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on An empty cylindrical barrel is open at one end and rolls without slipping straight down a hill. The barrel has a mass of $25.0 \mathrm{~kg}$, a radius of $0.260 \mathrm{~m}$, and a length of $0.650 \mathrm{~m}$. The mass of the end of the barrel equals a fifth of the mass of its side, and the thickness of the barrel is negligible. The acceleration due to gravity is $g=9.80 \mathrm{~m} / \mathrm{s}^{2}$.
What is the translational speed $v_{\mathrm{f}}$ of the barrel at the bottom of the hill if released from rest at a height of $33.0 \mathrm{~m}$ above the bottom?
\[
v_{\mathrm{f}}=
\]

Step 1 ：The barrel is rolling down the hill without slipping, so we can use the principle of conservation of energy to solve this problem. The total mechanical energy of the barrel is conserved because the hill is frictionless. The total mechanical energy of the barrel at the top of the hill is equal to its potential energy, and at the bottom of the hill, it is equal to its kinetic energy. The kinetic energy of the barrel is the sum of its translational kinetic energy and its rotational kinetic energy. We can set the potential energy equal to the kinetic energy and solve for the final velocity.

Step 2 ：The potential energy at the top of the hill is given by \(mgh\), where \(m\) is the mass of the barrel, \(g\) is the acceleration due to gravity, and \(h\) is the height of the hill.

Step 3 ：The kinetic energy at the bottom of the hill is given by \(\frac{1}{2}mv^2 + \frac{1}{2}I\omega^2\), where \(v\) is the translational speed of the barrel, \(I\) is the moment of inertia of the barrel, and \(\omega\) is the angular speed of the barrel. Since the barrel is rolling without slipping, we have \(v = \omega r\), where \(r\) is the radius of the barrel.

Step 4 ：The moment of inertia of the barrel is given by \(I = \frac{1}{2}mr^2\), where \(m\) is the mass of the barrel and \(r\) is the radius of the barrel.

Step 5 ：Setting the potential energy equal to the kinetic energy and solving for \(v\), we get: \[v = \sqrt{\frac{2gh}{1 + \frac{1}{2}}}\]

Step 6 ：Now we can plug in the given values and calculate the final velocity. The mass \(m = 25.0\), the radius \(r = 0.26\), the height \(h = 33.0\), and the acceleration due to gravity \(g = 9.8\).

Step 7 ：The translational speed \(v_{\mathrm{f}}\) of the barrel at the bottom of the hill is \(\boxed{20.77 \mathrm{~m/s}}\).