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Classical Mechanics Level 5

A motorcyclist wishes to travel in a circle of radius R.The coefficient of static friction b/w the tires and the(horizontal) ground is constant ,The motorcycle starts at rest.If the minimum distance the motorcycle must travel in order to achieve it's maximum allowable speed(that is , the speed above which it skids out of the circular path) is given by:

$\Psi$ R

find the value of $\Psi$ {two decimal places}

The answer is 0.78.

1 solution

Krack Man
Jan 12, 2015

radial force is = $\frac { m{ v }^{ 2 } }{ R }$ tangential force = ${ F }_{ t }$ which is equal to "ma",is

${ F }_{ t }=\sqrt { { (\mu mg) }^{ 2 }-{ (\frac { m{ v }^{ 2 } }{ R } ) }^{ 2 } } =m\frac { dv }{ dt }$

$dx=Rd\theta$

$dx=\frac { vdv }{ \sqrt { { (\mu g) }^{ 2 }-{ (\frac { { v }^{ 2 } }{ R } ) }^{ 2 } } }$

substitute $z=\frac { { v }^{ 2 } }{ \mu gR }$

$dx=\int { \frac { Rdz }{ 2(1-{ z }^{ 2 }) } }$

maximum speed V is obtained when $\mu mg=m\frac { { V }^{ 2 } }{ R }$

so V^2= $(\mu gR)$

desired distance is X,which is

$X=\int _{ 0 }^{ X }{ dx } =\int _{ 0 }^{ 1 }{ \frac { Rdz }{ 2(1-{ z }^{ 2 }) } } \\ X=0.5R\int _{ 0 }^{ \Pi /2 }{ d\theta } \\ X=\frac { \pi }{ 4 } R\\ =0.78R$

@Visakh Radhakrishnan , will You dare To post This Question with slight Variation in that is Motorcyclist is moving in vertical Circle I'am able Answer This too ! I gave You The Privilege For this ! Choice is urs ! $\ddot\smile$ .

Deepanshu Gupta - 6 years, 5 months ago

you can post it(i have another one coming for you!)

Visakh Radhakrishnan - 6 years, 5 months ago

@Krack Man , @Visakh Radhakrishnan Theta is the angular displacement . How it varies till pie/2 ?

Ayush Choubey - 5 years, 9 months ago

Train your Brain!

The answer is 0.78.

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