Question

A uniform disk with mass 41.7kg and radius 0.290m is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 28.0N is applied tangent to the rim of the disk.


What is the magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.120 revolution?

What is the magnitude a of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.120 revolution?

Answer 1

Strategy:
1) Find angular acceleration alpha first.
2) Find angular velocity omega after 0.120 rev
3) Find tangential velocity from omega found in (2)
4)

Answer 2

oh okay ill equate it

Answer 3

@michelle092294 : you find the answer?

Answer 4

no -_- i keep having the wrong answer. then i got 1 more chance. and ill have zero. -_- can you send me the equations?

Answer 5

tangential velocity = radius * omega

Answer 6

What is the moment oh inertia of disk with mass m radius r?

Answer 7

\[V=r* \omega\] where omega= angular velocity

Answer 8

now find omega first

Answer 9

thank you!!!! :D

Answer 10

\[torque= inertia* \omega \]

Answer 11

inertia= m r^2

Answer 12

thank you jason :)

Answer 13

:) You're Welcome michi

Answer 14

\[(\omega _{f})^2=2*\alpha*\theta\] initial angular velocity = 0 okay?

Answer 15

okay :D now il solve it :)

Answer 16

great

Answer 17

\[\alpha= equation ?\]

Answer 18

\[\tau=I*\alpha=F*radius\] where I is moment of inertia and F= force

Answer 19

okay?

Answer 20

michi , find inertia , you have force and radius ..you'll get alpha ..after that you can calculate angular velocity ...

Answer 21

yes got the inertia value. now need to get alpha :)