Question

A rod of mass $m$ is kept on a cylinder and sphere each of radius $R$. The masses of the sphere and cylinder are $m_1 = 4m$ and $m_2 = 5m$ respectively. If the speed of the rod is $v$, find the KE of the system. Assume that the surfaces do not slide relative to each other.

Answer 1

$\frac{171}{80}mv^2$

Answer 2

The system consists of a rod, a sphere, and a cylinder. All surfaces do not slide relative to each other, implying pure rolling for the sphere and cylinder.

1. Kinetic Energy of the Rod ($KE_{rod}$): The rod of mass $m$ moves with a translational speed $v$. $KE_{rod} = \frac{1}{2} m v^2$

2. Kinetic Energy of the Sphere ($KE_s$): The sphere has mass $m_1 = 4m$ and radius $R$. It is a solid sphere, so its moment of inertia about its center of mass is $I_s = \frac{2}{5} m_1 R^2$. Let $v_s$ be the translational speed of the sphere's center of mass and $\omega_s$ be its angular speed. The rod is moving at speed $v$. The point of contact between the rod and the sphere on the top of the sphere must also have speed $v$ (no slipping). The velocity of the top point of the sphere is $v_s + R\omega_s$. So, $v_s + R\omega_s = v$ (Equation 1). The sphere is rolling without slipping on the ground (assumed stationary). The velocity of the bottom point of the sphere is $v_s - R\omega_s$. Since there is no slipping, this velocity must be zero. So, $v_s - R\omega_s = 0 \Rightarrow v_s = R\omega_s$ (Equation 2). Substituting (Equation 2) into (Equation 1): $R\omega_s + R\omega_s = v \Rightarrow 2R\omega_s = v \Rightarrow \omega_s = \frac{v}{2R}$. Then, $v_s = R \left(\frac{v}{2R}\right) = \frac{v}{2}$. The kinetic energy of the sphere is the sum of its translational and rotational kinetic energies: $KE_s = \frac{1}{2} m_1 v_s^2 + \frac{1}{2} I_s \omega_s^2$ $KE_s = \frac{1}{2} m_1 \left(\frac{v}{2}\right)^2 + \frac{1}{2} \left(\frac{2}{5} m_1 R^2\right) \left(\frac{v}{2R}\right)^2$ $KE_s = \frac{1}{2} m_1 \frac{v^2}{4} + \frac{1}{5} m_1 R^2 \frac{v^2}{4R^2}$ $KE_s = \frac{1}{8} m_1 v^2 + \frac{1}{20} m_1 v^2 = \left(\frac{5+2}{40}\right) m_1 v^2 = \frac{7}{40} m_1 v^2$ Substitute $m_1 = 4m$: $KE_s = \frac{7}{40} (4m) v^2 = \frac{7}{10} m v^2$

3. Kinetic Energy of the Cylinder ($KE_c$): The cylinder has mass $m_2 = 5m$ and radius $R$. It is a solid cylinder, so its moment of inertia about its center of mass is $I_c = \frac{1}{2} m_2 R^2$. Let $v_c$ be the translational speed of the cylinder's center of mass and $\omega_c$ be its angular speed. Similar to the sphere, the no-slip condition at the rod-cylinder contact means the top point of the cylinder has speed $v$: $v_c + R\omega_c = v$ (Equation 3). And the no-slip condition at the cylinder-ground contact means the bottom point of the cylinder has zero speed: $v_c - R\omega_c = 0 \Rightarrow v_c = R\omega_c$ (Equation 4). Substituting (Equation 4) into (Equation 3): $R\omega_c + R\omega_c = v \Rightarrow 2R\omega_c = v \Rightarrow \omega_c = \frac{v}{2R}$. Then, $v_c = R \left(\frac{v}{2R}\right) = \frac{v}{2}$. The kinetic energy of the cylinder is the sum of its translational and rotational kinetic energies: $KE_c = \frac{1}{2} m_2 v_c^2 + \frac{1}{2} I_c \omega_c^2$ $KE_c = \frac{1}{2} m_2 \left(\frac{v}{2}\right)^2 + \frac{1}{2} \left(\frac{1}{2} m_2 R^2\right) \left(\frac{v}{2R}\right)^2$ $KE_c = \frac{1}{2} m_2 \frac{v^2}{4} + \frac{1}{4} m_2 R^2 \frac{v^2}{4R^2}$ $KE_c = \frac{1}{8} m_2 v^2 + \frac{1}{16} m_2 v^2 = \left(\frac{2+1}{16}\right) m_2 v^2 = \frac{3}{16} m_2 v^2$ Substitute $m_2 = 5m$: $KE_c = \frac{3}{16} (5m) v^2 = \frac{15}{16} m v^2$

4. Total Kinetic Energy of the System ($KE_{total}$): The total kinetic energy is the sum of the kinetic energies of the rod, sphere, and cylinder: $KE_{total} = KE_{rod} + KE_s + KE_c$ $KE_{total} = \frac{1}{2} m v^2 + \frac{7}{10} m v^2 + \frac{15}{16} m v^2$ Factor out $mv^2$: $KE_{total} = mv^2 \left(\frac{1}{2} + \frac{7}{10} + \frac{15}{16}\right)$ Find a common denominator for 2, 10, and 16, which is 80: $KE_{total} = mv^2 \left(\frac{40}{80} + \frac{7 \times 8}{80} + \frac{15 \times 5}{80}\right)$ $KE_{total} = mv^2 \left(\frac{40}{80} + \frac{56}{80} + \frac{75}{80}\right)$ $KE_{total} = mv^2 \left(\frac{40 + 56 + 75}{80}\right)$ $KE_{total} = \frac{171}{80} mv^2$

The final answer is $\boxed{\frac{171}{80}mv^2}$.

Explanation of the solution:

1. Rod's KE: Calculated as pure translational KE: $KE_{rod} = \frac{1}{2}mv^2$.
2. Sphere's KE: Calculated as combined translational and rotational KE. No-slip conditions at the rod-sphere interface ($v_{top} = v$) and sphere-ground interface ($v_{bottom} = 0$) were used to find the sphere's center of mass velocity ($v_s = v/2$) and angular velocity ($\omega_s = v/(2R)$). Moment of inertia for a solid sphere $I_s = \frac{2}{5}m_1R^2$ was used. $KE_s = \frac{7}{10}mv^2$.
3. Cylinder's KE: Calculated similarly to the sphere. No-slip conditions yielded cylinder's center of mass velocity ($v_c = v/2$) and angular velocity ($\omega_c = v/(2R)$). Moment of inertia for a solid cylinder $I_c = \frac{1}{2}m_2R^2$ was used. $KE_c = \frac{15}{16}mv^2$.
4. Total KE: Sum of individual kinetic energies, resulting in $\frac{171}{80}mv^2$.