Answered

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 :

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)

[tex]\omega=\dfrac{2\pi N}{60}[/tex]

[tex]127.64=\dfrac{2\pi N_2}{60}[/tex]

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)

[tex]\omega=\dfrac{2\pi N}{60}[/tex]

[tex]333.32=\dfrac{2\pi N_1}{60}[/tex]

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

(a) the maximum angular speed is 332.32 rad/s and the minimum is 127.65 rad/s

(b) the maximum angular speed is at the inner diameter and the minimum angular speed is at the outer diameter.

(c) The average angular acceleration is 7.1 x 10⁻³ rad/s²

Angular speed:

Given information:

outer diameter, D = 11.75 cm

inner diameter, d = 4.5 cm

The laser maintains a constant linear speed of v = 7.5 m/s

(a) The angular speed and linear speed are related as:

v = ω × r

ω = v/r

where r is the radius and ω is the angular speed.

maximum angular speed:

[tex]\omega_{min}=\frac{v}{D/2}\\\\ \omega_{min}=\frac{7.5}{0.1175/2}\\\\\omega_{min}=127.65\;rad/s[/tex]

[tex]\omega_{min}=2\pi f_{min}\\\\f_{min}=127.65/2\pi \;rps\\\\f_{min}=127.65\times60/2\pi\;rpm\\\\f_{min}=1219.4 \;rpm[/tex]

maximum angular speed:

[tex]\omega_{max}=\frac{v}{d/2}\\\\ \omega_{max}=\frac{7.5}{0.045/2}\\\\\omega_{max}=333.32\;rad/s[/tex]

[tex]\omega_{max}=2\pi f_{max}\\\\f_{max}=333.32/2\pi \;rps\\\\f_{max}=332.32\times60/2\pi\;rpm\\\\f_{max}=3184.58 \;rpm[/tex]

(b) the maximum angular speed is at the inner diameter and the minimum angular speed is at the outer diameter.

(c) the average angular acceleration is :

[tex]\omega_{max}=\omega_{min}+\alpha t\\\\332.32=127.65+\alpha\times8\times60\times60\\\\\alpha= 7.1 \times 10^{-3} \;rad/s^2[/tex]

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