Question

A light rod of length / is pivoted at the upper end. A point mass m is attached at the free end of the rod. What horizontal velocity must be imparted to the lower end mass, so that tension in the rod is 2mg when it is at the topmost position?​

Answer 1

Given:

A light rod of length "l" is pivoted at the upper end. A point mass m is attached at the free end of the rod.

To find:

Horizontal velocity must be imparted to the lower end mass, so that tension in the rod is 2mg when it is at the topmost position.

Calculation:

Let velocity at top most position be v ;

$\therefore \: T + mg = \dfrac{m {v}^{2} }{l}$

$= > \: 2mg + mg = \dfrac{m {v}^{2} }{l}$

$= > \: 3mg = \dfrac{m {v}^{2} }{l}$

$= > \: v = \sqrt{3gl}$

Let Velocity applied at lower end be u ;

Applying Conservation of Mechanical Energy:

$\therefore \: 0 + \dfrac{1}{2} m {u}^{2} = mg(2l) + \dfrac{1}{2} m {v}^{2}$

$= > \: \dfrac{1}{2} m {u}^{2} = mg(2l) + \dfrac{1}{2} m {v}^{2}$

$= > \: \dfrac{1}{2} m {u}^{2} = mg(2l) + \dfrac{1}{2} m {( \sqrt{3gl}) }^{2}$

$= > \: \dfrac{1}{2} {u}^{2} = g(2l) + \dfrac{1}{2} {( \sqrt{3gl}) }^{2}$

$= > \: \dfrac{1}{2} {u}^{2} = 2gl + \dfrac{3gl}{2}$

$= > \: {u}^{2} = 4gl + 3gl$

$= > \: {u}^{2} =7gl$

$= > \: u = \sqrt{7gl}$

So, final answer is:

$\boxed{ \red{ \bold{ \huge{ u = \sqrt{7gl}}}}}$

Answer 2

Explanation:

Given:

A light rod of length "l" is pivoted at the upper end. A point mass m is attached at the free end of the rod.

To find:

Horizontal velocity must be imparted to the lower end mass, so that tension in the rod is 2mg when it is at the topmost position.

Calculation:

Let velocity at top most position be v ;

\therefore \: T + mg = \dfrac{m {v}^{2} }{l}∴T+mg=

l

mv

2

= > \: 2mg + mg = \dfrac{m {v}^{2} }{l}=>2mg+mg=

l

mv

2

= > \: 3mg = \dfrac{m {v}^{2} }{l}=>3mg=

l

mv

2

= > \: v = \sqrt{3gl}=>v=

3gl

Let Velocity applied at lower end be u ;

Applying Conservation of Mechanical Energy:

\therefore \: 0 + \dfrac{1}{2} m {u}^{2} = mg(2l) + \dfrac{1}{2} m {v}^{2}∴0+

2

1

mu

2

=mg(2l)+

2

1

mv

2

= > \: \dfrac{1}{2} m {u}^{2} = mg(2l) + \dfrac{1}{2} m {v}^{2}=>

2

1

mu

2

=mg(2l)+

2

1

mv

2

= > \: \dfrac{1}{2} m {u}^{2} = mg(2l) + \dfrac{1}{2} m {( \sqrt{3gl}) }^{2}=>

2

1

mu

2

=mg(2l)+

2

1

m(

3gl

)

2

= > \: \dfrac{1}{2} {u}^{2} = g(2l) + \dfrac{1}{2} {( \sqrt{3gl}) }^{2}=>

2

1

u

2

=g(2l)+

2

1

(

3gl

)

2

= > \: \dfrac{1}{2} {u}^{2} = 2gl + \dfrac{3gl}{2}=>

2

1

u

2

=2gl+

2

3gl

= > \: {u}^{2} = 4gl + 3gl=>u

2

=4gl+3gl

= > \: {u}^{2} =7gl=>u

2

=7gl

= > \: u = \sqrt{7gl}=>u=

7gl

So, final answer is:

\boxed{ \red{ \bold{ \huge{ u = \sqrt{7gl}}}}}

u=

7gl