Question

a uniform cylindrical shell rotates at a constant rate of 2.46 \\( \frac{\mathrm{rad}}{\mathrm{s}} \\) around its axis (its axis is marked with a dot), while a uniform solid sphere of equal mass rolls without slipping in place inside the cylinder at its bottom (left side of diagram). suddenly the outer surface of the sphere becomes locked with inner surface of the cylinder, and the two objects begin to rotate as a single rigid object. find the rotational speed of the system in radians per second at the moment when they become locked (i.e. before the sphere has moved up at all). provide at least two decimal places

Explanation:

Step1: Calculate the moment of inertia of the cylindrical shell

The moment of inertia of a cylindrical shell about its axis is \(I_{shell}=mr^{2}\).

Step2: Calculate the moment of inertia of the solid sphere

The moment of inertia of a solid sphere about an axis tangent to its surface is \(I_{sphere}=\frac{7}{5}mr^{2}\) (using the parallel - axis theorem \(I = I_{cm}+md^{2}\), where \(I_{cm}=\frac{2}{5}mr^{2}\) and \(d = r\)).

Step3: Apply the law of conservation of angular momentum

The initial angular momentum \(L_{i}=I_{shell}\omega_{0}\) (since the sphere is rolling without slipping initially, its angular momentum about the cylinder's axis is considered in the combined system's initial state. But when we consider the locking - up, we can also use \(L_{i}=I_{shell}\omega_{0}\) because before locking, the sphere's motion (rolling without slipping) and the cylinder's rotation. After locking, \(L_{f}=(I_{shell}+I_{sphere})\omega\).
By conservation of angular momentum \(L_{i}=L_{f}\), so \(mr^{2}\omega_{0}=(mr^{2}+\frac{7}{5}mr^{2})\omega\).
Cancel out \(mr^{2}\) from both sides of the equation: \(\omega_{0}=(1 +\frac{7}{5})\omega\).
We know \(\omega_{0}=2.46\frac{rad}{s}\), and \(1+\frac{7}{5}=\frac{5 + 7}{5}=\frac{12}{5}\).
Then \(\omega=\frac{5}{12}\omega_{0}\).
Substitute \(\omega_{0}=2.46\frac{rad}{s}\) into the formula: \(\omega=\frac{5\times2.46}{12}\).
\(\omega=\frac{12.3}{12}=1.025\approx1.03\frac{rad}{s}\)

Answer:

\(1.03\)