Question

A spherical ball rolls on a table without slipping, then the fraction of its total energy associated with rotation is \(\\{\mathrm{A}\\} 2 / 5\) \\{B \\} \(3 / 5\) \(\\{\mathrm{C}\\} 2 / 7\) \(\\{\mathrm{D}\\} 3 / 7\)

Short answer

The fraction of the total energy associated with rotation of the spherical ball rolling without slipping on the table is \(\boxed{\frac{2}{7}}\), which corresponds to the option **C**.

Step-by-step solution

Identify given variables

In this problem, we are only given the information that the spherical ball is rolling without slipping on the table. The relationship between linear and angular quantities in rolling without slipping is given by: \(v = Rω\) Where: - \(v\) is the linear velocity of the ball - \(R\) is the radius of the spherical ball - \(ω\) is the angular velocity of the spherical ball We don't know the specific values for \(v\) or \(ω\), but we can still find the fraction of the total energy of the ball associated with rotation.

Calculate the linear kinetic energy (KE)

The kinetic energy due to the linear (translational) motion is given by: \(KE = \frac{1}{2}mv^2\) Where: - \(m\) is the mass of the ball - \(v\) is the linear velocity of the ball

Calculate the rotational energy (RE)

The rotational energy of the rotating spherical ball can be calculated using the formula: \(RE = \frac{1}{2}Iω^2\) For a spherical ball, the moment of inertia (I) is given by: \(I = \frac{2}{5}mR^2\) Where: - \(m\) is the mass of the ball - \(R\) is the radius of the ball Substitute the moment of inertia and the relationship \(v = Rω\) into the formula for rotational energy: \(RE = \frac{1}{2}\cdot\frac{2}{5}mR^2\frac{v^2}{R^2}\) Which simplifies to: \(RE = \frac{2}{5}\cdot\frac{1}{2}mv^2\)

Calculate the fraction of rotational energy

The total energy (TE) of the ball is the sum of the linear kinetic energy (KE) and the rotational energy (RE): \(TE = KE + RE\) Substitute KE and RE's expressions in the equation: \(TE = \frac{1}{2}mv^2 + \frac{2}{5}\cdot\frac{1}{2}mv^2\) Now, divide both sides by \(\frac{1}{2}mv^2\) \(\frac{TE}{\frac{1}{2}mv^2} = \frac{KE + RE}{\frac{1}{2}mv^2}\) Now, we want to find the fraction of the rotational energy in the total energy: \(\frac{RE}{TE} = \frac{\frac{2}{5}\cdot\frac{1}{2}mv^2}{\frac{1}{2}mv^2 + \frac{2}{5}\cdot\frac{1}{2}mv^2}\) By canceling \(mv^2\) terms, we get: \(\frac{RE}{TE}=\frac{\frac{2}{5}}{1 + \frac{2}{5}}\) Solve the fraction: \(\frac{RE}{TE} = \frac{2}{5+2} = \frac{2}{7}\)

Select the correct answer

The fraction of the total energy associated with rotation of the spherical ball rolling without slipping on the table is \(\boxed{\frac{2}{7}}\), which corresponds to the option **C**.