Q.A pendulum consisting of a massless string of length 20 cm and a tiny bob of mass 100 g is set up as a conical pendulum. Its bob now performs 75 rpm. Calculate the kinetic energy and the increase in the gravitational potential energy of the bob. (Use $g = \pi^2$ m/s^2)

With $n=1.25$ rps ($\omega=2.5\pi$ rad/s) and $g=\pi^2$ m/s^2, $\cos\theta=\dfrac{g}{L\omega^2}=0.8$, giving $\sin\theta=0.6$; then KE $=\frac{1}{2}mv^2\approx0.044$ J and $\Delta$PE $=mgL(1-\cos\theta)\approx0.0 …

n = 75 rpm = 1.25 rps, so $\omega=2\pi n=2.5\pi$ rad/s. Using $\omega^2=\dfrac{g}{L\cos\theta}$ with $g=\pi^2$ m/s^2 and L = 0.2 m: $\cos\theta=\frac{g}{L\omega^2}=\frac{\pi^2}{0.2\times(2.5\pi)^2}=\frac{\pi^2}{0.2\times6.25\pi^2}=\frac{1}{1.25}=0.8$ so $\sin\theta=\sqrt{1-0.8^2}=\sqrt{0.36}=0.6$.

Radius of the circular path: $r=L\sin\theta=0.2\times0.6=0.12$ m. Speed of the bob: $v=r\omega=0.12\times2.5\pi=0.3\pi\approx0.9425$ m/s, so $v^2\approx0.8886$ m^2/s^2. Kinetic energy: $K.E.=\frac{1}{2}mv^2=\frac{1}{2}(0.1)(0.8886)\approx0.0444\text{ J}$ (this platform's own computation gives approximately 0.044 J; a scanned/printed source value quoting a different figure at this last digit is not treated as authoritative here, since the underlying formula and every other quantity in this problem check out consistently). …

Use $\omega^2=g/(L\cos\theta)$ to find $\cos\theta$ (and hence $\sin\theta$) from the given frequency, then find the bob's radius and speed to get its kinetic energy …

Using g = 10 m/s^2 (the chapter's usual default) instead of the $g=\pi^2$ m/s^2 convention this problem needs to make $\omega^2$ (which involves $\pi^2$ from $\omega=2\pi n$) simplify cleanly against g; or forgetting to co …