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Rotating bodies, Car around a corner

 
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Nov13-09, 10:53 AM   #1
 

Rotating bodies, Car around a corner


1. The problem statement, all variables and given/known data
A car turns a corner with a radius of curvature of 11.1 m while braking to reduce its speed. If the brakes generate an angular deceleration of 0.5 rad/s2 what is the magnitude of the acceleration of the car half way through the corner when the car's linear speed is 9.6 m/s?



2. Relevant equations
Tangential velocity= wr
arc lenth=r(theta)
equation: w^2=(w0)^2 + 2 (angular acceleration)(theta)



3. The attempt at a solution
What I did was I converted the linear speed into angular speed by using the first formula, then I found the time, and halved it, but the answer I'm getting for acceleration HALF WAY is not correct, I have no clue what I did wrong.
Please help
1. The problem statement, all variables and given/known data
A square sheet with a uniform density and total mass m is pivoted about an axis A in one corner of the sheet and perpendicular to the plane of the sheet as shown below. If the moment of inertia of the sheet about this axis is \frac{8}{3}ma^2, what is the sheet's moment of inertia about a parallel axis, B, at the mid-point of one of its sides?

http://moodle.phys.ualberta.ca/file....rallelaxis.png
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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Nov13-09, 11:41 AM   #2
 
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Welcome to PF, Sakura.
I can't tell what you did wrong - you haven't shown your work!
Nov13-09, 12:26 PM   #3
 
Quote by Delphi51 View Post
Welcome to PF, Sakura.
I can't tell what you did wrong - you haven't shown your work!
Hey! Sorry about that..but here it goes..
a(tangential) = wr
= 11.1m x 0.5 rad/s^2
=5.55 m/s^2
Then I used the formula
v=v0 + at
v= at
v= 5.55 m/s^2 x t
t= 9.6 m/s^2 / 5.55 m/s^2
t= 1.73
t/2 because it asks for the deceleration half way through the curve
Then I used the v=at again with half the time = 0.865 ..
then v=at
a=v/t --> 9.6 m/s / 0.865 seconds
= 11.098 m/s^2

Please tell me what I did wrong..thanks again for your help.
Nov13-09, 12:29 PM   #4
 

Rotating bodies, Car around a corner


For the 2nd question I have no clue, so please give some hints, so I can get something going in my head.
Nov13-09, 01:48 PM   #5
 
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Quote by Sakura22 View Post
a(tangential) = wr
= 11.1m x 0.5 rad/s^2
=5.55 m/s^2
You found the tangential component of the acceleration. So far, so good!
Then I used the formula
v=v0 + at
v= at
v= 5.55 m/s^2 x t
t= 9.6 m/s^2 / 5.55 m/s^2
t= 1.73
t/2 because it asks for the deceleration half way through the curve
Then I used the v=at again with half the time = 0.865 ..
then v=at
a=v/t --> 9.6 m/s / 0.865 seconds
= 11.098 m/s^2
Not sure what you're doing here. You need the radial component of the acceleration. Note that they tell you the speed, so no need for any kinematics. (Hint: The motion is circular.)
Nov13-09, 02:05 PM   #6
 
What is the radial component? I don't understand. Is that the centripetal acceleration?
Nov13-09, 02:10 PM   #7
 
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Quote by Sakura22 View Post
Is that the centripetal acceleration?
Eactly! (The "radial" direction is along the radius, thus perpendicular to the tangential direction.)
Nov13-09, 02:22 PM   #8
 
Quote by Doc Al View Post
Eactly! (The "radial" direction is along the radius, thus perpendicular to the tangential direction.)
I am not sure about the formula BUT..here is what I think should work..please tell me if its correct a(centripital)= w^2 r
Nov13-09, 02:33 PM   #9
 
Also, please give me some hints about the second problem, thanks
Nov13-09, 02:33 PM   #10
 
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Quote by Sakura22 View Post
I am not sure about the formula BUT..here is what I think should work..please tell me if its correct a(centripital)= w^2 r
That's perfectly correct. You can also use a different formula for centripetal acceleration (equivalent of course) expressed in terms of tangential speed v, instead of ω.
Nov13-09, 02:40 PM   #11
 
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Quote by Sakura22 View Post
A square sheet with a uniform density and total mass m is pivoted about an axis A in one corner of the sheet and perpendicular to the plane of the sheet as shown below. If the moment of inertia of the sheet about this axis is \frac{8}{3}ma^2, what is the sheet's moment of inertia about a parallel axis, B, at the mid-point of one of its sides?

http://moodle.phys.ualberta.ca/file....rallelaxis.png
The diagram is not viewable. Hint: Make use of the parallel axis theorem.

Try posting the diagram to a publically accessible image hosting site.
Nov13-09, 02:50 PM   #12
 
Nov13-09, 02:52 PM   #13
 
I missed the class on The parallel axis theorem, and now I looked it on the wikipedia website, I don't understand it.
Nov13-09, 02:56 PM   #14
 
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Quote by Sakura22 View Post
I missed the class on The parallel axis theorem, and now I looked it on the wikipedia website, I don't understand it.
Your textbook should describe it. Also read this: Parallel Axis Theorem
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