Question

A sphere of mass 20 kg is suspended by a metal wire of unstretched length 4 m and diameter 1 mm. When in equilibrium, there is a clear gap of 2 mm between the sphere and the floor. The sphere is gently pushed aside so that the wire makes an angle θ with the vertical and is released. Find the maximum value of θ so that the sphere does not rub the floor. Young modulus of the metal of the wire is 2.0 × 10¹¹ N m⁻². Make appropriate approximations. 

Answer 1

Given:
Mass of sphere (m) = 20 kg
Length of metal wire (L) = 4 m
Diameter of wire (d = 2r) = 1 mm
⇒ r = 5 × 10⁻⁴ m
Young's modulus of the metal wire = 2.0 × 10¹¹ N m⁻²
Tension in the wire in equilibrium = T
T= mg
When it is moved at an angle θ and released, let the tension at the lowest point be T'​. 

\[\Rightarrow \text{ T' = mg} + \frac{\text{mv}^2}{\text{r}}\]

The change in tension is due to the centrifugal force.

∴​ \[∆ \text{ T = T' - T }\]

\[∆ \text{ T }= \frac{\text{mv}^2}{\text{r}} . . . \left(\text{i} \right)\]

Now, using work energy principle:

\[\frac{1}{2}\text{mv}^2 - 0 = \text{mgr}\left( 1 - \cos\theta \right)\]

\[ \Rightarrow \text{v}^2 = 2\text{gr}\left( 1 - \cos\theta \right) . . . \left( 2 \right)\]

Applying the value of v² in (i):

\[∆ T = \frac{\text{m} \left[ 2\text{gr}\left( 1 - cos\theta \right) \right]}{r}\]
\[ = 2\text{mg}\left( 1 - cos\theta \right)\]

\[\text{ Now, F = ∆ T}\]

\[\text{ Also, F }= \frac{\text{ YA ∆ L}}{\text{L}}\]

\[ \Rightarrow \frac{\text{ YA ∆ L}}{\text{L}} = 2\text{mg}\left( 1 - \cos\theta \right)\]

\[ \Rightarrow \cos\theta = 1 - \frac{\text{ YA ∆ L}}{\text{L}\left( 2\text{mg} \right)}\]

\[ \Rightarrow cos\theta = 1 - \left[ \frac{2 \times {10}^{11} \times 4 \times 3 . 14 \times \left( 5 \right)^2 \times {10}^{- 8} \times 2 \times {10}^{- 3}}{4 \times 2 \times 20 \times 10} \right]\]

\[ \Rightarrow \cos\theta = 0 . 80\]

\[\text{ Or}, \theta = 36 . 4^\circ\]

Hence, the required maximum value of θ is 35.4˚.