A $25 \mathrm{~mm}$ diameter glass marble with a mass of $17 \mathrm{~g}$ rolls without slipping down a track toward a vertical loop of radius $R=13 \mathrm{~cm}$, 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.

Step 1 ：We are given a glass marble of diameter 25 mm and mass 17 g that rolls without slipping down a track towards a vertical loop of radius 13 cm. We are asked to find the minimum translational speed the marble must have in order to complete the loop without falling off the track when it is a height 23 cm above the bottom of the loop.

Step 2 ：The marble must have enough kinetic energy at the top of the loop to overcome the gravitational potential energy and the centripetal force required to keep it on the track. The kinetic energy is given by \(KE = \frac{1}{2}mv^2\) and the gravitational potential energy is given by \(PE = mgh\), where \(m\) is the mass of the marble, \(v\) is its speed, \(g\) is the acceleration due to gravity, and \(h\) is the height.

Step 3 ：The centripetal force required to keep the marble on the track is given by \(F_c = \frac{mv^2}{r}\), where \(r\) is the radius of the loop. At the top of the loop, the marble's weight also acts as a centripetal force, so we have \(F_c = mg + \frac{mv^2}{r}\).

Step 4 ：Solving this equation for \(v\) gives us the minimum speed required for the marble to stay on the track. Substituting the given values, we get \(m = 0.017\), \(g = 9.81\), \(r = 0.13\), and \(h = 0.23\).

Step 5 ：Calculating the minimum speed, we get \(v_{min} = 2.657668150842012\).

Step 6 ：Rounding off to two decimal places, the minimum translational speed the marble must have in order to complete the loop without falling off the track when it is a height 23 cm above the bottom of the loop is approximately \(\boxed{2.66 \, m/s}\).