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.

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

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

## Answer

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

**KE _{i} + PE_{i} = KE_{t} +**

PE_{t}

0 + mgh = (0.5) m V^{2} + mg (2R)

2gh = V^{2} + 4gR

**V = sqrt (2gh -
4gR)**

eq-1

Kinetic energy is given as

KE = (0.5) m V^{2}

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

F_{n} + mg = m

V^{2}/R

F_{n} = normal force , mg = weight

for F_{n}

=0

force losing contact F_{n} =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**