Question

An optical disk drive in your computer can spin a disk up to 10,000 rpm (about 1045 rad / s). If a particular disk is spun at 558.2 rad / s while it is being read, and then is allowed to come to rest over 0.435 seconds, what is the magnitude of the average angular acceleration of the disk?

____rad/s^2




If the disk is 0.12 m in diameter, what is the magnitude of the linear acceleration of a point 1/11 of the way out from the center of the disk?


______m/s^2

Answer 1

Step-by-step explanation:

initial angular velocity, ωo = 558.2 rad/s

final angular velocity, ω = 0

time, t = 0.435 s

Use first equation of motion

ω = ωo + αt

where, α is the angular acceleration

0 = 558.2 + α x 0.435

α = - 1283.22 rad/s²

diameter of disc = 0.12 m

radius of disc, r = 0.06 m

distance, d = 0.06 / 11 m

linear acceleration, a = d x α = - 0.06 x 1283.22 / 11

a = - 7 m/s²

Answer 2

Step-by-step explanation:

Given

initial angular velocity
`\omega _0=558.2 rad/s`

Time taken to stop
`t=0.435 s`

using kinematic relation

`\omega =\omega _0+\alpha \cdot t`

here
`\omega =`final angualr velocity is zero because it is coming to rest

`\alpha =`angular acceleration

thus
`\alpha =-(\omega _0)/(t)`

`\alpha =-(558.2)/(0.435)=-1283.21 rad/s^2`

Magnitude is 1283.21 rad/s

If the diameter of disk
`d=0.12 m`

then

radius
`r=0.06 m`

Linear acceleration is givenby
`\alpha * r`

here radial distance is
`(1)/(11)` from center

thus
`r'=(r)/(11)=(0.06)/(11)=0.00545`

`a_L=1283.21* 0.00545=6.99 m/s^2`