A $21 \mathrm{~mm}$ diameter glass marble with a mass of $15 \mathrm{~g}$ rolls without slipping down a track toward a vertical loop of radius $R=11 \mathrm{~cm}$, as shown in the figure.
Approximate the minimum translational speed $v_{\min }$ the marble must have in order to complete the loop without falling off the track when it is a height $H=23 \mathrm{~cm}$ above the bottom of the loop.
Figure is not to scale.
\[
v_{\min }=
\]
$\mathrm{m} / \mathrm{s}$

Step 1 ：First, we need to understand that the marble needs to have enough kinetic energy at the top of the loop to overcome the gravitational potential energy and the centripetal force required to keep it in circular motion. This is the key to solving this problem.

Step 2 ：We start by calculating the gravitational potential energy at the top of the loop, which is given by \(mgh\), where \(m\) is the mass of the marble, \(g\) is the acceleration due to gravity, and \(h\) is the height. Substituting the given values, we get \(mgh = 0.015 \times 9.8 \times 0.23 = 0.03411 \, \text{J}\).

Step 3 ：Next, we calculate the centripetal force required to keep the marble in circular motion at the top of the loop. This is given by \(\frac{mv^{2}}{r}\), where \(v\) is the speed of the marble and \(r\) is the radius of the loop. At the top of the loop, this force must be equal to the weight of the marble, \(mg\), for the marble to just stay in contact with the track. Therefore, we have \(\frac{mv^{2}}{r} = mg\), which simplifies to \(v = \sqrt{rg}\). Substituting the given values, we get \(v = \sqrt{0.11 \times 9.8} = 1.04 \, \text{m/s}\).

Step 4 ：Finally, we equate the kinetic energy at the bottom of the loop to the sum of the gravitational potential energy and the kinetic energy at the top of the loop. The kinetic energy is given by \(\frac{1}{2}mv^{2}\), so we have \(\frac{1}{2}mv_{\min}^{2} = \frac{1}{2}mv^{2} + mgh\). Solving for \(v_{\min}\), we get \(v_{\min} = \sqrt{v^{2} + 2gh}\). Substituting the values we calculated earlier, we get \(v_{\min} = \sqrt{(1.04)^{2} + 2 \times 9.8 \times 0.23} = 2.23 \, \text{m/s}\).

Step 5 ：So, the minimum translational speed the marble must have in order to complete the loop without falling off the track when it is a height \(H=23 \, \text{cm}\) above the bottom of the loop is approximately \(\boxed{2.23 \, \text{m/s}}\).