# Maximum permissible speed while going down a ramp

So, I was playing hill climb racing and I noticed that if we move with high speeds towards a ramp going down we just jump it off. While lower speeds, help us to stay in contact with the ramp.

Sticks :

Off it goes :

So, for fun, I decided to calculate maximum permissible speed which allows us to stay in contact with the ramp. Here's my working :

Let's consider the case of a cylinder of mass $M$ and radius $R$. Consider sufficient friction so as to not allow any slipping.

Lets start....

$u = w_0 R$

$v = w R$

$I = \frac 1 2 MR^2$

As there is no slipping, Friction does no work. Initial Energy : Taking initial level to be U=0 for gravitational potential energy : $\frac 1 2 M u^2 + \frac 1 2 I w_0^2 +MgR =\frac 3 4 M u^2 + MgR$

Final Energy :

$\frac 1 2 M v^2 + \frac 1 2 I w^2 + MgRcos\theta =\frac 3 4 M v^2 + MgRcos\theta$

As Energy is conserved,

$v^2 = \frac 4 3 (gR(1-cos\theta) + u^2)$

Now we have to find N in terms of v and put it $\geqslant 0$ But I cannot see how to find N? Please help.

Suppose the ramp wasn't there, then the trajectory of the object would the same as if it fell off a cliff:

To get the equation of motion you simply note that the horizontal and vertical coordinates are given by (neglecting air resistance):

$$x = ut$$

$$y = \tfrac{1}{2} g t^2$$

So you can get the trajectory by substituting for $t$ to get:

$$y = \frac{g}{2u^2} x^2$$

This is taking the edge of the cliff as the origin $(0, 0)$ and for convenience we take $y$ to be positive in the downwards direction. If we put the ramp back and zoom in on the take off point you'll see:

Because the slope of the trajectory is zero at the takeoff point, and the slope of the ramp is non-zero (i.e. it has a sharp edge) there will always be a period after the edge where the car leaves the ground. Reducing the sped or making the ramp shallower will reduce the length of the jump, but it will always be present. The only way to avoid a jump is for the gradient of the ramp to be everywhere less than or equal to the gradient of the trajectory.

• If we made the ramp of same shape as trajectory as suggested, won't then there will be a speed where cylinder will fly off? Feb 25, 2014 at 10:58
• @Awesome: the trajectory depends on the speed. As you make the speed faster the trajectory will become flatter. If you make the ramp the same shape as the trajectory at some speed $u_0$ then for all speeds $u \le u_0$ the car won't jump and for all speeds $u > u_0$ it will jump. Feb 25, 2014 at 12:02
• I recently thought of this:Take the centre of mass of cylinder to perform circular motion about "that" point and write $Mgcos\theta-N=\frac{Mv^2}{R}$ if I put $v$ I obtained earlier it will give me condition for $N\geqslant 0$ What is wrong with it? Feb 25, 2014 at 12:29
• And that y coordinate should not be negative... Feb 25, 2014 at 12:38
• Anyone?????????? Feb 25, 2014 at 14:15

Consider centre of mass of cylinder to perform circular motion about that point :

$\frac{Mv^2} R = Mgcos\theta - N$

$\frac 2 3 M((gR(1-cos\theta) + u^2) = Mgcos\theta -N$

$N=\frac 2 3 M((gR(1-cos\theta) + u^2) - Mgcos\theta$

Condition : $\frac 2 3 M((gR(1-cos\theta) + u^2) - Mgcos\theta \geqslant 0$