Answer to Question #97850 in Mechanics | Relativity for Stefanie

Question

A bead with a hole through it slides on a wire track. The wire is threaded through the hole in the bead, and the bead slides without friction around a loop-the-loop (see figure below). The bead is released from rest at a height
h = 3.00R.
a. What is its speed at point A? (Use the following as necessary: the acceleration due to gravity g, and R.)
b. How large is the normal force on the bead at point A if its mass is 4.90 grams? magnitude and direction?
c. What If? What is the minimum height h from which the bead can be released if it is to make it around the loop? (Use any variable or symbol stated above as necessary.)

Answer 1

a.)

$$mg(3R) = 0.5mv^2 + mg(2R)\\or\\v^2=2gR\\or\\v=\sqrt{2gR}$$

It is the velocity at A

b.)

Normal force at A = $\frac{mv^2}{R}=\frac{m\times2gR}{R}=2mg=0.09604\ N$

Direction will be radially outward

c.)For completing the loop

$$mg=\frac{mv^2}{R}$$ $$v=\sqrt{gR}$$

$$0.5mv^2+mg(2R)=mgh$$

$$or\\0.5gR+2gR=gh\\or\\h=\frac{5R}{2}$$