91影视

A toy car of mass $m$is speeding up in a circular track.

The net force is parallel to the track or it requires centripetal acceleration.

The net force has to point to the center of the circle.

Centripetal acceleration and net force acting when toy car is down the track. We can analyze the situation quantitatively by writing the r-component of Newton鈥檚 second law. At the bottom of the circle, with the r-axis pointing upward, we have

$\underset{}{鈭�}{F}_r=n_r+{\left(F_G\right)}_r=n-mg=m{a}_r=\frac{m{\left(v_{bot}\right)}^2}{r}$

From the above equation, we find,

$n=mg+\frac{m{\left(v_{bot}\right)}^2}{r}$

The normal force at the bottom is larger than mg. The normal force of the track pushes up when the car is at the bottom of the circle.

The toy car is still moving in a circle, so there must be a net force toward the center of the circle. The r-axis, which points toward the center of the circle, now points downward. Consequently, both forces have positive components. Newton's second law at the tope of the circle is

$\underset{}{鈭�}{F}_r=n_r+{\left(F_G\right)}_r=n+mg=\frac{m{\left(v_{top}\right)}^2}{r}$

Thus at the top the normal force of the track on the toy car is

$n=\frac{m{\left(v_{top}\right)}^2}{r}-mg$

The normal force at the top can exceed $mg$if $v_{top}$is large enough. Our interest, however, is in what happens as the car goes slower and slower. As $v_{top}$decreases, there comes a point when n reaches zero. "No normal force" means "no contact," so at that speed, the track is not pushing against the car. Instead, the toy car which is shown in the above figure is able to complete the circle because gravity alone provides sufficient centripetal acceleration.