Question

A bead slides on a wire without friction around a loop the loop of radius 1.2 m. The bead is released with a speed 0.2 m/s from a height h= 4.6 m where h is measured from the bottom of the loop. How large is the normal force on the bead at the bottom of the loop, if its mass is 5 g?

Answer 1

In this case, the normal force on the bead at the bottom of the loop is approximately 0.049 N.

Step-by-step explanation:

To determine the normal force on the bead at the bottom of the loop, we need to consider the forces acting on the bead. At the top of the loop, the force of gravity acting downwards and the normal force acting upwards provide the centripetal force for the bead to move in a circle. At the bottom of the loop, the force of gravity still acts downwards, but the normal force now acts upwards, providing the centripetal force. Since the bead is sliding without friction, the normal force will be equal to the force of gravity.

First, let's find the gravitational force on the bead. The weight of the bead can be calculated using the formula:

Weight = mass × gravitational acceleration

Given that the mass of the bead is 5 g = 0.005 kg and the gravitational acceleration is 9.8 m/s², we can calculate:

Weight = 0.005 kg × 9.8 m/s² = 0.049 N

Therefore, the normal force on the bead at the bottom of the loop is also 0.049 N.