What is the maximum velocity of a motorbike on a curved track?

[Saitama]

Homework Statement

Homework Equations

The Attempt at a Solution

I am not sure which equations to start with here.

Let $\mu$ be the coefficient of friction.
The maximum velocity is attained when it is just about to slip i.e
\frac{mv_{max}^2}{R}=\mu mg
where m is the mass of motorbike.
\Rightarrow v_{max}=\sqrt{\mu gR}
Next I think I have to use the energy conservation but I don't know the force which helps the bike to accelerate.

 

[Basic_Physics]

I think you need to take into consideration that the cyclist is leaning inwards.

 

[tia89]

I think you need to take into consideration that the cyclist is leaning inwards.

I don't think that it would change anything being the circuit flat (at least there are no indications it is inclined). You may see the motorbike as a "point particle", or in any way gravity has no component in the horizontal direction so it doesn't seem to me it would change things.

As for the rest, I think you can assume constant acceleration to reach total maximum speed, and simply use circular accelerated motion for the tangential part.

 

[Basic_Physics]

http://cnx.org/content/m42086/latest/?collection=col11406/1.7
Chapter 6 Centripetal force Exercise 6

 

[tia89]

http://cnx.org/content/m42086/latest/?collection=col11406/1.7
Chapter 6 Centripetal force Exercise 6

I see what you mean, but I think that here it does not enter. It is right of course that the angle depends on the speed you are going, but this not to let the cyclist fall down. I don't think that this would permit a higher speed, as the forces are anyway the same. I mean the normal force is anyway only vertical and therefore has to be the same as weight, otherwise it would jump. Also the friction force is just given by the normal and therefore nothing changes. This is why I was thinking that in the problem in question you could consider the cyclist a point particle.

I could be wrong of course, my two cents...

 

[tia89]

Saying it perhaps better, to lean is the only way to be able to reach the maximal velocity allowed without killing yourself, but does not influence the value of this maximal velocity

 

[NascentOxygen]

I see friction playing a rôle in two places here. Below a certain speed it stops the cyclist sliding off the track, but it also limits him to a certain acceleration which if he attempts to exceed he'll spin his back wheel.

 

[tia89]

it also limits him to a certain acceleration which if he attempts to exceed he'll spin his back wheel.

No, it is explicitly stated in the assumptions that the motorbike never slip as it accelerates

 

[TSny]

... but I don't know the force which helps the bike to accelerate

It's the same force that let's you start walking from rest.

 

[Basic_Physics]

The angle at which he leans will change from 0 degrees to a maximum as he accelerates around the track. This is because the speed around the track changes. One should therefore not assume that the centripetal acceleration is constant. It is the problem exercise 6 I am referring to in the link http://cnx.org/content/m42086/latest/?collection=col11406/1.7

 

[NascentOxygen]

No, it is explicitly stated in the assumptions that the motorbike never slip as it accelerates

You are re-stating precisely what I said.

 

[tia89]

The angle at which he leans will change from 0 degrees to a maximum as he accelerates around the track. This is because the speed around the track changes. One should therefore not assume that the centripetal acceleration is constant.http://cnx.org/content/m42086/latest/?collection=col11406/1.7

You are completely right, but I do not see where it comes in in the problem. Here we are trying to find the maximum velocity it can reach... no doubt the centripetal will have to increase with velocity (by definition also, without speaking of angles which are just a consequence of that) but here we concentrate in the moment we actually reach this maximum velocity. To do so you impose to be at maximum friction allowable, which is $F=\mu N$, and compute the velocity when this force is the centripetal force. And we are done in this way. Then of course we can consider that to manage not to fall, at this speed we will have to lean a certain angle, but this is not essential in finding the maximum speed.

Then we want to find the space the man needs to reach such maximum velocity. We have initial velocity, final velocity (i.e. $v_{max}$) and the time it needs, then also here (assuming constant TANGENTIAL acceleration and no sliding as stated in the problem) we have all the data... using uniformly accelerated circular motion we can find the space needed to reach $v_{max}$. And the whole problem is solved.

 

[TSny]

assuming constant TANGENTIAL acceleration...

I don't think we want to assume constant tangential acceleration. Friction will allow only a certain maximum amount of net acceleration. As the motorbike speeds up, more and more of the acceleration will need to be centripetal, leaving less acceleration for tangential.

 

[tia89]

Friction will allow only a certain maximum amount of net acceleration. As the motorbike speeds up, more and more of the acceleration will need to be centripetal, leaving less acceleration for tangential.

I can agree with you if you said it need more and more power to speed up due to friction, but centripetal and tangential acceleration are not related, so I don't see why you can't have a constant tangential acceleration... also because if then tangential acceleration is not constant, then it would be loads more difficult to solve the problem without an expression of acceleration as function of time...

 

[TSny]

so I don't see why you can't have a constant tangential acceleration...

Well, we'll see. I think we should wait for Pranav-Arora to return to the discussion.

