Question

A Smooth circular tube of radius R is fixed in a vertical plane .A particle is projected from the 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 :
$\bold{\sqrt{ \frac{R}{g} } log(1 + \sqrt{2} ) }$
​

Answer 1

Answer:

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 :

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

Assumption :

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

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.

$\frac{1}{2} m {v}^{2} = mg(2R)$

$= > v = \sqrt{4gR}$

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

So change in height will be :

$R - R \cos( \theta)$

$= R \{1 - \cos( \theta) \}$

Let Velocity be V2 at that height.

Applying Conservation of Mechanical energy :

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

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

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

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

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

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

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

So, we can say that :

$dx = (v2) \: dt$

Integrating on both sides :

$= > \int \: dx = \int \: (v2) \: dt$

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

$= > \int \:R \: d \theta = \int \: (v2) \: dt$

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

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

Putting the limits :

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

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

Putting the limits we get :

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

Answer 2

Answer:

Answer:

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 :

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

g

R

{log(1+

2

)}

Assumption :

\int \: \sec( \theta) d \theta = \log \{\sec( \theta) + \tan( \theta) \}∫sec(θ)dθ=log{sec(θ)+tan(θ)}

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.

\frac{1}{2} m {v}^{2} = mg(2R)

2

1

mv

2

=mg(2R)

= > v = \sqrt{4gR}=>v=

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 :

R - R \cos( \theta)R−Rcos(θ)

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

Let Velocity be V2 at that height.

Applying Conservation of Mechanical energy :

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

2

1

mv

2

=

2

1

m(v2)

2

+mgR{1−cos(θ)}

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

2

=(v2)

2

+2gR{1−cos(θ)}

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

2

+2gR{1−cos(θ)}

= > ( {v2)}^{2} = 2gR \{1 + \cos( \theta) \}=>(v2)

2

=2gR{1+cos(θ)}

= > v2 = \sqrt{2gR \{1 + \cos( \theta) \}}=>v2=

2gR{1+cos(θ)}

= > v2 = 2 \cos( \dfrac{ \theta}{2} ) \sqrt{gR}=>v2=2cos(

2

θ

)

gR

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

So, we can say that :

dx = (v2) \: dtdx=(v2)dt

Integrating on both sides :

= > \int \: dx = \int \: (v2) \: dt=>∫dx=∫(v2)dt

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

= > \int \:R \: d \theta = \int \: (v2) \: dt=>∫Rdθ=∫(v2)dt

= > \int \:R \: d \theta = \int \: \{2 \cos( \frac{ \theta}{2} ) \sqrt{gR} \} \: dt=>∫Rdθ=∫{2cos(

2

θ

)

gR

}dt

\displaystyle = > \int \: dt = \frac{1}{2} \sqrt{ \frac{R}{g} } \int \: \sec( \frac{ \theta}{2} ) d \theta=>∫dt=

2

1

g

R

∫sec(

2

θ

)dθ

Putting the limits :

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

0

t

dt=

2

1

g

R

∫

0

2

π

sec(

2

θ

)dθ

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

2

1

g

R

{

2

1

log{sec(

2

θ

)+tan(

2

θ

)}

}

Putting the limits we get :

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

g

R

{log(1+

2

)}