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 Dl between the sphere and the floor. The sphere is gently pushed aside so that it makes an angle q with the vertical. Find q_max so that sphere fails to rub the Floor. Young's modulus of the wire is Y –

• A. sin^–1$\left( 1–\frac{Y\pi d^{2}\Delta\mathcal{l}}{8MgL} \right)$
• B. tan^–1 $\left( 1–\frac{Y\pi d^{2}\Delta\mathcal{l}}{8MgL} \right)$
• C. cos^–1$\left( 1–\frac{Y\pi d^{2}\Delta\mathcal{l}}{8MgL} \right)$
• D. None

Answer 1

Answer: (C)

cos^–1$\left( 1–\frac{Y\pi d^{2}\Delta\mathcal{l}}{8MgL} \right)$

Answer 2

Y = $\frac{F\mathcal{l}}{A\Delta\mathcal{l}}$ = $\frac{2Mg(1–\cos\theta)L}{\pi\frac{d^{2}}{4}\Delta\mathcal{l}}$

[Q$\frac{Mv^{2}}{2}$ = Mgl(1 – cos q)

Ž $\frac{Mv^{2}}{\mathcal{l}}$ = 2Mg (1 – cos q)]

1 – cos q = $\frac{Y\pi d^{2}\Delta\mathcal{l}}{8Mg\mathcal{l}}$ Ž cos q = 1 – $\frac{Y\pi d^{2}\Delta\mathcal{l}}{8Mg\mathcal{l}}$