# Find an expression for the kinetic energy of the car at the top of the loop.

A-Find an expression for the kinetic energy of the car at the top
of the loop. Express in terms of m,g,h, and R.

B- find the minimum initial height h at which the car can be
released that still allows the car to stay in contact with the
track at the top of the loop. Express the minimum height in terms
of R

A.

initial height = h

final height at top = 2R

velocity at top = V

using conservation of energy

Total energy at height h = total energy at top of loop

KEi + PEi = KEt +
PEt

0 + mgh = (0.5) m V2 + mg (2R)

2gh = V2 + 4gR

V = sqrt (2gh -
4gR)

eq-1

Kinetic energy is given as

KE = (0.5) m V2

KE = (0.5) m (sqrt (2gh - 4gR)
)2

KE = (0.5) m (2gh - 4gR)

KE = mgh - 2mgR

b)

at the top of loop , force equation is given as

Fn + mg = m
V2/R
Fn = normal force , mg = weight

for Fn
=0
force losing contact Fn =0

V=
sqrt(gR)
eq-2

from eq-1 and eq-2

sqrt (2gh - 4gR)   = sqrt(gR)

2gh - 4gR = gR

2h = 5R

h = 2.5 R