# Answer only part B) A small smooth particle P of mass m is free to move under gravity in a thin smooth circular tube of radius r and centre O, fixed in a vertical plane. The particle is projected...

**Answer only part B)**

A small smooth particle P of mass m is free to move under gravity in a thin smooth circular tube of radius r and centre O, fixed in a vertical plane. The particle is projected horizontally from the lowest point of the tube with speed `sqrt (3gr)` .

A) If v is the speed of the particle when OP makes an angle `theta ` with the downward vertical show that `v^2 = gr (1 +2 cos theta)` .

B) Hence show that the reaction of the tube on the particle changes its direction when `theta = cos^-1 (-1/3)` and find the speed of the particle at that point

**Answer only part B**

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### 1 Answer

Please refer the image attached.

From part A) we have found that the velocity v of the particle at a point at the downward vertical has the relationship;

`v^2 = gr(1+2costheta)`

When the particle goes upward due to the weight of the particle acting downward it will have a deceleration. So the reaction of the tube on the particle which acts towards O will get reduced. At one moment this reaction will become zero hence the direction will be changed.

Using Newton’s second law (F = ma) at the direction towards center O;

`R-mgcostheta = mv^2/r`

`R = m/r(gr(1+2costheta))+mgcostheta`

When R = 0;

`0 = 1+2costheta+costheta`

`costheta = -1/3`

`theta = cos^(-1)(-1/3)`

*So when `theta = cos^(-1)(-1/3)` the reaction of the tube on the particle changes direction.*

`v^2 = gr(1+2costheta)`

`v^2 = gr(1+2xx(-1/3))`

`v^2 = (1/3)gr`

`v = sqrt((gr)/3)`

** So when the reaction changes its direction particle will have a velocity of **`sqrt((gr)/3)`

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