Please anyone help me out ASAP!!!Circular motion Questions

[saltrock]

Hey guys,i was given about 50 numericals to do related to circular motion .I did most of it but got stuck in the following questions:

1)The gravitational field near the equator is less than that at the poles.It is partially accounted for by the fact that earth rotates about polar axis.At poles the earth does not move.At the equator,the gravitational pull has to also provide a centripetal acceleration.Calculate the acceleration which represents some of the difference in the gravitational field?(r=6.4 times 10^6)

2)A car travels over a humpback bridge at the speed of 20 m/s.
a)calculate the minium radius of the brifge if the car's road wheels are to remain in contact with the bridge?
b)what happens if the radius is less than this limiting value?

3)At what angle should ta road with 150 m curvature be banked for travel at 75 km/hr?

If you guys can answer any of these questions please please answer me back.I'd really be grateful to you.Thanyou very much. :smile:

[Doc Al]

Welcome to PF saltrock!

You'll get plenty of help here, but you have to show some work. Here are a few hints:

Problem 1: What's the formula for centripetal acceleration? I assume you know the period of rotation of the earth!

Problem 2: Draw yourself a diagram of the car going over the bridge. It's going in a circle, so what must be its acceleration? The tighter the circle (smaller the radius) the greater the centripetal acceleration. What forces act on the car to provide that acceleration?

Problem 3: What provides the centripetal force on the car? To get the proper banking angle, ignore friction.

[saltrock]

Hi DOC.Cheers for your help.Heres my working

a) given,
radius= 6.4 times 10^6
using a= rw^2 w=2pie/T
a=6.4 times 10^6.5.2 times 10^-9=0.34 m/s^2

[saltrock]

2) I think i should use this forumla
v=rw i have got v , i need to calculate r, i also need to get the value of w which i can calculate by using w=2pie/T but i dont know T so this formula is useless .I dont know any other method od doing this .so i dont know how to do it.can you help me do this please.Thank you very much.

[Doc Al]

a) given,
radius= 6.4 times 10^6
using a= rw^2 w=2pie/T
a=6.4 times 10^6.5.2 times 10^-9=0.34 m/s^2
Your method is fine, but check your arithmetic. You're off by a factor of 10.

[Doc Al]

2) I think i should use this forumla
v=rw i have got v , i need to calculate r, i also need to get the value of w which i can calculate by using w=2pie/T but i dont know T so this formula is useless .I dont know any other method od doing this .so i dont know how to do it.can you help me do this please.Thank you very much.
The weight provides the centripetal force holding the car onto the road as it goes over the curved bridge. The maximum force is the weight of the car, so use F = ma for centripetal acceleration. Note that centripetal acceleration is give by a_c = \omega^2 r = v^2/r, the two versions being related by v = \omega r. So, F = ma leads to mg = mv^2/r. Solve this for r.

[saltrock]

on question no theree i dont know what formula i should be using..if you can give me some tips i'll try to solve it.thanks in advance

[Doc Al]

on question no theree i dont know what formula i should be using..if you can give me some tips i'll try to solve it.thanks in advance
First, as always, draw a picture of the car on the inclined road. Identify all the forces on the car and their directions: weight of car (downwards), normal force of road (perpendicular to road surface).

Then consider these facts:
(1) The centripetal force is provided by the horizontal component of the normal force.
(2) The weight must be balanced by the vertical component of the normal force.

Express these facts mathematically and you'll get a relationship between road angle, speed, and curve radius. Give it a shot.