Calculating Loops and Radius for a Cart

[wolves5]

A cart slides down a frictionless inclined track to a circular loop of radius R = 13 m. In order for the cart to negotiate the loop safely, the normal force acting on the cart at the top of the loop, due to the track, must be at least equal to the cart's weight. (Note: This is different from the conditions needed to just negotiate the loop.)

a) What must be the minimum speed |vmin| of the cart at the top of the loop?

For this question, I did sq rt. (13 x 9.81) and got 11.293, but that's not the right answer.

b) How high h above the top of the loop must the cart be released?

c) When the car is descending vertically in the loop (point (c) in the picture), what is its speed |v|?

d) At the bottom of the loop, on the flat part of the track, the cart must be stopped in a distance of d = 10 m. What returning acceleration |a| is required?

I don't think I can do the rest without knowing a first.

 

[member 553083]

The minimum speed |vmin| of the cart at the top of the loop can be calculated using the equation vmin = sqrt(2gr), where g is the acceleration due to gravity (9.81 m/s2) and r is the radius of the loop (13 m). Therefore, the minimum speed at the top of the loop is vmin = sqrt(2 x 9.81 x 13) = 22.88 m/s.b) The height h above the top of the loop that the cart must be released from can be determined by using the equation h = (vmin^2)/(2g), where vmin is the minimum speed of the cart at the top of the loop (22.88 m/s) and g is the acceleration due to gravity (9.81 m/s2). Therefore, the height h above the top of the loop that the cart must be released from is h = (22.88^2)/(2 x 9.81) = 25.31 m.c) The speed of the cart when it is descending vertically in the loop can be calculated using the equation v = sqrt(2gh), where g is the acceleration due to gravity (9.81 m/s2) and h is the height above the top of the loop that the cart must be released from (25.31 m). Therefore, the speed of the cart when it is descending vertically in the loop is v = sqrt(2 x 9.81 x 25.31) = 42.98 m/s.d) The returning acceleration |a| required to stop the cart in a distance of d = 10 m can be calculated using the equation a = (v^2)/(2d), where v is the speed of the cart when it is descending vertically in the loop (42.98 m/s) and d is the distance the cart must be stopped in (10 m). Therefore, the returning acceleration |a| required to stop the cart in a distance of d = 10 m is a = (42.98^2)/(2 x 10) = 859.93 m/s2.