Newton's 2nd law in circular motion

AI Thread Summary
In the discussion about Newton's 2nd law in circular motion, participants focus on the dynamics of a car negotiating a banked curve at Daytona International Speedway. The key calculations yield a centripetal acceleration of 5.89 m/s² and a speed of 43.1 m/s. The confusion arises regarding the use of cosine for the normal force in circular motion, as opposed to sine in linear motion. It is clarified that the angle of the bank affects the components of the forces, with cosine corresponding to the vertical component due to the orientation of the banking. Understanding the diagram and the angle labeled is crucial for determining which trigonometric function to use for future problems.
miamiphysics42
Messages
4
Reaction score
0

Homework Statement



Daytona International Speedway in Florida. Both of its courses feature four story, 31.0 degree banked curves with max radius of 316 if a car negotiates the curve too slowly it tends to slip down the incline of the turn whearas if it going too fast it may begin to slide up the incline. find the centripetal acceleration on the curve so won't slip up or down and calculate the speed (neglect friction

Homework Equations



F=ma
v^2/r

The Attempt at a Solution



I have the solution it is 5.89 m/s^2 and 43.1 m/s, but i don't understand why the normal force uses cos. It says use the y-component of Newton's second law to solve for the normal force n...so why am I supposed to use cos instead of sin and how can I know for future problems that are similar?

ma= F = n + mg
ncostheta -mg = 0
n= mg/ costheta
F = mg tan theta
centripetal acceleration = g*tan*theta

then v^2/r = 43.1
 
Last edited:
Physics news on Phys.org
I am having trouble understanding why cos is the y-component of New's second law in regards to solving for the normal force of a problem regarding circular motion (car on a banked racetrack) and in linear motion the y-component of Newton's second law was sin. How do I know whether the y component pertains to cos or sin for a circular problem or if it is always cos then why??
 
You have to look at which angle you have labelled. Do you have a specific question?
 
Thanks to whoever combined these.

miamiphysics42 said:
I have the solution it is 5.89 m/s^2 and 43.1 m/s, but i don't understand why the normal force uses cos. It says use the y-component of Newton's second law to solve for the normal force n...so why am I supposed to use cos instead of sin and how can I know for future problems that are similar?
If you draw a diagram it should become apparent. The curve is banked at 30 degrees, so calculating the y component of the weight you will need to use cosine.

Try drawing the diagram, then if you are still stuck ask away.
 
I can't seem to upload the paint picture I drew, but the components as they are drawn in in the book are a vertical line with n cos theta at the top arrow and -mg straight down from it and n sin theta horizontal and pointing to the left and the angle of the road is to the left of the picture. If I was to put these components on an axis it seems like the components of the normal force should be switched with sin corresponding to the y direction. (In linear motion sin was always in the y direction why does it seem different, what am I missing or what I'm I not seeing?) Thanks for your help!
 
http://books.google.com/books?id=wD...ts=fnlv24PLT5&sig=GaX-fnjm4Nv6qe0MtuNgwZEzEVo
has the problem, i don't understand the component diagram to the right of the car, it seems like if you line the axis up with the gravity component on the vertical axis then the normal force you should be using is n sin theta...before without the circular motion it was always mg sin of theta for the y-direction and ramps why do we now use cos for the y-direction doesn't sin always correspond to the y direction?
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top