Question

A small car with mass 0.800 kg travels at constant speed on the inside of the track that is a vertical circle with radius 5.00 m. If the normal force exerted by the track on the car when it is at the top of the track (point B) is 6.00 N. What is the normal force on the car when it is at the bottom of the track what is the speed of the car

Answer 1

Answer:

$N=21.690\,N$

Explanation:

The equation of equilibrium at the top of the vertical circle is:

$\Sigma F = - N - m\cdot g = - m \cdot \frac{v^{2}}{R}$

The speed experimented by the car is:

$\frac{N}{m}+g=\frac{v^{2}}{R}$

$v = \sqrt{R\cdot (\frac{N}{m}+g) }$

$v = \sqrt{(5\,m)\cdot (\frac{6\,N}{0.8\,kg} +9.807\,\frac{kg}{m^{2}} )}$

$v\approx 9.302\,\frac{m}{s}$

The equation of equilibrium at the bottom of the vertical circle is:

$\Sigma F = N - m\cdot g = m \cdot \frac{v^{2}}{R}$

The normal force on the car when it is at the bottom of the track is:

$N=m\cdot (\frac{v^{2}}{R}+g )$

$N = (0.8\,kg)\cdot \left(\frac{(9.302\,\frac{m}{s} )^{2}}{5\,m}+ 9.807\,\frac{m}{s^{2}} \right)$

$N=21.690\,N$