A bead slides without friction around a loop-the-loop. The bead is released from a
height h = 3.5R. A is at the highest point of the loop and R is the radius of the loop
(a) What is its speed at point A?
(b) How large is the normal force on it if its mass is 5g?
on a hot summer night
Solution and explain
To answer this question, we can use the principle of conservation of mechanical energy and the concept of centripetal acceleration.
(a) To find the speed at point A, we can equate the potential energy at the initial height to the kinetic energy at point A. The potential energy (PE) at the initial height is given by PE = mgh, where m is the mass of the bead and g is the acceleration due to gravity. Since the bead is released from a height of 3.5R, the potential energy at the initial height is PE = (5g)(3.5R).
At point A, all the potential energy is converted into kinetic energy. The kinetic energy (KE) of the bead at point A is given by KE = (1/2)mv^2, where v is the speed at point A. Equating these two expressions, we have:
mgh = (1/2)mv^2
Simplifying, we get:
gh = (1/2)v^2
Solving for v, we get:
v = √(2gh)
Substituting the value of h = 3.5R, we have:
v = √(2g(3.5R))
(b) To find the normal force on the bead at point A, we need to consider the forces acting on the bead. At the highest point of the loop, the normal force (N) will be equal to the sum of the weight of the bead (mg) and the centripetal force (Fc). The centripetal force is given by Fc = mv^2/R, where R is the radius of the loop.
So, the normal force can be calculated as:
N = mg + mv^2/R
Substituting the given values (m = 5g and v from part (a)), we can calculate the normal force.
height at top of loop = 2 R
so A is (3.5 - 2) = 1.5 R below drop point
so
(1/2) m v^2 = m g (1.5 R)
v^2 = 3 g R
F =m (v^2/R-g) = m (3g-g) = 2 m g
m = .005
g = 9.81