Solving Rotational Motion Problem: Find Acceleration

Click For Summary
To find the acceleration in a rotational motion problem, both centripetal and tangential components need to be calculated. The tangential acceleration is derived using the formula a_t = rα, where α is angular acceleration. Angular velocity can be determined from revolutions by converting them to radians and applying kinematic equations. The calculations for centripetal acceleration involve using a_n = rω², while ensuring unit consistency is crucial. The discussion emphasizes careful arithmetic and unit conversion to arrive at accurate results.
Jason03
Messages
161
Reaction score
0
Heres the problem I am workin on

http://img294.imageshack.us/img294/9166/rotatesm3.jpg

Now in order to find the magnitude of acceleration I am 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 I am not exactly sure how to get tangential...i need angualr velocity...

how to I get angular velocity from revolutions?
 
Last edited by a moderator:
Physics news on Phys.org
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.
 
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
 
Last edited:
Your method looks good, but check your arithmetic here:
Jason03 said:
a_{n} = r\omega^2 = (.6)(20.16) = 27.18
 
ohhh thanks...im having trouble reading my calculators display!

that changes the answer to 12.09 rad/s^2
 
Jason03 said:
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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 14 ·
Replies
14
Views
1K
Replies
12
Views
2K
Replies
26
Views
1K
  • · Replies 12 ·
Replies
12
Views
965
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
1
Views
826