A marble slides without friction in a vertical plane around the inside of a smooth, 21 cm-diameter horizontal pipe. The marble's speed at the bottom is 4.5 m/s this is fast enough so that the marble makes a complete loop, never losing contact with the pipe.What is its speed at the top?

At the top, it needs to have centripetal acceleration of v^2/r>g

So is v^2/r greater than g?

Now its speed at the top is determined by the bottom speed minus losses to gravity.

speed at top^2 =speedbottom^2-2g*.21
so figure speed at top, and see if it is greater than required.

2.2

To find the marble's speed at the top of the loop, we will apply the principle of conservation of mechanical energy. At any point along the loop, the total mechanical energy of the marble (sum of kinetic and potential energies) remains constant, assuming no energy is lost due to friction.

The total mechanical energy of the marble at the bottom of the loop can be expressed as the sum of its kinetic energy (KE) and potential energy (PE):

Total Mechanical Energy at Bottom = KE + PE.

Since the marble is sliding without friction, there is no work done by friction, and the total mechanical energy is conserved. Therefore, we can write:

Total Mechanical Energy at Bottom = Total Mechanical Energy at Top.

At the bottom of the loop, the marble's speed is given as 4.5 m/s. The potential energy at the bottom of the loop is zero because the marble is at the lowest point of the loop. Thus, we have:

KE at Bottom + PE at Bottom = KE at Top + PE at Top.

The kinetic energy of the marble is given by the equation:

KE = (1/2) * m * v^2,

where m is the mass of the marble, and v is its velocity.

At the bottom of the loop, we can write the kinetic energy as:

KE at Bottom = (1/2) * m * (4.5 m/s)^2.

At the top of the loop, the marble's height is equal to the diameter of the pipe, which is 21 cm. We can calculate the potential energy at the top using the equation:

PE = m * g * h,

where g is the acceleration due to gravity (9.8 m/s^2) and h is the height.

At the top of the loop, we have:

PE at Top = m * g * (21 cm).

Now, equating the total mechanical energy at the bottom and top:

(1/2) * m * (4.5 m/s)^2 + 0 = (1/2) * m * v^2 + m * g * (21 cm).

We can set up this equation to find the speed (v) at the top of the loop. Simplifying the equation and solving for v, we get:

(1/2) * (4.5 m/s)^2 = g * (21 cm) + v^2.

Now, let's convert the units to meters:

(1/2) * (4.5 m/s)^2 = (9.8 m/s^2) * (0.21 m) + v^2.

Simplifying further:

(1/2) * 20.25 = 2.058 + v^2.

10.125 = v^2 + 2.058.

v^2 = 10.125 - 2.058.

v^2 = 8.067.

Taking the square root of both sides:

v ≈ √8.067.

v ≈ 2.841 m/s.

Therefore, the marble's speed at the top of the loop is approximately 2.841 m/s.