Aditya's challenges in Mechanics 3

$\bullet$ Find out equation of parabola. $\frac{-x^2}{100} +100$ $\bullet$ Write down equations of motion. $mg \cos \theta -\text{N}=\frac{mv^2}{R}$ $\text{F\_r} = mg \sin \theta$

Where,

$\theta$ is the angle of inclination of the curve at the point with $+x$ axis.

$\text{N}$ is normal reaction between hill and bicycle.

$\text{F\_r}$ is frictional force on bicycle by hill.

$R$ is the radius of curvature at that point.

$R=\frac{\left( 1+(f'(x))^2\right)^{\frac{3}{2}}}{f''(x)}$

$\bullet$ Net force is resultant of frictional force and normal force.

i suggest you to draw a rough diagram to understand solution better.

Firstly we consider the point of start to be the origin. now we can analyze the motion of the particle to be like a projectile .

Writing equation of the trajectory using alternate form of trajectory of projectile

$y$ = $x$ tan(q) (1-x/R) Where q denotes the angle of projection and R Represents the range of projectile .

We Already Know R = 200 m .

For getting q we have been also given the maximum height attained . We have a relation between R And Maximum

Height of projectile which says H/R = tan(q) / 4 . Which can be easily derived using standard results for maximum

height and range of projectile.

Plugging in the values we get tan(q) = 2 .

At Any Point draw the Free body diagram of the particle . The forces acting on it will be The Normal reaction and Weight .

Let the tangent to the path At A Make an angle Of "t" with the horizontal .

Resolving forces and writing equilibrium along the direction normal to the path (apply centrifugal force )

we get

mgcos(t)- N = ma

Where a is the centripetal acceleration .

Also we know that a = v^2 / R . where R is the radius of curvature .

For any curve the radius of curvature is given by a general result which can be seen at

http://physics.stackexchange.com/questions/190262/radius-of-curvature

Solving For R At point A We Obtain a = 5*root(2) / 2

Also Slope of tangent at point A Can be obtained by differentiating the equation wrt x and plugging in the value of x .

Finally to find The total force applied by path we need to take the resultant of tangential force (mgsin(t) ) and The

above obtained normal reaction

After all numerical calculations we get Answer As 500 newton ! :)

Nice problem

A solution using no calculus,no trig functions and no trig identities. Just the Pythagorean theorem.

The hill is in the shape of a parabola.

$Y=100\left( 1-{ \left( \frac { x }{ 100 } \right) }^{ 2 } \right) \\$

Find the slop of the parabola at the 50-meter mark. The slope of the parabola can be determined by comparing a few points near the point of interest using simple substitution for the value of x

Slope of parabola at x = 50 meters is converging on one (ie change in rise = change in run).

At X= -50 meters $slope=1=\frac { \Delta Y }{ \Delta X }$ $\Delta Y= \Delta X$
Let t be a unit of time

$\frac{\Delta Y}{t} = \frac{\Delta X}{ t}$
$\\ { V }_{ Y\quad }=\quad { V }_{ X }$ velocity horizontal = velocity vertical

Slope of the curve at any point is a representation of the relationship of Vy to Vx $\\ { V }_{ B }=velocity\quad of\quad bicycle\quad =\quad 10\sqrt { 5 }$ $\\ { { V }_{ B } }^{ 2 }=\quad { { V }_{ X } }^{ 2 }+{ { V }_{ Y } }^{ 2 }$ ${ V }_{ Y }={ V }_{ X }=\sqrt { 250 } m/s$ Vertical and horizontal components of bicycle velocity at the instant of interest.

$\frac { \Delta Y }{ { t }^{ 2 } } =\quad acceleration\quad Y\quad =\quad { A }_{ Y }$

change in the slope of the curve is a representation of (+/-) acceleration in both X and Y axis; from substitution shown in the above table $\Delta t\quad =\quad time\quad interval\quad of\quad \Delta Y$ ${ A }_{ Y }=\quad \frac { { V }_{ Y } }{ 100\quad \Delta t }$ Velocity of bicycle is constant, net accelerations of bicycle = 0, Ab =0 ${ { A }_{ B } }^{ 2 }=\quad { { A }_{ X } }^{ 2 }+{ { A }_{ Y } }^{ 2 }$ $A_Y = -A_X$

acceleration in Y axis

${ A }_{ Y }=\quad \frac { \Delta { V }_{ Y } }{ \Delta t } \quad =\quad \frac { \frac { \sqrt { 250 } }{ 100 } }{ \frac { 1 }{ \sqrt { 250 } } } =\quad 2.50\quad ㎨$ Ax =2.50 meters /second^2; acceleration horizontal do to shape of hill

Ay = -2.50 meters /second^2; acceleration verticaldo to shape of hill

acceleration do to gravity $g=\quad 10\quad ㎨$ net acceleration in the vertical axis ${ A }_{ total\quad V }=\quad 10.0-2.5\quad ㎨$

acceleration vector on hill vertical and horizontal $\\ { A }_{ all }=\quad \sqrt { { 7.5 }^{ 2 }+{ 2.5 }^{ 2 } } \quad =\quad \sqrt { 62.5 } ㎨$

$f=ma$ $\\ m\quad =\quad 100\quad \sqrt { \frac { 2 }{ 5 } }$

$\\f\quad =\quad 100\quad \sqrt { \frac { 2 }{ 5 } } \quad \sqrt { 62.5 } =\quad 500\quad Neutons$

Hey sorry, This is not the solution, though i did crack the problem(Answer:500). I just wanted to ask you how did you manage to get a Level 5 rating on your question? Please help. I'll delete this "solution" afterwards. Thanks a Lot