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 = 3.15R.
A bead is on a loop of wire at height h. The wire curves down and to the right, then loops into a complete circle of radius R before continuing on to the right. The highest point of the circle is labeled A. Position A is lower than the initial position of the bead.
(a)
What is its speed at point A?
(b)
How large is the normal force on the bead at point A if its mass is 4.85 g?
(a) At point A, the bead will have converted all its potential energy into kinetic energy, so we can use the conservation of energy principle to find its speed. The potential energy at height h is given by mgh, where m is the mass of the bead, g is the acceleration due to gravity, and h is the height. The kinetic energy at point A is given by (1/2)mv^2, where v is the speed.
Setting the potential energy equal to the kinetic energy, we have:
mgh = (1/2)mv^2
Solving for v, we get:
v = sqrt(2gh)
Plugging in the values, we get:
v = sqrt(2 * 9.8 * 3.15R) = sqrt(62.28R) = 7.89 m/s
(b) The normal force on the bead at point A can be found using Newton's second law, which states that the net force on an object is equal to its mass times its acceleration. At point A, the net force acting on the bead is the centripetal force required to keep it moving in a circle.
The centripetal force is given by:
F = (mv^2)/R
where R is the radius of the loop. The normal force at point A will be equal in magnitude but opposite in direction to the centripetal force, so the normal force will be:
N = mv^2/R
Plugging in the values, we get:
N = (4.85g * 7.89^2) / R = 299.86 / R
So, the normal force at point A is 299.86/R.