Linear speed of sphere as it passes through lowest point

[dl447342]

Homework Statement

A slender rod is 0.8 m long and has mass 0.12 kg. A small 0.02 kg sphere is welded to one end of the rod, and a small 0.05 kg sphere is welded to the other end. The rod, pivoting about a stationary, frictionless axis at its center, is held horizontal and released from rest. What is the linear speed of the 0.05 kg sphere as it passes through its lowest point?

Relevant Equations

Moment of inertia of slender rod of mass M and length L: $\frac{1}{12} ML^2$, work done by constant torque through an angular displacement $\Delta \theta = \tau \Delta\theta$, work-kinetic energy theorem, work done by gravity when an object of mass m falls a vertical distance h is $mg h$. Kinetic energy = $\frac{1}2 I \omega^2$.

Let $m_s = 0.05, m_{s_1} = 0.02, m_r = 0.12, L = 0.8.$ be the masses of the two spheres, mass of the rod, and length of the rod. Then the work done by gravity when the rod reaches the vertical position is $(m_s(L/2) - m_{s_2}(L/2))g$ and the kinetic energy equals $\frac{1}2 (\frac{1}{12} m_rL^2 + \frac{L^2}4 (m_s+m_{s_2}))(\omega_f^2)$, where $\omega_f$ is the final angular speed. If I treat the work done by torque as zero, then I get the correct answer of $\omega_f = 3.656 rad/s\Rightarrow v_f = 1.46 m/s$, but I thought the work done by torque should be $\frac{L(m_s-m_{s_2})}2\cdot 9.8$. Why would the torque be zero or is some other quantity different from what I calculated?

 

[Doc Al]

Realize that gravity provides the torque on this system. So the work done by gravity is the work done by the torque. Don't count it twice.

 

[dl447342]

Thanks. I was thinking I was counting something twice.