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 235.6 rad/s while it is being read, and then is allowed to come to rest over 0.167 seconds , what is the magnitude of the average angular acceleration of the disk? average angular acceleration: rad s 2 If the disk is 0.12 m in diameter, what is the magnitude of the tangential acceleration of a point 1 / 3 of the way out from the center of the disk? tangential acceleration

This question involves the concepts of equations of motion for angular motion and tangential acceleration.

a) The magnitude of the average angular acceleration is "1410.78 rad/s²".

b) The magnitude of the tangential acceleration of a point 1/3 of the way out from the center of the disk is "28.22 m/s²".

a)

We will use the first equation of motion for the angular motion to find out the angular acceleration:

where,

= angular acceleration = ?

= final angular speed = 0 rad/s

= initial angular acceleration = 235.6 rad/s

t = time = 0.167 s

Therefore,

b)

The tangential acceleration can be found using the following formula:

where,

= tangential acceleration = ?

r = distance of point from center =

Therefore,

Learn more about the angular motion here:

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The attached picture shows the angular equations of motion.

Part(a): The angular acceleration is .

Part(b): The linear acceleration is .

Explanation:

Given:

The angular velocity,

Time taken,

Diameter of the disk,

Radius of the concerned point, .

Part(a):

The angular acceleration () is given by

Part(b):

The radius () of the concerned point is given by

Linear acceleration () is give by

answergravity;

it would stay the same