Solving Rotational Kinetics Homework: Angular Acceleration of Flywheel

[ohhi]

Homework Statement

The flywheel of a steam engine runs with a constant angular speed of 176 rev/min. When steam is shut off, the friction of the bearings and the air brings the wheel to rest in 2.8 hours. What is the magnitude of the constant angular acceleration of the wheel? (Use units of rev/min^2. )

What is the magnitude of the tangential component of the linear acceleration of a particle that is located at a distance of 53 cm from the axis of rotation when the flywheel is turning at 88 rev/min? QUESTION 1 ABOVE DESCRIBES THE MOTION OF THIS FLYWHEEL.

What is the magnitude of the net linear acceleration of the particle in the above question?

Homework Equations

a=(wr)^2 * 1/r

rev/min = 2rad/min
rad/min = rad/60s
rad/s = (3.14d)m/s

The Attempt at a Solution

I've attempted the problem by doing the conversion and plugging it into the equation, but I cannot get the right answer. Can someone please explain the fault in my reasoning?

 

[physixguru]

omega=176 rev /min =176/60 rev/sec
time taken = 2.8 hours.= 2.8*3600 sec
final omega=0
>>0=176/60 + alpha *2.8*3600
solve for alpha yourself.
you will get the answer in rev/sec^2.
divide it by 60 to get the desired result.

tangential acceleration=v/t
v is given by omega * radius of rotation
>>(176/60 * .53) this gives you the velocity of the particle. in rev/sec.

now tangential accln= v/t
>> [(176/60)*.53] / 2.8*3600

 

[ohhi]

I'm confused as to what numbers to use. Am I using the numbers from the first problem or the second?

 

[ohhi]

I can't get it to work out.

 

[ohhi]

w =(88*.53)/60 *2pi = 4.88 rad/s

a_tan= wr = (4.88 rad/s) * (.53m) = 2.59 rad/s^2 = (pi)(d)(2.59) m/s^2 = 8.61 m/s^2