A thin rod MN, free to rotate in the vertical plane about the fixed end N, is held horizontal. When the end M is released the speed of this end, when the rod makes an angle α with the horizontal, will be proportional to : (see figure)

  • Option 1)

    \sqrt{sin }\alpha

  • Option 2)

    sin\alpha

  • Option 3)

    \sqrt{cos}\: \alpha

  • Option 4)

    cos\alpha

 

Answers (2)
A Anuj Pandey
D Divya Saini

As we have learnt

Kinetic energy of rotation -

K=\frac{1}{2}Iw^{2}

- wherein

I = moment of inertia about axis of rotation

w = angular velocity

 

 

From energy conservation,

\frac{mg.L}{2}sin\alpha =\frac{1}{2}\left ( \frac{mL^{2}}{3} \right ).\omega ^{2}

\omega ^{2}=\frac{3g}{l}sin\alpha or   \omega=\sqrt{\frac{3g}{l}}\sqrt{sin }\alpha

speed of end m=\omega l

=\sqrt{3gl}\sqrt{sin }\alpha

\therefore v\: \alpha \sqrt{sin }\: \alpha

 

 


Option 1)

\sqrt{sin }\alpha

Option 2)

sin\alpha

Option 3)

\sqrt{cos}\: \alpha

Option 4)

cos\alpha

Exams
Articles
Questions