Question

A bead slides without friction around a loop–the–loop (see figure below). The bead is released from rest at a height h = 3.40R.

(a) What is its speed at point ? (Use the following as necessary: the acceleration due to gravity g, and R.)
v =

(b) How large is the normal force on the bead at point if its mass is 4.60 grams?
magnitude N
direction

Answer 1

since it's frictionless the easy way to solve a is conservation of energy. the bead changes height = to 1.40R (that's 3.40init height - diam of circle). sochange of energy from potential (mg1.4R) = kinetic energy (1/2mv^2). m's cancel and solve for v in terms of R and g.

For b) normal force comes from F= m v^2 / R pushing it up and gravity pulling it down.