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

`a =29.54\ m/s^2`

Step-by-step 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 * 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 = (v^2)/(r)`

`a_r = (1.8^2)/(0.12)`

`a_r =27\ m.s^2`

now

acceleration

`a = √(a_r^2+a_t^2)`

`a = √(27^2+12^2)`

`a =29.54\ m/s^2`