Question

a small model car with mass m travels at constant speed on the inside of a track that is a vertical circle with radius r = 3.00 m. the normal force exerted by the track on the car when it is at the bottom of the track (point a) is equal to 2.50mg. part a how much time does it take the car to complete one revolution around the track. express your answer with the appropriate units. t = value units

Explanation:

Step1: Apply Newton's second - law at the bottom of the circle

At the bottom of the vertical - circular track, the net force acting on the car is $F_{net}=N - mg$, and this net force provides the centripetal force $F_c$. Given $N = 2.50mg$, then $F_{net}=2.50mg−mg = 1.50mg$. Since $F_c=\frac{mv^{2}}{r}$, we have $1.50mg=\frac{mv^{2}}{r}$.
$1.50mg=\frac{mv^{2}}{r}$

Step2: Solve for the speed $v$

Cancel out the mass $m$ from both sides of the equation $1.50mg=\frac{mv^{2}}{r}$. We get $1.50g=\frac{v^{2}}{r}$. Given $r = 3.00m$ and $g = 9.8m/s^{2}$, then $v=\sqrt{1.50gr}=\sqrt{1.50\times9.8\times3.00}=\sqrt{44.1}\approx6.64m/s$.
$v=\sqrt{1.50\times9.8\times3.00}$

Step3: Find the time $t$ for one - revolution

The distance traveled in one - revolution is the circumference of the circle $s = 2\pi r$. The speed $v=\frac{s}{t}$, so $t=\frac{2\pi r}{v}$. Substitute $r = 3.00m$ and $v\approx6.64m/s$ into the formula.
$t=\frac{2\pi\times3.00}{6.64}\approx2.85s$
$t=\frac{2\pi\times3.00}{6.64}$

Answer:

$t = 2.85s$