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 = 3.80R.
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? (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 A if its mass is 4.80 grams?
magnitude
N
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.)
h =
(a) The speed of the bead at point A can be found using conservation of energy. The bead starts with gravitational potential energy at height h and converts it to kinetic energy at point A.
At point A:
Potential energy = Kinetic energy
mgh = 1/2 mv^2
Solving for v:
v = sqrt(2gh)
Given h = 3.80R and g = 9.81 m/s^2:
v = sqrt(2 * 9.81 * 3.80R) = sqrt(74.676R) = 8.64R
So, the speed of the bead at point A is 8.64R.
(b) The normal force on the bead at point A can be found using Newton's second law.
At point A, the only forces acting on the bead are its weight (mg) and the normal force (N). Since the bead is moving in a circular path at point A, there is a centripetal force acting towards the center of the loop. The net force in the vertical direction must be equal to the centripetal force.
Net force in the vertical direction:
N - mg = mv^2/R
Substitute the values:
N - (0.00480 kg * 9.81 m/s^2) = (0.00480 kg * (8.64R)^2)/R
N - 0.047088 N = 0.037824R
N = 0.047088 N + 0.037824R
N = 0.047088 N + 0.037824 * R
So, the magnitude of the normal force on the bead at point A is 0.047088 N + 0.037824R. The direction of force is towards the center of the circle.
(c) To find the minimum height h from which the bead can be released in order to make it around the loop, we need to consider the point where the normal force becomes zero. At this point, the bead just loses contact with the track.
At the top of the loop (point A), the normal force is equal to zero. We already calculated the normal force at point A - N = 0.047088 N + 0.037824R. Setting this equal to zero:
0.047088 + 0.037824R = 0
Solving for R:
0.037824R = -0.047088
R = -1.246R
Since R cannot be negative, the normal force cannot be zero. This means that the bead will always require some minimum height h to make it around the loop.