Deriving the equations of Rotational Motion a graphical explanation of where they come from. By J. Jaffrey at Long Bay College, NZ

The 5 Variables symbol variable unit initial angular velocity s-1 ωf final angular velocity α angular acceleration s-2 θ displacement radians t time s

A washing machine starts to speed up in the spin cycle of the wash A washing machine starts to speed up in the spin cycle of the wash. It starts at 60rpm and after 5.0s ends up at 900rpm. We can plot this motion on a graph with angular velocity on the vertical axis and time on the horizontal axis. The starting 60rpm will give us the initial angular velocity ωi 60 revolutions needs to be converted to radians. 60 x 2π =377radians This then needs to be changed to s-1 Notice the unit of ω is s-1. Radians are simply a ratio of arc length to radius so have no dimension. So ωi = 6.3 s-1 (2π) Similarly the final 900rpm will give us ωf So ωf = 94 s-1 Time t = 5.0s Click for next ►

ωf = ωi + αt α = 18 s-2 (2sf) Initial velocity ωi = 6.3s-1 Final velocity ωf = 94s-1 Time t = 5.0s ω(ms-1) What is the angular acceleration? 25 50 75 100 ωf Angular acceleration is rate of change of angular velocity in s-2. It is the gradient of the line. This is given by: ∆ω/∆t. ∆ω =(ωf- ωi). So here: ∆ω = ∆ω ωav 88s-1 87.96s-1 Acceleration a=∆ω/∆t So now we have: α=88/5 α = 18 s-2 (2sf) ∆t=5 ωi 1 2 3 4 5 t (s) What is the average angular velocity? Average angular velocity: ωav = ½ (ωi + ωf) ωav = 50s-1 The change in angular velocity ∆ω is given by the angular acceleration (α) multiplied by time (t). ωf = ωi + αt So ∆ω = αt and Click for next ►

θ = ωit θ = ωit + ½ αt2 + ½ αt2 = 250 ω(ms-1) αt θ = 31.5radians ½ αt2 ωav = ½ (ωi + ωf) ωf = ωi + αt ω(ms-1) θ = ½(ωf + ωi )t 25 50 75 100 Can we calculate θ another way? The displacement for a steady ωi (6.3s-1) for t (5.0s) gives: αt θ = 31.5radians The displacement because of the increase in velocity is the average extra velocity times t or: ½ αt2 θ = ωavt ½ x αt x t (½ x 88) x 5 = 220rad or ½ αt2 ωit 1 2 3 4 5 t (s) This is the area of the triangle under the line. Can we calculate the angular displacement over 5.0s? So the total displacement is: Displacement is average velocity multiplied by time so here: θ = ½ (ωf + ωi )t θ = ωit + ½ αt2 θ = ωit + ½ αt2 = 32 + 220 = 250 θ= 50 x 5 θ = 250radians Click for next ►

θ = ωft - ½ αt2 ½ αt2 = ½ αt2 250 = 470 - 220 ω(ms-1) αt = 220 t (s) ωav = ½ (ωi + ωf) A third way to calculate θ=250radians is by looking at the displacement in the 5.0s of t at the final velocity ωf = 94s-1 ωf = ωi + αt θ = ½(ωf + ωi )t ω(ms-1) θ = ωit + ½ αt2 t 25 50 75 100 ωf ½ αt2 This is too large so we must subtract enough displacement to allow for the acceleration. αt ωft We must subtract the area of the triangle above the line. = ½ αt2 = 220 Finally the displacement is the area under the line given by: 1 2 3 4 5 t (s) At ωf for 5.0s we have: θ = 94 x 5 θ = θ = ωft - ½ αt2 470 radians 250 = 470 - 220 This is the area of the rectangle on the graph

s-2 ωf2 = ωi2 + 2αθ ωf = ωi + αt ωf2 = (ωi + αt)2 ωav = ½ (ωi + ωf) t=5.0s ωi= 6.3s-1 ωf = 94s-1 θ=250radians ωf = ωi + αt The last equation cannot be shown easily from a graph. However it can be derived using algebra. θ = ½(ωf + ωi )t θ = ωit + ½ αt2 ωf = ωi + αt square both sides θ = ωit - ½ αt2 ωf2 = (ωi + αt)2 expand ωf2 = ωi2 + 2ωiαt + α2t2 extract 2α ωf2 = ωi2 + 2α (ωit + ½αt2 ) But this term in brackets is simply θ so simplifying we get our final equation of rotational motion: A numerical check gives us: 942 = 6.32 + (2x17.6x250) 8836 = 40 + 8800 ωf2 = ωi2 + 2αθ s-2 (s-1)2 (s-1)2 s-2 All three terms give: We can check the units both sides are the same. Click for next ►

The Equations of Rotational Motion ωf = ωi + αt ωav = ½ (ωi + ωf) θ = ½(ωf + ωi )t θ = ωit + ½ αt2 θ = ωit - ½ αt2 ωf2 = ωi2 + 2αθ Click to finish