Question 7

The point $$A$$ moves with a uniform speed along the circumference of a circle of radius 0.36 m and covers 30° in 0.1 s. The perpendicular projection $$P$$ from $$A$$ on the diameter $$MN$$ represents the simple harmonic motion of $$P$$. The restoration force per unit mass when $$P$$ touches $$M$$ will be:

Solution

$$\omega = \frac{\theta}{\Delta t} = \frac{\pi / 6}{0.1} = \frac{5\pi}{3}\text{ rad s}^{-1}$$

When projection $$P$$ touches extreme point $$M$$, displacement equals the radius: $$x = R = 0.36\text{ m}$$

$$a = \omega^2 R \implies a = \left(\frac{5\pi}{3}\right)^2 \times 0.36 = \frac{25\pi^2}{9} \times 0.36 = 25\pi^2 \times 0.04 = \pi^2\text{ N kg}^{-1}$$

$$a \approx 9.87\text{ N kg}^{-1}$$

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