when it starts

to slow down. After making 5.00

revolutions, it is rotating at

1.11 rad/s. How much time did

that take?

(Unit = s)

can someone please answer this thourgly?

Answer 1

This is a standard application of the angular kinematic equations

Δθ=ω₀t+1/2 αt²

ω_f=ω₀+αt

ω_f²=ω₀²+2αΔθ

(A helpful hint: if you can't remember what these equations are, just take the kinematics for linear motion with x, v, and a and replace each variable with its angular counterpart. You'll get a valid rotational kinematic!)

Note that these only apply if the angular acceleration α is a constant. The problem doesn't say that explicitly, but we must assume it to make progress.

From the statement of the problem, we can deduce that ω₀=2.3 rad/s, ω_f=1.11 rad/s, and Δθ=5*2π=10π rad (5 revolutions is 10π rads). We are tasked with finding t. Unfortunately, None of these kinematics relates our known quantities to our unknown; the first two involve t but also the unknown α, and the last equation only has α. But this gives us an avenue forward; we can use the third kinematic to find α, and then either of the first two to find t.

ω_f²=ω₀²+2αΔθ -> α=(ω_f²-ω₀²)/(2Δθ)

And

ω_f=ω₀+αt -> t=(ω_f-ω₀)/α

I like to work symbollically (In part because it helps reduce errors when substituting in numbers), so I'll go ahead and do that, but you could just stop here, put in the numbers to get α, and then use the number for α to get t.

t=(ω_f-ω₀)/α = (ω_f-ω₀) (2Δθ) / (ω_f²-ω₀²) = 2Δθ/(ω_f+ω₀)

And now, the answer:

t=2Δθ/(ω_f+ω₀)=2(10π)/(1.11+2.3)=20π/3.41=18.43s

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It's worth noting that there is a fourth kinematic that is not used as much as the others

Δθ=1/2 (ω_f+ω₀)t

Which technically I derived above. If you already know this kinematic exists, you can skip straight to here and solve things much faster.