### Rotational kinematics

#### Problem:

A centrifuge in a medical laboratory rotates at an angular speed of 3600 rev/min.  When switched off, it rotates 50 times before coming to rest.  What is the constant angular acceleration of the centrifuge?

Solution:

• Concepts:
Rotational kinematics
• Reasoning:
We have motion with constant angular acceleration.
• Details of the calculation:
Formulas for rotational kinematics:
ω = dθ/dt,  α = dω/dt,  Δω = αΔt,  ωf = ωI + αΔt.
θf = θi + ωi(tf - ti) + ½α(tf - ti)2.
Δθ = <ω>Δt.  <ω> = (ωi + ωf)/2
Given:
θi = 0,  θf = 50 * 2π (50 revolutions)
ωi = (3600 * 2π/min) *(1 min/60 s) = 377/s,  ωf = 0
α = constant
α = ?
Find Δt:  <ω> = (ωi + ωf)/2 = 188.5/s,  Δt = Δθ /<ω> = 1.667 s
Find α:  α = (ωf - ωi)/ Δt = -(377/s)/(1.666 s) = -226.2/s2

#### Problem:

An automobile drag racer drives a car with constant acceleration ai.  The tires with radius r0 are not slipping.  At t = 0 the car starts from rest.
(a)   Derive an equation for the acceleration of a point P(θ) on the outer surface of the tire when the instantaneous velocity of the car is vi.
(b)   For which point P(θ) on the outer surface of the tire is the magnitude of the acceleration the greatest?  What is this magnitude of the acceleration?

Solution:

• Concepts:
Rigid body motion
• Reasoning:
Let ai denote the acceleration of the point P.
ai = a + arel, where a is the acceleration of the CM and arel is the acceleration of P in the CM frame
• Details of the calculation:
(a)  ai = ai + atangential + acentripetal = ai + a(-cosθ j + sinθ i) - (v2/r0) (cosθ i + sinθ j)
= (a + a sinθ - (v2/r0)cosθ)i - (a cosθ  + (v2/r0) sinθ)j.
(b)  ai is largest when the acceleration of the CM and the acceleration of P in the CM frame both point in the same direction.  This can happen for a point P in the second quadrant, when a cosθm + (v2/r0) sinθm = 0, tanθm = -ar0/v2, tan(π - θm) = ar0/v2.
This identifies the point of maximum acceleration.
Substituting a cosθm = -(v2/r0) sinθm yields aim = (a + a sinθm - (v2/r0)cosθm)i.
aim  = a + a sinθm - (v2/r0)cosθm = a + [a2 + (v2/r0)2 ]½.
(Since at P(θm) the acceleration of the CM and the acceleration of P in the CM frame point in the same direction, their magnitudes just add.)