Question

A smooth circular tube of radius $R$ is fixed in a vertical plane. A particle is projected from its lowest point with a velocity just sufficient to carry it to the highest point. Show that the time taken by the particle to reach the end of the horizontal diameter is $\sqrt{ \dfrac{R}{g} } \: ln \: (1 + \sqrt{2} )$.​

Answer 1

⊙ S O L U T I O N ⊙

Given:

Given:Velocity of particle at starting point is adequate just to reach the top . Radius of track is given as

• R

To show :

❏ Time taken to reach a position of horizontal diameter is :

⇝ $\bf t = \sqrt{ \dfrac{R}{g} } \{ log(1 + \sqrt{2} ) \}$

Assumption :

⇝ $\bf\int \: \sec( \theta) d \theta = \log \{\sec( \theta) + \tan( \theta) \}$

❏ Calculation:

• First of all we will calculate the velocity at bottom necessary to reach the top . Let the velocity be v , such that all the Kinetic energy will be converted to Potential energy at top.

⇝ $\bf\frac{1}{2} m {v}^{2} = mg(2R)$

⇝ $\bf v = \sqrt{4gR}$

Now , let's consider a position on the track such that it is angle of θ with vertical.

❏ So change in height will be :

⇝ $\bf R - R \cos( \theta)R−Rcos(θ)$

⇝ $\bf= R \{1 - \cos( \theta) \}=R{1−cos(θ)}$

• Let Velocity be V2 at that height.

• Applying Conservation of Mechanical energy :

$⇝ \bf\frac{1}{2} m {v}^{2} = \frac{1}{2} m {(v2)}^{2} + mgR \{1 - \cos( \theta) \}$

$\bf⇝ {v}^{2} = {(v2)}^{2} + 2gR \{1 - \cos( \theta) \}$

$\bf ⇝ 4gR = {(v2)}^{2} + 2gR \{1 - \cos( \theta)$

$\bf⇝ ( {v2)}^{2} = 2gR \{1 + \cos( \theta) \}$

$\bf ⇝ v2 = \sqrt{2gR \{1 + \cos( \theta) \}}$

$\bf ⇝ v2 = 2 \cos( \dfrac{ \theta}{2} ) \sqrt{gR}$

❏ Now , let's consider movement of a small distance on the track (dx) in time (dt)

❏ So, we can say that :

$\bf⇝ dx = (v2) \: dtdx=(v2)dt$

• Integrating on both sides :

$\bf⇝ \int \: dx = \int \: (v2)$

The small distance can be written in form of angle and radius :

$\bf⇝ \int \:R \: d \theta = \int \: (v2)$

$\bf⇝ \int \:R \: d \theta = \int \: \{2 \cos( \frac{ \theta}{2} ) \sqrt{gR} \}$

$\sf\displaystyle ⇝ \int \: dt = \frac{1}{2} \sqrt{ \frac{R}{g} } \int \: \sec( \frac{ \theta}{2} ) d \theta$

❏ Putting the limits :

$\displaystyle ⇝ \bf \int_{0}^{t} \: dt = \frac{1}{2} \sqrt{ \frac{R}{g} } \int_{0}^{ \frac{\pi}{2}} \: \sec( \frac{ \theta}{2} ) d \theta$

$\displaystyle ⇝ \bf t = \frac{1}{2} \sqrt{ \frac{R}{g} } \: \bigg \{\dfrac{ \log \{\sec( \frac{ \theta}{2} ) + \tan( \frac{ \theta}{2} ) \}}{ \frac{1}{2} } \bigg \}$

Putting the limits we get :

$\bf⇝ t = \sqrt{ \dfrac{R}{g} } \{ log(1 + \sqrt{2} ) \}$