Question

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 = 3.70R.

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 Ndirection ---Select--- upward downward

(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 =

Answer 1

(a)
`\[ V = √(2gh) \]`

(b)
`\[ N = mg \cos \theta \]`

where
`\(\theta\)` is the angle of the wire at point A.

(c)
`\[ h = (5R)/(2) \]`

Step-by-step explanation:

In part (a), the speed at point A can be calculated using the conservation of mechanical energy. The potential energy at the initial height is converted into kinetic energy at point A. The formula
`\(V = √(2gh)\)` relates the final speed
`(\(V\))` to the height
`(\(h\))`, where
`\(g\)` is the acceleration due to gravity. Substituting the given values, we find the speed at point A.

For part (b), the normal force
`(\(N\))` at point A is influenced by the gravitational force and the angle of the wire. The normal force can be calculated using
`\(N = mg \cos \theta\)`, where
`\(m\)` is the mass of the bead,
`\(g\)` is the acceleration due to gravity, and
`\(\theta\)` is the angle of the wire at point A.

Finally, in part (c), the minimum height
`(\(h\))` from which the bead can be released to make it around the loop is found by considering the conservation of energy. At the highest point of the loop (point A), all initial potential energy is converted to kinetic energy, allowing the bead to complete the loop. The formula
`\(h = (5R)/(2)\)` gives the minimum height required for this condition.

Answer 2

The speed of the bead at point A can be found using the law of conservation of mechanical energy. The kinetic energy at point A is 0, so the potential energy at point A is equal to the initial potential energy. Therefore, using the equation PE + KE = PE_A + KE_A, we can solve for the speed at point A.

Step-by-step explanation:

According to the laws of conservation of energy, the total mechanical energy of the bead is conserved as it slides on the wire track. At point A, the bead is at its highest point on the loop and therefore has only potential energy, given by the equation PE = mgh, where m is the mass of the bead and h is its height above a reference point. The speed at point A can be found using the conservation of mechanical energy:

PE + KE = PEA + KEA

Since the initial speed is 0, the kinetic energy at point A is 0. We can then write:

mgh = ½ mvA2

Simplifying the equation gives:

vA = √(2gh)

Substituting the given height h = 3.70R into the equation, we get:

vA = √(2g(3.70R))

This gives the speed at point A, V = √(2g(3.70R)).