Answer
(a) The minimum value of $h$ is $2.5\times R$
(b) $v = \sqrt{(5.00)~gR}$
The radial acceleration is (5.00)g.
The tangential acceleration is g.
Work Step by Step
(a) We can assume that the normal force at the top of the loop is zero. Then gravity provides the centripetal force to keep the car moving in a circle.
$\frac{mv^2}{R} = mg$
$v^2 = gR$
$v = \sqrt{gR}$
This is the minimum speed such that the car will not fall off the loop. We can use this speed to find the minimum height $h$.
$K_1 + U_1 = K_2 + U_2$
$0 + mgh = \frac{1}{2}mv^2+ 2mgR$
$h = \frac{v^2}{2g}+ 2R$
$h = \frac{gR}{2g}+ 2R$
$h = 2.5~R$
The minimum value of $h$ is $2.5\times R$
(b) $K_2 + U_2 = K_1 + U_1$
$\frac{1}{2}mv^2 + mgR = 0+ (3.50)~mgR$
$\frac{1}{2}v^2 + gR = (3.50)~gR$
$\frac{1}{2}v^2 = (2.50)~gR$
$v^2 = (5.00)~gR$
$v = \sqrt{(5.00)~gR}$
We can use the speed to find the radial acceleration.
$a_R = \frac{v^2}{R} = \frac{(5.00)gR}{R} = (5.00)~g$
Since the only vertical force acting on the car at point C is gravity, the tangential acceleration is g.
