Struggling with Circular Motion & Magnetic Field Questions

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Homework Help Overview

The discussion revolves around problems related to circular motion and magnetic fields, specifically focusing on a motorcycle stunt known as the 'Wall of Death' and the behavior of an electron in opposing electric and magnetic fields.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants explore the forces acting on a rider and bike in circular motion, questioning how weight is supported and what additional forces are present when transitioning from a sloped to a vertical wall. They also discuss the relationship between electric and magnetic forces on an electron, with some suggesting the use of equations to find speed.

Discussion Status

Some participants have offered guidance on drawing free-body diagrams and understanding force relationships, while others are seeking further clarification on specific forces at play. Multiple interpretations of the forces involved are being explored, particularly regarding the role of friction and centripetal force.

Contextual Notes

Participants express uncertainty about the setup and the forces acting on the rider and bike, indicating a need for additional hints and clarification on the second part of the question regarding the electron's speed.

saltrock
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Circular Mtion and Magnetic field Question ...This is really giving me a hard time!

1) In the 'Wall of Death' motorcycle stunt,the rider gradually rides higher around a curving wall of a circular arena until he is traveling around a vertical wall.

a) show the forces acting on a rider and the bike and explain how their weight is supported.
b)Before reaching the vertical wall ,the rider travels more slowly round the sloping section of the arena.What additional force provides support at this stage?

2)A magnetic field of 0.5 T and an electric field of 5000V/m are arranged to produce opposing forces on an elcetron.calculate the speed at which electron is travel in a straight line through these fields.

I haven't got a clue how to do the above questions.Any hints will highly be appriciated.
 
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RE: 2 - set [itex]\vec E + \vec v \times \vec B = 0[/itex] and obtain

[tex]v = \frac {E}{B}[/tex]

You can figure out the directions
 
1. a) Draw a free-body diagram for the rider/bike: include vectors of all forces acting on the two. What forces are acting on the rider/bike while they approach the vertical wall? Once you do this you will understand how the weight is supported.
b) This question is easy once you get a.

2. This type of system is used as a velocity selector in most mass spectrometers. If the electron travels in a straight line, the electric force must be equal in magnitude to the magnetic force. Therefore:
[tex]|\vec{F}_{e}|=|\vec{F}_{m}|[/tex]
[tex]q|\vec{E}|=q|\vec{v}||\vec{B}|\sin{\theta}[/tex]
where [itex]q[/itex] is the charge on the electron, [itex]\vec{E}[/itex] is the electric field strength, [itex]\vec{B}[/itex] is the magnetic field strength, and [itex]\theta[/itex] is the angle between the velocity vector of the electron and the magnetic field vector. Since the electron is traveling in a direction perpendicular to the magnetic field vector, [itex]\theta[/itex] is 90 degrees. Can you find an expression for the speed of the electron? Plug in the values given in the question to get the numerical answer.

EDIT: I see Tide has developed the expression above. Now you know where it comes from.
 
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I have tried to draw the picture of the track ,the outer boundary(kinda like L shaped curve) is the track while @=the stunt man and ---- = the bike...so,where is the force acting??
I think the weight of both bike and the man is acting downward so,the horizontal component(I don't know which force is that is trying to act to the centre...i think i am wrong and i don't know how the weight is supported..can u please gimme some more hint..Any help would b highly appriciated
 
The man is also being accelerated by a normal force to the wall of the ring (otherwise he would be going in a straight line at constant velocity by Newton's laws). This is a centripetal force, which you should already know how to find the magnitude of. :smile:
Given that the man is experiencing a force normal to the surface, there is one more non-zero force that is present. Do you see what it is ?
 
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is it the frictional force between the tyre and the wall...but which direction is it acting on... refer to the picure above i have drawn..on that,centripetal force is to the right while weight is to the left...so should the frictional force be acting downward...is it how thw weight is supported?



can u please help me answer the second part of my question (question is on very top)

thaks for the help in advance
 

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