Centripetal Acceleration Problem

[sap_54]

Hello,
I am having the worst time figuring out how to solve this problem:

A student build and calibrates an acclerometer, which she uses to determine the speed of her car around a certain unbanked highway curve. The accelerometer is a plumb bob with a prtractor that she attaches to the roof of her car. A friend riding in the car with her observes that the plumb bob hangs at an angle of 15 degrees from the vertical when the car has a speed of 23m/s. What is the centripetal acceleration of the car?

Ok, I have my force equations:

The sum of the forces in the x=-Tcos75=-(mv^2)/r
The sum of the forces in the y=Tsin75-mg=0
I substituted T for mv^2/r, so mass cancels out, but I can't seem to figure out where to go from there.

Any help would be greatly appreciated!

 

[carlz]

I didnt use the speed.

resolve Tcos15 = m 9.81
Tsin15 = ma

divide

9.81tan15 = a

a = 2.628ms^-2

 

[quicknote]

Hi there,

Firstly, great job on setting up your force equations! It looks like you're on the right track. To solve for the centripetal acceleration, you will need to use the equation a = v^2/r, where v is the velocity and r is the radius of the curve. In this case, the velocity is given as 23m/s and the radius is the distance from the center of the curve to the car's position.

To find this radius, you will need to use trigonometry. Since the plumb bob hangs at an angle of 15 degrees from the vertical, you can use the sine function to find the ratio of the opposite side (the distance from the center of the curve to the car's position) to the hypotenuse (the distance from the center of the curve to the roof of the car). This will give you the value of r.

Once you have the value of r, you can plug it into the centripetal acceleration equation along with the given velocity of 23m/s to solve for the acceleration. I hope this helps. Keep up the good work!