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Hi everyone, i'm new to this forum and I'm having some problem with the question below.

A circular track is banked so that a motor cyclist who travels at 40 km/h in a

horizontal circle of radius 100 m experiences no tendency to side-slip. If the

coefficient of friction is 0.25 and the size of the cycle and cyclist can be ignored,

(a) show that the cyclist can ride on the track without slipping at all speeds

below 40 km/h, and

(b) find the greatest speed at which he can ride without slipping.

[tex]

r=100~m\\

\mu=0.25\\

\text{ideal speed} ~~v=\frac{100}{9}~ ms^{-1}[/tex]

As there's no tendency to side-slip at 40 km/h, I assume there is

no friction acting on the body. Then, I can find the banked angle [itex]\theta[/itex]

[tex]mg\cos\theta=N[/tex]

Consider the component parallel to the inclined plane

[tex]

\begin{align}

&~~~~~~μN - mg\sin\theta\\

&=μmg\cos\theta -mg\sin\theta\\

&=0.25×mg\cos\{\arctan(\frac{10}{81})\}-mg\sin\{\arctan(\frac{10}{81})\}\\

&≈0.126mg>0

\end{align}

[/tex]

therefore it will not slip?

Also, if I want to calculate the minimum speed for not sliding down.

I would consider the x-component and the y-components, then I have:

x-components

[tex]

\frac{mv^2}{r}=N\sin\theta-\mu N\cos\theta[/tex]

y-component

[tex]

N\cos\theta+\mu N\sin\theta=mg[/tex]

By solving, I have

[tex]

v^2=\frac{rg(\sin\theta-\mu\cos\theta)}{\cos\theta+\mu\sin\theta}[/tex]

By Resolving the forces. I have

[tex]

v^2=\frac{rg(\mu\cos\theta+\sin\theta)}{\cos\theta-\mu\sin\theta}\\

v≈19.6~ms^{-1}=70.7~km/h[/tex]

I think I have no problem with part B. Please help me with part A, thanks guys!

## Homework Statement

A circular track is banked so that a motor cyclist who travels at 40 km/h in a

horizontal circle of radius 100 m experiences no tendency to side-slip. If the

coefficient of friction is 0.25 and the size of the cycle and cyclist can be ignored,

(a) show that the cyclist can ride on the track without slipping at all speeds

below 40 km/h, and

(b) find the greatest speed at which he can ride without slipping.

[tex]

r=100~m\\

\mu=0.25\\

\text{ideal speed} ~~v=\frac{100}{9}~ ms^{-1}[/tex]

**2. The attempt at a solution***Part A*As there's no tendency to side-slip at 40 km/h, I assume there is

no friction acting on the body. Then, I can find the banked angle [itex]\theta[/itex]

[tex]N\sin\theta = \frac{mv^2}{r}\\

N\cos \theta=mg\\

\theta=\arctan(\frac{10}{81})[/tex]

N\cos \theta=mg\\

\theta=\arctan(\frac{10}{81})[/tex]

**So to show that there is no slip at all speeds below 40 km/h,**

Is it correct to just consider the force components like this?Is it correct to just consider the force components like this?

[tex]mg\cos\theta=N[/tex]

Consider the component parallel to the inclined plane

[tex]

\begin{align}

&~~~~~~μN - mg\sin\theta\\

&=μmg\cos\theta -mg\sin\theta\\

&=0.25×mg\cos\{\arctan(\frac{10}{81})\}-mg\sin\{\arctan(\frac{10}{81})\}\\

&≈0.126mg>0

\end{align}

[/tex]

therefore it will not slip?

**Is this the correct way to show the statement?**Also, if I want to calculate the minimum speed for not sliding down.

I would consider the x-component and the y-components, then I have:

x-components

[tex]

\frac{mv^2}{r}=N\sin\theta-\mu N\cos\theta[/tex]

y-component

[tex]

N\cos\theta+\mu N\sin\theta=mg[/tex]

By solving, I have

[tex]

v^2=\frac{rg(\sin\theta-\mu\cos\theta)}{\cos\theta+\mu\sin\theta}[/tex]

**And this is the weird part, if I substitute the value of [itex]\theta[/itex] I got,**

[itex]v^2<0[/itex]....Why is that?Can someone please point out my mistakes?[itex]v^2<0[/itex]....Why is that?

*Part B*By Resolving the forces. I have

[tex]

v^2=\frac{rg(\mu\cos\theta+\sin\theta)}{\cos\theta-\mu\sin\theta}\\

v≈19.6~ms^{-1}=70.7~km/h[/tex]

I think I have no problem with part B. Please help me with part A, thanks guys!

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