Question

The outside diameter of the playing area of an optical Blu-ray disc is 11.75 cm, and the inside diameter is 4.50 cm. When viewing movies, the disc rotates so that a laser maintains a constant linear speed relative to the disc of 7.50 m/s as it tracks over the playing area.
(a) What are the maximum and minimum angular speeds (in rad>s and rpm) of the disc?

(b) At which location of the laser on the playing area do these speeds occur?

(c) What is the average angular acceleration of a Blu-ray disc as it plays an 8.0-h set of movies?

Answer 1

Answer:

a)ω₂=127.64 rad/s      ( min)

ω₁= 333.32 rad/s    ( max)

b) At  d₂= 11.75 cm ,ω₂=127.64 rad/s

d₁=4.5 cm,ω₁= 333.32 rad/s

c)α = 7.1 x 10⁻³ rad/s²

Explanation:

Given that

r₂= 11.75 cm

r₁=4.5 cm

v= 7.5 m/s

a)

We know that

v=ω r

ω =Angular speed

r= radius

v= velocity

When d₂= 11.75 cm :

v=ω r

7.5 x 2 =ω₂ x 0.1175

ω₂=127.64 rad/s      ( min)

$\omega=\dfrac{2\pi N}{60}$

$127.64=\dfrac{2\pi N_2}{60}$

N₂=1219.4 rpm

When d₁=4.5 cm :

v=ω r

7.5 x 2=ω₁ x 0.045

ω₁= 333.32 rad/s    ( max)

$\omega=\dfrac{2\pi N}{60}$

$333.32=\dfrac{2\pi N_1}{60}$

N₁=3184.58 rpm

The average angular acceleration α given as

ω₁ = ω₂ + α t

333.32 = 127.64 + α x 8 x 60 x 60

α = 7.1 x 10⁻³ rad/s²