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**= d**ω**/dt, Δ**ω**=**α**Δt,**ω**_{f }=**ω**_{I }+**α**Δt.

θ_{f}= θ_{i}+ ω_{i}(t_{f}- t_{i}) + ½α(t_{f}- t_{i})^{2}.

Δθ = <ω>Δt. <ω> = (ω_{i}+ ω_{f})/2

Given:

θ_{i}= 0, θ_{f }= 50 * 2π (50 revolutions)

ω_{i}= (3600 * 2π/min) *(1 min/60 s) = 377/s, ω_{f}= 0

α = constant

Asked:

α = ?

Find Δt: <ω> = (ω_{i}+ ω_{f})/2 = 188.5/s, Δt = Δθ /<ω> = 1.667 s

Find α: α = (ω_{f}- ω_{i})/ Δt = -(377/s)/(1.666 s) = -226.2/s^{2}

An automobile drag racer drives a car with
constant acceleration a**i.
**The tires with radius r_{0} 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 v**i**.

(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**a**_{i}denote the acceleration of the point P.

**a**_{i}=**a**+**a**_{rel}, where**a**is the acceleration of the CM and**a**_{rel}is the acceleration of P in the CM frame - Details of the calculation:

(a)**a**_{i}= a**i**+**a**_{tangential}+**a**_{centripetal}= a**i**+ a(-cosθ**j**+ sinθ**i**) - (v^{2}/r_{0}) (cosθ**i**+ sinθ**j**)

= (a + a sinθ – (v^{2}/r_{0})cosθ)**i**– (a cosθ + (v^{2}/r_{0}) sinθ)**j**.

(b)**a**_{i}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 cosθ_{m}+ (v^{2}/r_{0})^{ }sinθ_{m}= 0, tanθ_{m}= -ar_{0}/v^{2}, tan(π - θ_{m}) = ar_{0}/v^{2}.

This identifies the point of maximum acceleration.

Substituting a cosθ_{m}= -(v^{2}/r_{0})^{ }sinθ_{m}yields**a**_{im}= (a + a sinθ_{m}– (v^{2}/r_{0})cosθ_{m})**i**.

a_{im}= a + a sinθ_{m}– (v^{2}/r_{0})cosθ_{m}= a + [a^{2}+ (v^{2}/r_{0})^{2}]^{½}.

(Since at P(θ_{m}) the acceleration of the CM and the acceleration of P in the CM frame point in the same direction, their magnitudes just add.)