How Do You Calculate a Car Wheel's Angular Acceleration and Total Acceleration?

[122760]

1. A cars wheels starts from rest and accelerate to 210 rpm in .73 seconds.
a.) Find angular acceleration in radians/sec^2
b.) Find the revolutions completed in the time interval
c.) Find the total acceleration when it is 12m from the rotational axis at 180 rpm.

3. 1. a.) 210x2pi= radians/ min
1319.47/60= 21.99 radians/sec giving omega
21.99/.73= 30.12 radians/sec^2 giving angular acceleration. I think I did this right the whole revolution thing is throwing me for a loop.

b.) I think i just plug in everything into the formula W=Wo + at? If i do that, I get 0+30.12(.73)=21.99rad/sec
Then, I have to covert back to revolutions right? So, 21.99/2pi = 3.5 revolutions? Does that sound right?

c.) I know total acceleration is Centripetal plus tangential but after that I'm completely lost. Any help would be greatly appreciate I just don't get this. =/

 

[Xerxes1986]

For the first part...you have 210 Revolutions Per Minute...converting this to radians/sec we have:

\frac{210 Revolutions}{Minute}*\frac{1 Minute}{60 Seconds}*\frac{2\pi}{1 Revolution}

Which i get about 21.99 rad/s, which is the same as yours so you did it right. It always helps to draw it out like in the above equation and strike through the units that cancel so you know what units you are left with.

For the second part, finding the revolutions, you did that part wrong. Think about it this way. If I gave you the acceleration of an object, give me an equation that determines it's position as a function of time. (Like a ball dropping). Those kinematic equations have angular equivalents.

And for the 3rd part, look at your book. They have an equation for the centripetal acceleration (most likely v^2 / r), but you can relate v (linear velocity) with w (omega, angular velocity). They give you angular velocity and r so you can calculate centripetal acceleration.

 

[122760]

Ok for the second part I used the equation \theta=\varpit + 1/2\alphat^2

So far that I got .5*30.12*(.73)^2= 8.03 radians
8.03/2\pi = 1.3 revs??

For the third part I have r=12m and 180 rpm which equals (180*2pi)/60= 18.85rads/sec then 18.85/.73= 25.82 rad/sec^2
That gives me omega so Ac=12*18.85^2 = 4263.87and At= 12(25.82) = 309.84
4263.87^2 + 309.84^2= square root answer and I get 4275.11 that feels miserably wrong.

 

[Doc Al]

Ok for the second part I used the equation \theta=\varpit + 1/2\alphat^2

So far that I got .5*30.12*(.73)^2= 8.03 radians
8.03/2\pi = 1.3 revs??

Sounds good.

Is part three part of the same problem? A car wheel with radius 12 m??

For the third part I have r=12m and 180 rpm which equals (180*2pi)/60= 18.85rads/sec then 18.85/.73= 25.82 rad/sec^2

Why did you divide by 0.73 seconds? You already found the angular acceleration in part one.

That gives me omega so Ac=12*18.85^2 = 4263.87and At= 12(25.82) = 309.84
4263.87^2 + 309.84^2= square root answer and I get 4275.11 that feels miserably wrong.

You'll have to correct At, but otherwise this seems correct. But these numbers seem wildly unrealistic.