Question

You have a pendulum clock made from a uniform rod of mass M and length L pivoting around one end of the rod. Its frequency is 1 radian/sec. The pivot breaks. Though the mass change is negligible, you need to rehang the rod halfway between the old pivot point and the middle of the rod. The new oscillation frequency is:
A. 0.88 rad/sec
B. 1.13 rad/sec
C. 1.07 rad/sec
D. 0.92 rad/sec

Answer 1

The new oscillation frequency of the pendulum clock is 1.14 rad/s.

     

The given parameters;

• Mass of the pendulum, = M
• Length of the pendulum, = L
• Initial angular speed, $\omega _i$ = 1 rad/s

The moment of inertia of the rod about the end is given as;

$I_i = \frac{1}{3} ML^2$

The moment of inertia of the rod between the middle and the end is calculated as;

$I_f = \int\limits^L_{L/2} {r^2\frac{M}{L} } \, dr = \frac{M}{3L} [r^3]^L_{L/2} = \frac{M}{3L} [L^3 - \frac{L^3}{8} ] = \frac{M}{3L} [\frac{7L^3}{8} ]= \frac{7ML^2}{24}$

Apply the principle of conservation of angular momentum as shown below;

$I _i \omega _i = I _f \omega _f\\\\\frac{ML^2}{3} (1 \ rad/s)= \frac{7ML^2}{24} \times \omega _f\\\\\frac{24 \times ML^2}{3 \times 7 ML^2} (1 \ rad/s)= \omega _f\\\\1.14 \ rad/s = \omega _f$

Thus, the new oscillation frequency of the pendulum clock is 1.14 rad/s.

Learn more about moment of inertia of uniform rod here: brainly.com/question/15648129