Angular Kinematics Problem
A flywheel is initially rotating with an angular velocity of 900 rpm, but a frictional torque is slowing that rotation at a rate of −5 rpm/s. (a) How long does it take the flywheel to come to rest? (b) How many times does the flywheel rotate before coming to rest? (c) What is the angular speed when the flywheel has completed exactly 1000 rotations.
D1: No diagram is needed here.
D2: Let ω0 = 900 rpm be the initial angular velocity, α = −5 rpm/s be the uniform angular acceleration,
ω = ω(t) be the angular velocity at time t, θ0 = 0 be the initial angular displacement and θ(t) be the
angular displacement at time t.D3: We wish to solve for τ, the elapsed time at which ω(τ) = 0, then we shall find the value of θ(τ).
BE: Since we have uniform angular acceleration here, we can use the kinematic relationships
ω(t) = ω0 + αt (1), θ(t) = θ0 + ω0t + ½αt2 (2), and ω2 = ω02 + 2α(θ − θ0) (3).S: To solve part (a) we use equation (1). Setting ω(τ) = 0 = ω0 + ατ, we obtain τ = −ω0/α (4). To
solve part (b) we use equation (2), setting t = τ = −ω0/α, to obtain θ(τ) = θ0 + ω0τ + ½ατ 2
= θ0 − ω02/α + ½ω02/α = θ0 − ½ω02/α (5). Part (c) can be solved directly from equation (3) by
simply taking the square root of both sides of the equation and plugging the given numerical data
into the result ω = [ω02 + 2α(θ − θ0)]½.N: We now plug data into the results in step S above obtaining:
(a) τ = −ω0/α = 900 rpm/(−5 rpm/s) = 180 s = 3 m.
(b) θ(τ) = θ0 − ½ω02/α = 0 − ½(900 rpm)2/[(−5 rpm/s)(60 s/m)] = 1350 revolutions.
(c) ω = [ω02 + 2α(θ − θ0)]½ = [(900 rpm)2 + 2(−5 rpm/s)(1000 rev)(60 s/m)] ½ = 458 rpm.
U: ü (We note that keeping track of units here was a bit tricky since the angular speed was given in
rpm rather than the conventional rad/s and similarly the angular acceleration was given in rpm/s
rather than rad/s2. There was never any need to convert revolutions to radians here, but since
both minutes and seconds were used as time units, the conversion factor 60 s/m was needed
twice in step N. An alternative approach would been to have started with the conversions
ω0 = 900 rev/m = (900 ev/m)/(60 s/m) = 15 rev/s and α = −5 rpm/s = (−5 rev/m×s)/(60 s/m)
= (−1/12) rev/s2.)B: ü