A 3.0g bead has been threaded onto a vertical, frictionless, circular hoop with a 20cm

radius. The bead is released from rest at 25º below the vertical.
(a) Find the speed of the bead when it is 70° below the vertical.
(b) Using the free-body diagram technique, find the normal force acting on the bead and the
radial and tangential components of its acceleration vector when it is 70º below the vertical.

So far, I have only come up with an equation where the velocity is defined as

a = (V^2)/R
aR = V^2
V = √(aR)

Need help relating angle and mass to the resultant acceleration of radial and tangential acceleration.

punk

To find the speed of the bead when it is 70° below the vertical, we need to use the concepts of conservation of energy and circular motion.

(a) To solve for the speed, we can start by defining the initial and final positions of the bead and the energies associated with those positions.

Let's assign the bottom-most position of the hoop as the reference level for potential energy.

At the starting position (25º below the vertical), the bead is at a height h given by:
h = R(1 - cos(25º)),

where R is the radius of the hoop.

The potential energy at this position is given by:
PE_start = mgh,
where m is the mass of the bead, g is the acceleration due to gravity, and h is the height.

At the final position (70º below the vertical), the bead is at a height h' given by:
h' = R(1 - cos(70º)).

The potential energy at this position is given by:
PE_final = mgh'.

Since there is no friction and no external forces doing work on the system, the total mechanical energy at any point along the circular motion is conserved. Therefore, we have:

PE_start + KE_start = PE_final + KE_final,

where KE_start and KE_final are the initial and final kinetic energies, respectively.

Since the bead is released from rest, the initial kinetic energy is 0 (KE_start = 0).

Therefore, we have:

0 + mgh = mgh' + KE_final,

mgh - mgh' = KE_final.

Substituting the expressions for h, h', and simplifying, we get:

mgR(cos(25º) - cos(70º)) = (1/2)mv^2,

where v is the speed of the bead at the final position.

Simplifying further, we have:

v^2 = 2gR(cos(25º) - cos(70º)).

Now we can calculate the speed of the bead at the final position.

(b) To find the normal force acting on the bead and the radial and tangential components of its acceleration vector when it is 70º below the vertical, we need to draw a free-body diagram for the bead at that position.

The forces acting on the bead are:

1. Weight (mg) acting downward.
2. Normal force (N) exerted by the hoop, perpendicular to the surface of the hoop.

The radial acceleration (ar) is directed towards the center of the hoop and is given by ar = v^2 / R.

The tangential acceleration (at) is directed tangent to the circular path and is given by at = Rα,

where α is the angular acceleration.

Since the bead is moving in a vertical plane, the net force acting on the bead must provide the necessary radial component of the acceleration.

The net radial force is given by:

F_net = ma_r = m(v^2/R) = N - mg.

Solving for N, we get:

N = mg + m(v^2/R).

Therefore, the normal force acting on the bead when it is 70º below the vertical is mg + m(v^2/R).

The radial and tangential components of the acceleration vector can be found using the formulas given above. The radial component is ar = v^2 / R, and the tangential component is at = Rα.