Question

etermine the equation of motion of the particle subjected to gravity and determine the norma force exerted by the surface to keep the trajectory on the curve. Find an expression for the velocity when the particle detaches from the surface (i.e., N-0) (5 pts). D l

Answer 1

Solution

To determine the equation of motion of the particle on the parabolic surface, it is necessary to graph a free-body diagram, to observe the forces acting on the particle.

Ey Ex Fixed smooth curve y-- and z0 FN ac

Therefore, the force that accelerates the particle is the centripetal force towards the center of cuvarte. In this case it is a component of the weight (radial component)

Applying Newton's second law to the radial part, we have:

$\sum F_{r}=m\ddot{r}$ (1)

$\vec{F_{N}}-mgcos\theta =m\ddot{r}$ (2)

$m\ddot{r}-\vec{F_{N}}+mgcos\theta =0$ (3)

Clearing the normal force of the equation (2):

$F_{N}=m\ddot{r}+mgcos\theta$ (4)

Now, to find the velocity of the particle, we focus on the equation of motion, at the moment of leaving the parabolic surface, the movement is clearly tangential, therefore:

For FN=0

(5)

mgcosi

But

    ( 6)

dv dt

(7)

U) =--goose dlt

Integrating the expression (7), with respect to v, we can obtain an expression for the velocity of the particle

(8)

gcosot

In turn, by the geometric relationship, we can link the height where the particle is at the moment of leaving the surface, with the angula theta.

$cos\theta =\frac{y}{r}$

where r is the radius of curvature.

Then

$v=\frac{-gyt}{r}$

But

$y=-x^{2}$

Finally

$v=\frac{gx^{2}t}{r}$