A wire, which passes through the hole in a small bead, is bent in the form of a quarter of a circle. The wire is fixed vertically on ground as shown in the figure. The bead is released from near the top of the wire and it slides along the wire without friction. As the bead moves from A to B, the force it applies on the wire is [2014]

Question

Smart Achievers · Accepted Answer

(4)

Suppose 'N' is acting radially outward.

Then, $m g \cos θ - N = \frac{m v^{2}}{R}$

$\Rightarrow N = m g \cos θ - \frac{m v^{2}}{R}$ ...(i)

And by energy conservation,

$\frac{1}{2} m v^{2} = m g [R - R \cos θ]$

$∴$ $\frac{v^{2}}{R} = 2 g (1 - \cos θ)$

Putting this value of $\frac{v^{2}}{R}$ in eqn. (i),

$N = m g \cos θ - m [2 g - 2 g \cos θ]$

$\Rightarrow N = m g \cos θ - 2 m g + 2 m g \cos θ$

$\Rightarrow N = 3 m g \cos θ - 2 m g \Rightarrow N = m g (3 \cos θ - 2)$

Clearly when $\cos θ > \frac{2}{3}$ , N is positive and acts radially outward.

So, force on wire is inward and if $\cos θ < \frac{2}{3}$ , N acts radially inward.

So, force on wire is outward.

Solution

1 always radially outwards

2 always radially inwards

3 radially outwards initially and radially inwards later

4 radially inwards initially and radially outwards later

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