Heya,
The time for a "I don't really know what I'm doing" questio has come again 😉
"A circular bike track is being constructed. So that the bikers won't be pushed out, the track will be higher on the outside. The radius of the track is 50 metres.
What is the optimal angle for the track for a biker going the consistent speed of 36 km/h so that he won't be pushed out even on a very slippery surface?
(Sorry if it sounds weird, I translated it from German)
My idea was that the normal force the biker exerts on the ground would have to be greater than the centrifugal force.
But somehow, I get an angle of 78.24°... which seems a bit too much, doesn't it?
Thanks for your help,
Angie
Originally posted by angie88take it to posers and puzzles spanky!
Heya,
The time for a "I don't really know what I'm doing" questio has come again 😉
"A circular bike track is being constructed. So that the bikers won't be pushed out, the track will be higher on the outside. The radius of the track is 50 metres.
What is the optimal angle for the track for a biker going the consistent speed of 36 km/h so that he won't ...[text shortened]... gle of 78.24°... which seems a bit too much, doesn't it?
Thanks for your help,
Angie
Originally posted by angie88Are you building a velodrome, or is this homework?
Heya,
The time for a "I don't really know what I'm doing" questio has come again 😉
"A circular bike track is being constructed. So that the bikers won't be pushed out, the track will be higher on the outside. The radius of the track is 50 metres.
What is the optimal angle for the track for a biker going the consistent speed of 36 km/h so that he won't ...[text shortened]... gle of 78.24°... which seems a bit too much, doesn't it?
Thanks for your help,
Angie
Originally posted by angie88Manchester velodrome is a max of 42.5° and that is classed as frighteningly steep!!!
Heya,
The time for a "I don't really know what I'm doing" questio has come again 😉
"A circular bike track is being constructed. So that the bikers won't be pushed out, the track will be higher on the outside. The radius of the track is 50 metres.
What is the optimal angle for the track for a biker going the consistent speed of 36 km/h so that he won't ...[text shortened]... gle of 78.24°... which seems a bit too much, doesn't it?
Thanks for your help,
Angie
http://www.manchestervelodrome.com/static_info/about_us.htm
Originally posted by cheshirecatstevensI do hope that if I was building a velodrome, I'd be able to answer the question myself 😉
Are you building a velodrome, or is this homework?
It's part of the weekly physics questionsheet which I have to turn in. Just fyi, I am not really taking physics because I like it. I want to take astronomy as a side subject for my maths diploma, but in order to do that, I have to take physics first. No getting around that 😕
Originally posted by angie88how slippery?
Heya,
The time for a "I don't really know what I'm doing" questio has come again 😉
"A circular bike track is being constructed. So that the bikers won't be pushed out, the track will be higher on the outside. The radius of the track is 50 metres.
What is the optimal angle for the track for a biker going the consistent speed of 36 km/h so that he won't ...[text shortened]... gle of 78.24°... which seems a bit too much, doesn't it?
Thanks for your help,
Angie
if there is no friction, it is imnpossible...
the question, I believe, is only possible if we know the coefficient of friction between the tires and the surface. also, the mass of the bike and rider must be known. also, centrifugal force doesn't hold the bike on the track... centrifugal force is a myth. centripetal force keeps the thing on the track. and centripetal force can't be determined without the coefficient of static friction. do not confuse it with kinetic friction!
Originally posted by rubberjaw30It's still possible, even in the case of a frictionless track. Imagine the case of a tilted track and a frictionless surface. Now you magically place a cyclist and his bike onto the track with giant invisible genie fingers (or Green Lanter's ring, etc), you place the cyclist aligned perpendicular to the surface of the track (neither leaning left nor right relative to the track surface) AND the cyclist is not moving at all... you just place the cyclist down, and let go. What happens? The cyclist will not slide up the track toward the outer edge, instead the cyclist will fall over and slide down the track toward the inner edge. I think you can see that as you experimentally increase the cyclists speed on this frictionless, angled track, you will find a point of equilibrium wh
how slippery?
if there is no friction, it is imnpossible...
the question, I believe, is only possible if we know the coefficient of friction between the tires and the surface. also, the mass of the bike and rider must be known. also, centrifugal force doesn't hold the bike on the track... centrifugal force is a myth. centripetal force keeps the ...[text shortened]... etermined without the coefficient of static friction. do not confuse it with kinetic friction!
