A small bead of mass m is constrained to move along a frictionless track as shown. The track consists of a semicircular portion of radius R followed by a straight part. At the end of the straight portion there is a horizontal spring of spring constant k attached to a fixed support. At the top of the circular portion of the track, the bead is pushed with an unknown speed vo. The bead comes momentarily to rest after compressing the spring a distance d. The magnitude of the acceleration due to the gravitational force is g.
What is the magnitude of the normal force exerted by the track on the bead at the point A, a height R above the base of the track? Express your answer in terms of m, k, R, d, and g but not in terms of vo.
N=
(k*d^2/m-2*g*R)*(m/R)
To find the magnitude of the normal force exerted by the track on the bead at point A, we need to consider the forces acting on the bead at that point.
1. Gravitational Force:
The gravitational force on the bead is equal to its weight. The weight can be calculated as the product of mass (m) and acceleration due to gravity (g).
Weight = m * g
2. Centripetal Force:
At point A, the bead is moving in a circular path. The centripetal force required to keep the bead moving in a circle is provided by the normal force.
Centripetal Force = m * (velocity)^2 / R
3. Normal Force:
The normal force at point A will be equal to the sum of the gravitational force and the centripetal force.
Normal Force = Weight + Centripetal Force
Let's calculate the normal force step-by-step:
1. Gravitational Force:
Weight = m * g
2. Centripetal Force:
At the top of the circular portion, the bead is momentarily at rest. At this point, the centripetal force required is zero.
0 = m * (vo)^2 / R
(vo)^2 = 0
vo = 0
3. Normal Force:
Normal Force = Weight + Centripetal Force
Normal Force = m * g + 0
Normal Force = m * g
Therefore, the magnitude of the normal force exerted by the track on the bead at point A is equal to m * g.
To find the magnitude of the normal force exerted by the track on the bead at point A, we need to consider the forces acting on the bead.
1. Gravity force: The weight of the bead is given by the mass (m) multiplied by the acceleration due to gravity (g). So, the weight of the bead is m * g, and it acts vertically downward.
2. Centripetal force: At point A, the bead is moving in a circular path, so there must be a centripetal force acting towards the center of the circle. This force is provided by the normal force exerted by the track.
Since the bead is momentarily at rest at point A, the net force acting on it must be zero. This implies that the upward normal force must balance the downward gravitational force.
Therefore, we can write the equation:
Normal force (N) - Weight (m * g) = 0
Solving for N, we find:
N = m * g
So, the magnitude of the normal force exerted by the track on the bead at point A is simply equal to the weight of the bead, which is m multiplied by the acceleration due to gravity (g).