Rotational Motion

[Jason03]

Heres the problem Im workin on

http://img294.imageshack.us/img294/9166/rotatesm3.jpg [Broken]

Now in order to find the magnitude of accleration Im assuming I would need to find the two components of acceleration first, Centripetal and Tangential......

I found tangetial by the formula

$$a_{t} = r\alpha$$


I converted the 2 revolutions to

$$4\pi radians$$

but im not exactly sure how to get tangential...i need angualr velocity......

how to I get angular velocity from revolutions?

 

[Doc Al]

You'll need to do a little kinematics. You have the angle in radians,the angular acceleration, and the initial angular speed--use kinematic formulas to find the final angular speed. Hint: Angle is the analog to distance, just like angular speed and acceleration are the analogs to linear speed and acceleration.

 

[Jason03]

ok how does this look

$$\omega^2 = \omega_{o} + 2\alpha(\theta-\theta_{o})$$

$$= 0 + 2(.8)(4\pi radians)$$

$$\omega = 4.48 rad/s$$



$$a_{n} = r\omega^2 = (.6)(20.16) = 27.18$$

$$a_{t} = r\theta = (.6)(.8) = .480$$

$$A = \sqrt{(9.480)^2 + (27.18)^2} = 27.18$$

 

[Doc Al]

Your method looks good, but check your arithmetic here:

$$a_{n} = r\omega^2 = (.6)(20.16) = 27.18$$

 

[Jason03]

ohhh thanks....im having trouble reading my calculators display!

that changes the answer to 12.09 rad/s^2

 

[Doc Al]

that changes the answer to 12.09 rad/s^2

That looks better, but be careful with units. The acceleration is in m/s^2.