A couple of angular accelaration problems

[backseatgunner]

I'm having many a woes with angular accelaration in my DE Physics class. Could you guys explain to me how to do these 3 problems. (If I can get one of them by logical deduction from the other problems don't go out of your way.) Thanks.

A merry-go-round accelerating uniformly from rest achieves its operating speed of 2.5rpm in five revolutions. What is the magnitude of its angular accelaration?

A 33 1/3-rpm record on a turntable uniformly reaches its operating speed in 2.45s once the record player is turned on. (a) What is the angular distance traveled during this time? (b) What is the corresponding arc length in feet on the circumfrence of a 12in. diamater record?

A bicycle being repaired is turned upside down, and one wheel is rotated at a rate of 60 rpm. If the wheel slows uniformly to a stop in 15s, how many revolutions does it take during this time?

 

[Päällikkö]

You haven't showed what you've tried.

$$\theta = \theta _0 + \omega t + \frac{1}{2}\alpha t^2$$

 

[quicknote]

Angular acceleration is the rate of change of angular velocity. It is measured in radians per second squared (rad/s^2). In order to solve these problems, we need to use the equation for angular acceleration:

α = (ωf - ωi)/t

Where α is the angular acceleration, ωf is the final angular velocity, ωi is the initial angular velocity, and t is the time interval.

For the first problem, we are given the final angular velocity (2.5 rpm) and the time interval (5 revolutions). We can convert the final angular velocity to radians per second by multiplying it by 2π (since 1 revolution = 2π radians):

ωf = 2.5 rpm * 2π = 5π rad/s

We can also convert the time interval to seconds by multiplying it by the period of one revolution (1/2.5 rpm = 0.4 s/rev):

t = 5 rev * 0.4 s/rev = 2 s

Plugging these values into the equation, we get:

α = (5π rad/s - 0 rad/s)/2 s = 2.5π rad/s^2

So the magnitude of the angular acceleration is 2.5π rad/s^2.

For the second problem, we are given the final angular velocity (33 1/3 rpm) and the time interval (2.45 s). We can convert the final angular velocity to radians per second by multiplying it by 2π:

ωf = 33 1/3 rpm * 2π = 66 2/3π rad/s

We can also convert the time interval to seconds by multiplying it by the period of one revolution (1/33 1/3 rpm = 0.03 s/rev):

t = 2.45 s * 0.03 s/rev = 0.0735 s

Plugging these values into the equation, we get:

α = (66 2/3π rad/s - 0 rad/s)/0.0735 s = 907.2π rad/s^2

For part (a) of the problem, we can use the equation for angular displacement:

θ = ωi * t + 1/2 * α * t^2

Where θ is the angular displacement, ωi is the initial angular velocity, α