A Beginners Physics Question

A steam engine runs with a constant angular velocity of 110 rev/min.
When steam is shut off, the friction of the bearings and of the air stops the wheel in 2.4 h.
1) How many revolutions does the wheel make before stopping?

---So, i figured out the angular acceleration which is -0.7638rev/min^2
---I got stuck when using this: θ = ωt + 1/2αt^2
---do i convert ω to rev/s and α to rev/s^2

Thank you for your time.

I would convert everything to seconds and radians:

ω₀ = 11/6 rev/s = 11π/3 rad/s
t = 8640 s

Solving for α:

ω = ω₀ + αt
0 = 11π/3 + α(8640)

α = -11π/25920 rad/s²


Now you can solve for θ:

θ = ω₀t + ½αt² = (11π/3)(8640) - ½(11π/25920)(8640)² = 15840π rad

There are 2π radians / revolution, so:

15840π / 2π = 7920 revolutions.

A steam engine runs with a constant angular velocity of 110 rev/min.
When steam is shut off, the friction of the bearings and of the air stops the wheel in 2.4 h.
1) How many revolutions does the wheel make before stopping?

The friction force is constant, so the angular deceleration is constant. The decrease of the angular velocity divided by time = angular acceleration.

Time = 2.4 hr * 60 min/hr = 144 minutes

ωf – ωi = α * t
0 – 110 = α * 144
α = (-110 ÷ 144) rev/min^2

θ = ωi * t + ½ * α * t^2
ωi = 110 rev/min

θ = 110 * 144 + ½ * (-110 ÷ 144) * 144^2
θ = (110 * 144) – (½ * 110 * 144)
θ = 15840 – 7920 = 7920 revolutions

When an objects velocity increases from 0 to a specific velocity, at a constant rate of acceleration, the graph is the right side of a parabola.
You can see the graph at the web site below!
http://www.physicsclassroom.com/class/1d…

When an objects velocity decreases from the same specific velocity to 0, at a constant rate of deceleration, the graph is the left side of the same parabola.