Physics Question

Question

A bead slides without friction around a loop-the-loop. The bead is released from a height of 28.2 m from the bottom of the loop-the-loop which has a radius 9 m. The acceleration of gravity is 9.8 m/s^2 .

 

What is its speed at point A? Answer in units of m/s.

Point A is at the top of the loop-the-loop.

Answer 1

To find the speed of the bead at point A, we can use the conservation of energy principle, which states that the total mechanical energy (kinetic energy + potential energy) of a system remains constant if only conservative forces (like gravity) are acting on it.

At point A (the top of the loop), the bead has only potential energy (PE) and no kinetic energy (KE) because its speed is momentarily zero. The potential energy at point A is given by:

PE_A = mgh_A

Where:

- m is the mass of the bead,

- g is the acceleration due to gravity, and

- h_A is the height of point A above the bottom of the loop.

Given:

- h_A = 28.2 m,

- g = 9.8 m/s².

We need to find the mass of the bead to calculate its potential energy. To do this, we can use the conservation of energy principle at the bottom of the loop, where the bead has only kinetic energy (KE) and no potential energy (PE). The kinetic energy at the bottom of the loop is given by:

KE_B = 1/2 * m * v^2

Where:

- v is the speed of the bead at the bottom of the loop.

At the bottom of the loop, the bead is at the same height as the bottom of the loop, so its potential energy is zero:

PE_B = 0

Using the conservation of energy principle, we can equate the total mechanical energy at the bottom of the loop (KE_B + PE_B) to the total mechanical energy at the top of the loop (KE_A + PE_A):

KE_B + PE_B = KE_A + PE_A

Since PE_B = 0, this simplifies to:

KE_B = KE_A + PE_A

Substituting the expressions for KE_B and PE_A:

1/2 * m * v^2 = mgh_A

Solving for m:

m = 2 * mgh_A / v^2

Now that we have the mass of the bead, we can calculate its speed at point A using the potential energy expression:

PE_A = mgh_A

v = sqrt(2 * g * h_A)

Substituting the given values:

v = sqrt(2 * 9.8 m/s² * 28.2 m)

v ≈ 23.6 m/s

Therefore, the speed of the bead at point A is approximately 23.6 m/s.