PHYSCS HELP

Question

A bead slides without friction around a loop-the-loop as shown in the figure below. The bead is released from a height h = 3.00R.

 

 

http://www.webassign.net/pse/p8-15.gif

 

(a) What is the bead's speed (V) at point A? Answer in terms of R and g, the acceleration of gravity.

 

 

 

 

(b) How large is the normal force on the bead if its mass is 5.90 g?

N (downward)

Answer 1

(a)

 

Use energy conservation.  The sum of kinetic and potential energies is the same at all points along the path.

 

(1/2)mv₁² + mgh₁ = (1/2)mv₂² + mgh₂

 

v₁ = 0 (released from rest)

h₁ = 3R (given)

 

v₂ = v (the unknown we want to solve for)

h₂ = 2R (height at point A)

 

Substitute, simplify, and get

 

g(3R) = (1/2)v² + g(2R)

 

v = √(2gR)

 

(b)

 

Note that at point A, the normal force F_n is downward.  The sum of the normal force and weight equals the centripetal force at point A.

 

F_n + mg = mv²/R

 

F_n = mv²/R - mg = m(v²/R - g) = m(2gR/R - g) = mg

 

F_n = (0.00590 kg)(9.81 m/s²) ≅ 0.0579 N