Tangential and Centripetal Acceleration

[kwenz21]

Homework Statement

A bike is travel up and over a hill with a radius at the top of the hill of 10 meters. The bike is traveling at 10 m/s at position A and 5 m/s at position. (and the problem comes with a picture showing an angle of 60* between point A and B.

a. Determine the average tangential acceleration while slowing down
b. Determine the centripetal acceleration
c. Determine the net acceleration and its direction

Homework Equations

I used a(centripetal)=v^2/r and I was unsure of how to get tangential

The Attempt at a Solution

I evantually came out with a being equal to -2.5 m/s^2 (-5^2/10) and I was unable to get tangential just putting down my best guess which made it impossible for me to get a and c correct

 

[Andrew Mason]

Homework Statement

A bike is travel up and over a hill with a radius at the top of the hill of 10 meters. The bike is traveling at 10 m/s at position A and 5 m/s at position. (and the problem comes with a picture showing an angle of 60* between point A and B.

a. Determine the average tangential acceleration while slowing down
b. Determine the centripetal acceleration
c. Determine the net acceleration and its direction

Homework Equations

I used a(centripetal)=v^2/r and I was unsure of how to get tangential

The Attempt at a Solution

I evantually came out with a being equal to -2.5 m/s^2 (-5^2/10) and I was unable to get tangential just putting down my best guess which made it impossible for me to get a and c correct

Is the cyclist applying force to the pedals as he goes up the hill? In other words, is the cyclist adding energy to the bike or is the bike coasting up the hill?

I think you will have to provide the diagram or provide some more details of where A and B are in terms of horizontal and vertical distance from the top and base of the hill.

AM

 

[kwenz21]

This is the exact problem that was given to me. Sorry if some parts are hard to see the only relevant things I can think of that are in the picture and not the descriptionis that the radius of the circle is 10m and the angle between A and B is 60*

 

[ehild]

Average acceleration means that constant acceleration which would result in the same change of velocity and distance traveled in the same time. Assuming constant tangential acceleration and the path taken as pi/3 R, and knowing the initial and final speeds, how do you determine the acceleration?

As for the centripetal acceleration, the text does not specify if it is at A or B or at the maximum height. ehild

 

[Nickie]

.

Thank you for providing your attempt at a solution. It seems like you are on the right track with using the equation a(centripetal) = v^2/r to solve for the centripetal acceleration. However, in order to find the tangential acceleration, you will need to use the equation a(tangential) = (vf - vi)/t, where vf is the final velocity, vi is the initial velocity, and t is the time.

To solve for a(tangential), you will need to first determine the time it takes for the bike to travel from position A to position B. This can be found using the equation vf = vi + at, where vf is the final velocity (5 m/s), vi is the initial velocity (10 m/s), and a is the unknown acceleration. Rearranging the equation, we get t = (vf - vi)/a.

Substituting in the known values, we get t = (5 - 10)/a = -5/a. Now, we can plug in this value of t into the equation a(tangential) = (vf - vi)/t to solve for a(tangential). This gives us a(tangential) = (5 - 10)/(-5/a) = -a. Therefore, the average tangential acceleration while slowing down is also equal to -a.

To find the net acceleration, we can use the Pythagorean theorem to combine the tangential and centripetal accelerations. This gives us a(net) = √(a(tangential)^2 + a(centripetal)^2). We can also find the direction of the net acceleration by using the equation tanθ = a(centripetal)/a(tangential), where θ is the angle between the net acceleration and the tangential acceleration.

I hope this helps and clarifies any confusion you had. Keep up the good work!