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In a CD player, a CD starts from rest and accelerates at a rate of​​ . Suppose the CD has radius 1212 cm. At time 0.15 sec after it started spinning, what is the magnitude of the linear acceleration∣ for a point on its outer rim?

Respuesta :

Answer:

[tex]a =29.54\ m/s^2[/tex]

Explanation:

given,

radius of CD player = 12 cm

assume rate of acceleration = 100 rad/s²

times = 0.15 s

now,

tangential acceleration  

  [tex]a_t = \alpha r[/tex]

  [tex]a_t = 100 \times 0.12[/tex]

  [tex]a_t = 12 m/s^2[/tex]

now using equation

v = v₀ + a_t x t

v =0+ 12 x 0.15

v = 1.8 m/s

now, radial acceleration

[tex]a_r = \dfrac{v^2}{r}[/tex]

[tex]a_r = \dfrac{1.8^2}{0.12}[/tex]

[tex]a_r =27\ m.s^2[/tex]

now

acceleration

[tex]a = \sqrt{a_r^2+a_t^2}[/tex]

[tex]a = \sqrt{27^2+12^2}[/tex]

[tex]a =29.54\ m/s^2[/tex]

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