A conical pendulum of length 1 m makes an angle  \theta=45\degree w.r.t. Z-axis and moves in a circle in the XY plane.  The radius of the circle is 0.4 m and its center is vertically below O.  The speed of the pendulum, in its circular path, will be : (Take g=10 ms−2)

 

  • Option 1)

    0.4 m/s

  • Option 2)

    4 m/s

  • Option 3)

     0.2 m/s

  • Option 4)

     2 m/s

     

 

Answers (1)
A Aadil Khan

As we learnt in

If friction is also present in banking of road -

\frac{V^{2}}{rg}=\frac{\mu+tan\theta}{1-\mu tan \theta}

\theta= angle of banking

\mu= coefficient of friction

V = velocity

- wherein

Maximum speed on a banked frictional road

V=\sqrt{\frac{rg(\mu+tan\theta)}{1-\mu tan\theta}}

 

 

Tsin\theta=\frac{mv^{2}}{r}                                    (i)

Tcos\theta=mg                                        (ii)

tan\theta=\frac{v^{2}}{rg}=v^{2}=gr\ tan\theta

v^{2}=gr

v=\sqrt{gr}=\sqrt{0.4\times10}=2ms^{-1}

Correct option is 4.


Option 1)

0.4 m/s

This is an incorrect option.

Option 2)

4 m/s

This is an incorrect option.

Option 3)

 0.2 m/s

This is an incorrect option.

Option 4)

 2 m/s

 

This is the correct option.

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