Question

Consider two pendula consisting of a piece of clay stuck to a massless rigid rod, which is pivoted at its end. The two L- CEILING pendula are released from a horizontal orientation, as pivotV o pivot a drawn here, and allowed to swing down. On pendulum 1, the clay s stuck to the midpoint of the rod·On pendulum Pendulum 2, it's stuck to the end of the rod. In both cases, the rods have the same length, and the pieces of clay have the same mass. And in both cases, the pendulum swings down because a torque due to gravity acts an the clay. (Remember, the rod is essentially massless.) 11. 1 Pendulum 2 (a) What's different about pendulum 1 vs. pendulum 2: The rotational inertia, the torque, or both? denote the initial torque on pendulum 1 and pendulum 2, respectively, to τ? Is what's (b) Let τ| and the ratio of twice as big as τγ three times as big, or what? Explain. (c) Let 1 and 12 denote the rotational inertia of pendulum 1 and pendulum 2, respectively. What's the ratio of 12 to I? Explain. which pendulum reaches the bottom of its swing first. If you have time, create an experiment to test your answer (though the rod won't be infinitesimally light compared to the "clay," obviously (d) Using your answers to (b) and (c), figure out which pendulum swings down more quickly, i.e.,

Answer 1

(a) Both the Rotational Inertia and the Torques are different, Let's say is the length of the massless rod, then the rotational inertia is essentially due to the clay mass, so

and  

$\tau_{1} = (l/2)mgsin\theta$ and $\tau_{2} = lmgsin\theta$

(b) The ratio of the torques are

$\frac{\tau_{2}}{\tau_{1}}= \frac{l}{l/2}= 2$

(c) the ratio of the Inertias are

$\frac{I_{2}}{I_{1}}= \frac{l^{2}}{l^{2}/4}= 4$

(d) For the same angular displacement, we could calculate the ratio of angular accelerations, the one with the higher angular acceleration would reach the bottom first

$\frac{\alpha_{1}}{\alpha_{2}}= \frac{\tau_{1}}{I_{1}} \times \frac{I_{2}}{\tau_{2}}= \frac{4}{2} =2$

The pendulum 1 has twice the acceleration as that of 2, as expected the reason is torque acting is halved but the inertia is quartered giving pendulum 1 a higher angular acceleration.