Question

9.5.3. a thin - walled hollow cylinder is rotating about an axis that passes through the center of both ends. the radius of the cylinder is r. at what angular speed ω must this cylinder rotate to have the same total kinetic energy that it would have if it were moving horizontally with a speed v without rotation?
a) ( omega=\frac{v^{2}}{2 r} )
b) ( omega=\frac{v}{r} sqrt{2} )
c) ( omega=\frac{v}{r} )
d) ( omega=\frac{v}{2 r} )
e) ( omega=\frac{v^{2}}{r^{2}} )

Explanation:

Step1: Write the formula for translational kinetic energy

The translational kinetic energy \(K_{trans}=\frac{1}{2}mv^{2}\), where \(m\) is the mass of the cylinder and \(v\) is the linear speed.

Step2: Write the formula for rotational kinetic energy

For a thin - walled hollow cylinder, the moment of inertia \(I = mr^{2}\). The rotational kinetic energy \(K_{rot}=\frac{1}{2}I\omega^{2}\). Substituting \(I = mr^{2}\) into the formula for \(K_{rot}\), we get \(K_{rot}=\frac{1}{2}(mr^{2})\omega^{2}\).

Step3: Set \(K_{trans}=K_{rot}\)

Set \(\frac{1}{2}mv^{2}=\frac{1}{2}(mr^{2})\omega^{2}\). Cancel out \(\frac{1}{2}m\) from both sides of the equation. We are left with \(v^{2}=r^{2}\omega^{2}\).

Step4: Solve for \(\omega\)

Take the square root of both sides of the equation \(v^{2}=r^{2}\omega^{2}\). We get \(\omega=\frac{v}{r}\).

Answer:

C. \(\omega=\frac{v}{r}\)