How Does the Coriolis Effect Influence a Falling Object's Path at Latitude 44?

AI Thread Summary
The discussion focuses on understanding the Coriolis effect's impact on a falling object's trajectory at latitude 44, specifically in Florence, Italy. The key equations involved are modified to account for the Earth's rotation, emphasizing the use of the perpendicular component of velocity, expressed as v * cos lambda. Participants seek clarification on how to apply these equations to determine the horizontal deflection of a lead ball dropped from a height of 200 meters. The conversation highlights the need for a clear approach to solving angular momentum problems in the context of the Coriolis effect. Overall, the thread aims to enhance comprehension of the physics behind falling objects influenced by Earth's rotation.
giancoli
Messages
1
Reaction score
0
Angular Momentum Problem, Please Help!

Homework Statement


We can alter equations s=omega*v*t^2 and a_cor= 2omega*v for use on Earth by considering only the component of v perpendicular to the axis of rotation. From the figure (Intro 1 figure) , we see that this is v * cos lambda for a vertically falling object, where lambda is the latitude of the place on the Earth. If a lead ball is dropped vertically from a 200 m-high tower in Florence, Italy latitude 44 how far from the base of the tower is it deflected by the Coriolis force? Caption: Object of mass m falling vertically to Earth at a latitude lambda.


Homework Equations





The Attempt at a Solution


I am quite confused by the Coriolis effect and would appreciate any pointers on how to approach this problem.
 
Physics news on Phys.org


giancoli said:
alter equations s=omega*v*t^2 and ...

considering only the component of v perpendicular to the axis of rotation. ... this is v * cos lambda for a vertically falling object

Can you use the statement about "v * cos lambda" to modify this expression:
omega*v*t^2

p.s. Welcome to PF.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top