Starting from rest, a 12-cm-diameter compact disk takes 3.0 s to reach its operating angular velocity of 2000 rpm. Assume that the angular acceleration is constant. The disk's moment of inertia is 2.5 x 10^-5 kg m^2. How much torque is applied to the disk? How many revolutions does it make before reaching full speed?
- print Print
- list Cite
Expert Answers
jeew-m
| Certified Educator
calendarEducator since 2012
write1,657 answers
starTop subjects are Math, Science, and Social Sciences
you need to know the following equations inorder to do this calculation.
`omega = omega_0+alpha*t`
`tau= I*alpha`
For our question;
omega_0 = 0 because it starts from rest.
omega = 2000/60*2*pi/3 = 69.813rad/s
...
(The entire section contains 86 words.)
Unlock This Answer Now
Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.
Related Questions
- A torque of 12 N.m is applied to a solid , uniform disk of radius 0.50 m. If the disk accelerates...
- 1 Educator Answer
- An ultracentrifuge accelerates from rest to 100,000 rpm in 2.50 min. (a) What is its angular...
- 1 Educator Answer
- A wheel of radius R, mass M and moment of inertia I is mounted on a frictionless axle.A light...
- 1 Educator Answer
- A horizontal disk is rotating counter-clockwise about its axis of symmetry at `14 rps` . Its...
- 2 Educator Answers