Question

Review. A bead slides without friction around a loop-the loop (Fig. P8.5). The bead is released from rest at a height h=3.50R (b) How large is the normal force on the bead at point (A) if its mass is 5.00 g?

Answer 1

The normal force on the bead at point A is 1 N.

According to the Law Of Conservation of Energy, the total energy of the bead at the topmost point of height of 3.50 R is the same as its total energy at point A.

The energy at a height of 3.50R is equal to its potential energy which is

U= mgh= mg x 3.5R= 3.5mgR .........(i)

The energy at point A is,

E= mg(2R)+ 1/2 mv^2  ...........(ii)

Now, equating (i) and (ii)

3.5mgR = 2mgR + 1/2 mv^2

3.5mgR - 2mgR = 1/2mv^2

1.5gR = 1/2 v^2

3gR = v^2

V = √3gR

At point A, three forces are there weight of the bead(mg), normal reaction(N), and centripetal force(mv^2/r).

Balancing these forces,

mg + N = mv^2/r

N= mv^2/r - mg

Putting the given values in the above equation, we get the normal reaction

N= 1 N

Thus, the normal force on the bead at point A is 1N.

To know more about "Conservation of Mechanical Energy", refer to the following link:

brainly.com/question/11264649?referrer=searchResults

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