Missing picture in the problem. Found on internet:
https://gyazo.com/0415a6ba75f87f0229e08aab0f68fbe1
Solution:
part A)
we are at point A, where the velocity of the car is [tex]v_A = 20.0 m/s[/tex]. There are two forces acting on the car:
- its weight: [tex]W=mg[/tex], where m is the mass of the car and g is the gravitational acceleration
- the normal force exerted by the track on the car: [tex]N=2.06 \cdot 10^4 N[/tex]
The car is in circular motion, so the resultant of these 2 forces is equal to the centripetal force:
[tex]N-mg = m \frac{v_A^2}{R_1} [/tex]
If we re-arrange the equation and we plug the numbers into it, we can find m, the mass of the car:
[tex]m= \frac{N}{g+ \frac{v_A^2}{R_1} }= \frac{2.06 \cdot 10^4 N}{9.81 m/s^2 + \frac{(20.0m/s)^2}{10.0 m} }=413.6 kg [/tex]
Part B)
At point B, the equation of motion becomes
[tex]mg-N = m \frac{v_B^2}{R_2} [/tex]
where we changed the signs of mg and N, since now mg points toward the center of the circular trajectory.
The maximum speed that the car can achieve without flying off the track is the velocity at which the normal force exerted by the track is zero: N=0, so that the term on the left is maximum. If we put N=0, we find:
[tex]mg = m \frac{v_B^2}{R_2} [/tex]
from which we find the velocity of the car:
[tex]v_B = \sqrt{gR_2}= \sqrt{(9.81 m/s^2)(15 m)}=12.1 m/s [/tex]
