Question

A solid cube of side-length 10 cm is hanged from one of its vertices by a 1m long
thread. If the cube is allowed to oscillate, find its angular frequency. (Assume that,
the amplitude of oscillation is small, and, g=10m/s^-2

Answer 1

Answer on Question #72040-Physics-Other

A solid cube of side-length 10 cm is hanged from one of its vertices by a 1m long thread. If the cube is allowed to oscillate, find its angular frequency. (Assume that, the amplitude of oscillation is small, and, $g=10\mathrm{m}/\mathrm{s}^{\wedge}-2$

Solution

The angular frequency of physical pendulum is

$$\omega = \sqrt{\frac{mgL}{I}}$$

In our case:

$$L = 1 + \frac{0.1}{\sqrt{2}}$$ $$I = \frac{ma^2}{6} + mL^2.$$ $$\omega = \sqrt{\frac{mgL}{\frac{ma^2}{6} + mL^2}} = \sqrt{\frac{gL}{\frac{a^2}{6} + L^2}} = \sqrt{\frac{10\left(1 + \frac{0.1}{\sqrt{2}}\right)}{\frac{0.1^2}{6} + \left(1 + \frac{0.1}{\sqrt{2}}\right)^2}} = 3 \frac{rad}{s}.$$

Answer: $3 \frac{rad}{s}$

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