Rotating bodies, Car around a corner

AI Thread Summary
A car is turning a corner with a radius of 11.1 m while experiencing angular deceleration of 0.5 rad/s², and the task is to find the car's acceleration halfway through the turn at a speed of 9.6 m/s. The discussion reveals confusion regarding the calculation of acceleration, with initial attempts focusing on tangential acceleration rather than incorporating the necessary radial (centripetal) acceleration. Participants clarify that the radial component must be considered, as it is perpendicular to the tangential direction. Additionally, hints are provided for solving a separate problem involving the moment of inertia of a square sheet using the parallel axis theorem. Understanding both the centripetal acceleration and the parallel axis theorem is essential for solving these physics problems correctly.
Sakura22
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Homework Statement


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?



Homework Equations


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



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

Homework Statement


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.php/2/questions/images/rotation/rotation-parallelaxis.png
 
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Welcome to PF, Sakura.
I can't tell what you did wrong - you haven't shown your work!
 
Delphi51 said:
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.
 
For the 2nd question I have no clue, so please give some hints, so I can get something going in my head.
 
Sakura22 said:
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.)
 
What is the radial component? I don't understand. Is that the centripetal acceleration?
 
Sakura22 said:
Is that the centripetal acceleration?
Eactly! (The "radial" direction is along the radius, thus perpendicular to the tangential direction.)
 
Doc Al said:
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
 
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Also, please give me some hints about the second problem, thanks
 
  • #10
Sakura22 said:
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.
 
  • #11
Sakura22 said:
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.php/2/questions/images/rotation/rotation-parallelaxis.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.
 
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  • #12
http://moodle.phys.uAlberta.ca/file.php/2/questions/images/rotation/rotation-parallelaxis.png
 
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  • #13
I missed the class on The parallel axis theorem, and now I looked it on the wikipedia website, I don't understand it.
 
  • #14
Sakura22 said:
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: http://hyperphysics.phy-astr.gsu.edu/hbase/parax.html"
 
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