 

[tia89]

Yeah that's the best thing... and I could end out to be wrong, even if I am still convinced of what i was saying ;) anyway let's see what he says :)

 

[Saitama]

I am still lost on this one even after the replies.

Should I consider that motorbike leans at a certain angle and make the equations?

It's the same force that let's you start walking from rest.

Friction?

I don't think we want to assume constant tangential acceleration. Friction will allow only a certain maximum amount of net acceleration. As the motorbike speeds up, more and more of the acceleration will need to be centripetal, leaving less acceleration for tangential.

How both the accelerations are related?

 

[TSny]

Yes, friction is the force. What is the magnitude of the maximum acceleration that the bike can have? How would you express this magnitude of acceleration in terms of tangential and centripetal components?

 

[Saitama]

Yes, friction is the force. What is the magnitude of the maximum acceleration that the bike can have? How would you express this magnitude of acceleration in terms of tangential and centripetal components?

Maximum acceleration can be $\mu g$?

It can be expressed as:
\left(\frac{mv_{max}^2}{R}\right)^2+(ma_t)^2=(\mu g)^2
where $a_t$ is the tangential acceleration. Does this look correct?

 

[ehild]

Does this look correct?

NO. Check.

ehild

 

[Saitama]

NO. Check.

ehild

Oops, I think "m" won't come in there as I am dealing with accelerations. Right?

 

[ehild]

yes.

 

[Saitama]

yes.

Thanks but what am I supposed to do next?

 

[ehild]

The bike never slips. To reach the maximum speed in the shortest time, its tangential acceleration must be as high as possible. What does it mean for the tangential acceleration at any time?

 

[Saitama]

Find the distance traveled till v(s) becomes the maximum value allowed.

I am not sure but should I differentiate the above expression after expressing v in terms of a_t?

 

[ehild]

a_t=dv/dt and v=ds/dt where s is the arc length travelled. If you know the maximum allowed tangential acceleration at any time you get the arc length by integration.

 

[TSny]

See if you can express a_t in terms of dv/ds rather than dv/dt. Here, s is distance traveled. That way you won't need to worry with time. The trick is to use the chain rule. Think of v as a function of s while s is a function of t: v = v(s(t)). dv/dt = ?

 

[Saitama]

See if you can express a_t in terms of dv/ds rather than dv/dt. Here, s is distance traveled. That way you won't need to worry with time. The trick is to use the chain rule. Think of v as a function of s while s is a function of t: v = v(s(t)). dv/dt = ?

$a_t=v(dv/ds)$?

Do I have to plug this in the equation I wrote a few posts ago?

 

[ehild]

Which equation do you mean?

ehild

 

[Saitama]

Which equation do you mean?

ehild

This one:
\left(\frac{v_{max}^2}{R}\right)^2+(a_t)^2=(\mu g)^2
I think I am wrong about substituting $a_t=v(dv/ds)$ in this equation as it is only applicable for max acceleration.

 

[TSny]

This one:
\left(\frac{v_{max}^2}{R}\right)^2+(a_t)^2=(\mu g)^2
I think I am wrong about substituting $a_t=v(dv/ds)$ in this equation as it is only applicable for max acceleration.

You can always write $a_c^2 + a_t^2 = a^2$ at any time during the motion, not just at the time when you reach maximum speed. See what you get if you substitute $a_t = v dv/ds$ in this equation with $v$ not at max value.

 

[Saitama]

You can always write $a_c^2 + a_t^2 = a^2$ at any time during the motion, not just at the time when you reach maximum speed. See what you get if you substitute $a_t = v dv/ds$ in this equation.

I tried substituting it but I get a differential equation which is not solvable. :(

 

[TSny]

I tried substituting it but I get a differential equation which is not solvable. :(

What equation did you get?

 

[Saitama]

What equation did you get?

Sorry, I replied hastily before.

\left(\frac{v^2}{R}\right)^2+\left(v\frac{dv}{ds}\right)^2=(\mu g)^2
It can be rewriten as (I am substituting $\mu g=k$)
\frac{dv}{ds}=\frac{1}{vR}\sqrt{R^2k^2-v^4}
\frac{vR dv}{\sqrt{R^2k^2-v^4}}=dx

Integrating v from 0 to $v_{max}$ and s from 0 to $s_{min}$
\frac{R}{2}arctan\left(\frac{v_{max}^2}{\sqrt{R^2k^2-v^4}}\right)=s_{min}
What should I substitute from $v_{max}$?

 

[TSny]

At $v_{max}$ what is the direction of the total acceleration?

 

[Saitama]

At $v_{max}$ what is the direction of the total acceleration?

Towards the center? So $v_{max}=\sqrt{\mu gR}$?

 

[TSny]

Towards the center? So $v_{max}=\sqrt{\mu gR}$?

Yes.

 

[Saitama]

Yes.

Thank you TSny!

Substituting this value of $v_{max}$ in the final equation, I get $s_{min}=\pi R/4$.

 

[TSny]

I get $s_{min}=\pi R/4$.

I think that's right.