Car acceleration around a corner

[TonkaQD4]

A car is going around a quarter-circle from north to east. When halfway around the car's acceleration is (2m/s^2 , 15 degrees south of east). At this point, what is the COMPONENT of acceleration (1) tangent to the circle and (2) perpendicular to the circle?

Well I know that if I draw a line Tangent to the circle and then a line Perpendicular to the Tangent line I have created a X-Y coordinate system that is tilted. After this point I am stuck and not sure where to go.

Any help would be great.

 

[learningphysics]

Draw the circle... with the tangent and perpendicular drawn at the point of interest... draw a radius from the center to the point of interest... draw the vector 2m/s^2 at 15 degrees south of east... the two components of this vector along the perpendicular and the tangent are what you need...

so you need the angle between the tangent line to the circle and the 2m/s^2 line... what is this angle?

draw the horizontal line through the cente of the circle... what is the angle between this line and the radius to the point of interest?

 

[TonkaQD4]

The horizontal line that goes via the circle and the perpendicular line has an angle of 45 degrees. Is that what you were asking about?

 

[learningphysics]

The horizontal line that goes via the circle and the perpendicular line has an angle of 45 degrees. Is that what you were asking about?

yes. exactly. can you find the angle between the tangent to the circle and the 2m/s^2 at 15 degrees south of east vector?

 

[TonkaQD4]

Well yes, because the line that is perpendicular to the tangent line is 90 degrees therefore you could draw a line in both the south and east directions which would be 45 degrees. So, if directly east is 45 degrees, another 15 degrees south of east would make that vector 60 degrees from our tangent line.

 

[learningphysics]

Well yes, because the line that is perpendicular to the tangent line is 90 degrees therefore you could draw a line in both the south and east directions which would be 45 degrees. So, if directly east is 45 degrees, another 15 degrees south of east would make that vector 60 degrees from our tangent line.

wait... from the point of interest... we have a line going east and a line that is tangent... the angle between these two lines is 45 degrees. The 2m/s^2 line is in between these two...

 

[TonkaQD4]

I knew this would get confusing because you don't have a drawing in front of you. I am pretty sure that from our point of interest which lies on the tangent to circle line which is also are x coordinate TO the acceleration vector which is 15 degrees south of east is 60 degrees from the tangent line.

 

[learningphysics]

Have a look at this image I uploaded:

http://www.imagevimage.com/images/circle.GIF

 

[TonkaQD4]

well can u draw it in the second quadrant because that is how it was assigned to us. so draw the point of interest in the upper left hand corner so its like at 135 degrees

 

[learningphysics]

well can u draw it in the second quadrant because that is how it was assigned to us. so draw the point of interest in the upper left hand corner so its like at 135 degrees

Oh, I see. yes, you're right. Sorry about that. Then the angle is 60 degrees... so you should be able to get the 2 components using 2m/s^2 and 60 degrees...

 

[TonkaQD4]

Does this file show up for you

 

[learningphysics]

Does this file show up for you

it says attachment pending approval. it usually takes a while for attachments to get approved...

but I think you should be able to do the problem using the 60 degree angle.

 

[TonkaQD4]

The green line is 15 degrees south of the blue line, the red line is the line perpendicular to the black tangent line. So from the tangent line to the green line should be 60 degrees.

 

[TonkaQD4]

im not sure how though? please help

 

[learningphysics]

im not sure how though? please help

What is the component of the 2m/s^2 along the tangent?

What is the component of the 2m/s^2 along the perpendicular?

Use the 60 degree angle.

 

[TonkaQD4]

By using ...

A_x = A cos60
A_y = A sin60

and plugging 2 in for A

Is this how you solve for the components?

Help!

 

[learningphysics]

By using ...

A_x = A cos60
A_y = A sin60

and plugging 2 in for A

Is this how you solve for the components?

Help!

Yes. The 2 m/s^2 is the hypoteneuse of a right triangle with angle 60.

2cos60 is the tangential acc.
2sin60 is the perpendicular acc.

 

[TonkaQD4]

A_x = Acos60
A_y = Asin60

Where A is 2

Is this how you get the components??

Help!

 

[TonkaQD4]

Woops sorry

Should I be in radians or degrees?

 

[TonkaQD4]

2 cos 60 = 1
2 sin 60 = 1.73

 

[learningphysics]

2 cos 60 = 1
2 sin 60 = 1.73

looks good. units are m/s^2.

 

[TonkaQD4]

Now I am confused. Is this correct?

 

[learningphysics]

Now I am confused. Is this correct?

Yes. Why are you confused?

 

[TonkaQD4]

How should I write my answer?
Because the question is .. At this point, what is the component of a (a) tangent to the circle and (b) perpendicular to the circle?

 

[TonkaQD4]

(1m/s^2, 1.73m/s^2)?

 

[learningphysics]

How should I write my answer?
Because the question is .. At this point, what is the component of a (a) tangent to the circle and (b) perpendicular to the circle?

The component of a tangent to the circle is 1m/s^2. The component of a perpendicular to the circle is 1.73m/s^2.

Do you need the direction for each component?

 

[TonkaQD4]

or is it
(a) (1m/s^2, 60 degrees south of northeast)
(b) (1.73m/s^2, 30 degrees north of southeast)

??

 

[TonkaQD4]

I am not sure on how to label the directions because my original 15 degrees south of east was on a different coordinate system

 

[learningphysics]

1m/s^2 45 degrees north of east

1.73m/s^2 45 degrees south of east

 

[TonkaQD4]

Thank you so much!