# Rotating bodies, Car around a corner

1. Nov 13, 2009

### Sakura22

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.
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.php/2/questions/images/rotation/rotation-parallelaxis.png [Broken]
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: May 4, 2017
2. Nov 13, 2009

### Delphi51

Welcome to PF, Sakura.
I can't tell what you did wrong - you haven't shown your work!

3. Nov 13, 2009

### Sakura22

Hey! Sorry about that..but here it goes..
a(tangential) = wr
=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

4. Nov 13, 2009

### Sakura22

For the 2nd question I have no clue, so please give some hints, so I can get something going in my head.

5. Nov 13, 2009

### Staff: Mentor

You found the tangential component of the acceleration. So far, so good!
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.)

6. Nov 13, 2009

### Sakura22

What is the radial component? I don't understand. Is that the centripetal acceleration?

7. Nov 13, 2009

### Staff: Mentor

Eactly! (The "radial" direction is along the radius, thus perpendicular to the tangential direction.)

8. Nov 13, 2009

### Sakura22

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

Last edited: Nov 13, 2009
9. Nov 13, 2009

### Sakura22

10. Nov 13, 2009

### Staff: Mentor

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. Nov 13, 2009

### Staff: Mentor

The diagram is not viewable. Hint: Make use of the parallel axis theorem.

Try posting the diagram to a publically accessible image hosting site.

Last edited by a moderator: May 4, 2017
12. Nov 13, 2009

### Sakura22

http://moodle.phys.ualberta.ca/file.php/2/questions/images/rotation/rotation-parallelaxis.png [Broken]

Last edited by a moderator: May 4, 2017
13. Nov 13, 2009

### Sakura22

I missed the class on The parallel axis theorem, and now I looked it on the wikipedia website, I don't understand it.

14. Nov 13, 2009