Question

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?

Answer 1

Answer:

$a =29.54\ m/s^2$

Explanation:

given,

radius of CD player = 12 cm

assume rate of acceleration = 100 rad/s²

times = 0.15 s

now,

tangential acceleration  

  $a_t = \alpha r$

  $a_t = 100 \times 0.12$

  $a_t = 12 m/s^2$

now using equation

v = v₀ + a_t x t

v =0+ 12 x 0.15

v = 1.8 m/s

now, radial acceleration

$a_r = \dfrac{v^2}{r}$

$a_r = \dfrac{1.8^2}{0.12}$

$a_r =27\ m.s^2$

now

acceleration

$a = \sqrt{a_r^2+a_t^2}$

$a = \sqrt{27^2+12^2}$

$a =29.54\ m/s^2$