Question

A sphere of mass M kg is suspended by a metal wire of length L and diameter d. When in equilibrium, there is a gap of $\Delta l$ between the sphere and the floor. The sphere is gently pushed aside so that it makes an angle $\theta$ with the vertical. Find ${\theta }_{max}$ so that sphere fails to rub the Floor, Young's modulus of the wire is Y.

• A. ${\sin }^{-1}(1-\frac{Y\pi d^2\Delta l}{8MgL})$
• B. ${\tan }^{-1}(1-\frac{Y\pi d^2\Delta l}{8MgL})$
• C. ${\cos }^{-1}(1-\frac{Y\pi d^2\Delta l}{8MgL})$
• D. none

Answer 1

Answer: (C)

${\cos }^{-1}(1-\frac{Y\pi d^2\Delta l}{8MgL})$

(c) $Y=\frac{Fl}{A\Delta l}=\frac{2Mg(1-\cos \theta )L}{\pi \frac{d^2}{4}\Delta l}$

or $1-\cos \theta =\frac{Y\pi d^2\Delta l}{8MgL}$ or $\cos \theta =1-\frac{Y\pi d^2\Delta l}{8MgL}$

Here we have used the energy conservation

$KE=PE$

$\frac{m{v}^2}{2}=mgl(1-\cos \theta )$

or $\frac{m{v}^2}{l}=2mg(1-\cos \theta )$

Thus

$\theta ={\cos }^{-1}(1-\frac{Y\pi d^2\Delta l}{8MgL})$