I have said too much already.
Originally posted by Doctor Raton a frictionless surface, the tires have nothing to grip to, so the bike could never move, anyway.
It's still possible, even in the case of a frictionless track. Imagine the case of a tilted track and a frictionless surface. Now you magically place a cyclist and his bike onto the track with giant invisible genie fingers (or Green Lanter's ring, etc), you place the cyclist aligned perpendicular to the surface of the track (neither leaning left nor right ...[text shortened]... onless, angled track, you will find a point of equilibrium wh
I have said too much already.
however, assuming that this invisible hand gives the cycist a hove up to a certain speed.
and the cyclist will slide down the track even if moving tat half the speed of light! with no force to oppose gravity, it will move down the track. in real systems, friction opposes this downward force due to gravity, so it is possible, so long as the coefficient of friction is high enough.
Originally posted by rubberjaw30On a circular angled track, most assuredly, if pushed at half-the speed of light, the bicycle would definitely go over the high end of the track. Break the gravity and centripetal down into component forces vertical and horizontal, using your trusty sin and cos!, you will find
on a frictionless surface, the tires have nothing to grip to, so the bike could never move, anyway.
however, assuming that this invisible hand gives the cycist a hove up to a certain speed.
and the cyclist will slide down the track even if moving tat half the speed of light! with no force to oppose gravity, it will move down the track. in real systems, ...[text shortened]... force due to gravity, so it is possible, so long as the coefficient of friction is high enough.
I have said too much already.
Here's my solution, angie:
First draw a free body diagram for the cyclists on a slope. There are two forces present, the force of gravity and the normal force. We know that the force of gravity acts straight down, but the normal force acts perpendicular to the slope surface. These two forces sum to give a resultant force on the cyclist parallel to the slope surface and downward in direction (i.e. the biker will tend to fall down the slope if no friction or other forces oppose it). The magnitudes of these forces are:
F(g) = mg
F(n) = F(g)*cos(theta) = mg*cos(theta)
F(r) = F(g)*sin(theta) = mg*sin(theta)
Because the problem says "very slippery conditions", we can assume the worst case (frictional force = 0). Therefore, the only force available to balance the centripetal force is the component of F(r) that points straight inward, which is just the x-component of F(r). The magnitude of this force is:
F(r)x = F(n)*sin(theta) = mg*sin(theta)*cos(theta) = 0.5*mg*sin(2*theta)
This force must balance the centripetal force required to turn the cyclist in its path, which is given by:
F(c) = m*v^2/r
Therefore, the final equation to solve for "theta" will be:
F(c) = F(r)x
m*v^2/r = 0.5*mg*sin(2*theta)
2v^2/rg = sin(2*theta)
theta = 0.5*arcsin(2v^2/rg)
Plugging in the numbers, we get:
theta = 0.5*arcsin(2*(10 m/s)^2/(50 m)*(9.81 m/s^2))
Using Excel, I got:
theta =~ 0.210 radians = 12.0 degrees
So the optimal slope for frictionless conditions is a slope of 12.0 degrees.
Hope that's right, good luck angie!
Originally posted by PBE6angie's answer was 78 degrees, and PBE6's answer was 12 degrees, and I wondered if that was exactly how that was going to turn out. 78 + 12 = 90.
Here's my solution, angie:
First draw a free body diagram for the cyclists on a slope. There are two forces present, the force of gravity and the normal force. We know that the force of gravity acts straight down, but the normal force acts perpendicular to the slope surface. These two forces sum to give a resultant force on the cyclist parallel to the slo ...[text shortened]... frictionless conditions is a slope of 12.0 degrees.
Hope that's right, good luck angie!
I have said too much already.