Question

A small particle is moving in x-y plane on a smooth wire bent in curve shape as shown in figure. At a certain instant it is at P and moving
with speed vo. Here gravity is in negative y-direction and mass of particle is m At this instant y-component of acceleration is - 6 m/s2
Given vo = 2 m/s, m = 2kg, 0 = 45° then :-​

Answer 1

Answer:

Given:

$v_0=2$ m/s

$m=2$ kg

$\theta=45^\circ$

Acceleration $=-6\hat j$ m/s²

At the given instant, the velocity of the particle is given by

$v_P=v_0\cos\theta\hat i+v_0\sin\theta\hat j$

$\implies v_P=2\cos 45^\circ\hat i+2\sin 45^\circ\hat j$

$\implies v_P=2(\frac{1}{\sqrt{2}}\hat i+\frac{1}{\sqrt{2}}\hat j)$

$\implies v_P=\sqrt{2}(\hat i+\hat j)$

Given that acceleration due to gravity g is acting in the -y direction

Therefore, the force due to gravity mg is acting in the negative y direction

The component of mg perpendicular to the tangent at point P will be $mg\cos 45^\circ=\frac{mg}{\sqrt{2}}$

At point P, the centripetal force is $\frac{mv_0^2}{R}$, where R is the radius of curvature

Thus,

$\frac{mv_0^2}{R}=\frac{mg}{\sqrt{2}}$

$\implies \frac{2^2}{R}=\frac{g}{\sqrt{2}}$

$\implies R=\frac{4\sqrt{2}}{g}$ m

$\implies R=\frac{4\sqrt{2}}{10}$

$\implies R=\frac{2\sqrt{2}}{5}$ m

Since the y-component of acceleration is $-6\hat j$ m/s²

And there is $-10\hat j$ m/s² of acceleration due to gravity is acting too

Therefore, the additional acceleration in y-direction is $4\hat j$ m/s²

Thus, due to angle of 45°, the component of acceleration in x-direction will be $4\hat i$ m/s²

If normal reaction is N

Then

$N=mg\cos 45^\circ$

$\implies N=2\times 10\times\frac{1}{\sqrt{2}}$

$\implies N=10\sqrt{2}$ Newton

Hope this answer is helpful.