Advertisement Remove all ads

A closed circular wire hung on a nail in a wall undergoes small oscillations of amplitude 20 and time period 2 s. - Physics


A closed circular wire hung on a nail in a wall undergoes small oscillations of amplitude 20 and time period 2 s. Find (a) the radius of the circular wire, (b) the speed of the particle farthest away from the point of suspension as it goes through its mean position, (c) the acceleration of this particle as it goes through its mean position and (d) the acceleration of this particle when it is at an extreme position. Take g = π2 m/s2.

Advertisement Remove all ads


It is given that:
Time period of oscillation, T = 2 s
Acceleration due to gravity, g =\[\pi^2\]ms−2

Let I be the moment of inertia of the circular wire having mass m and radius r.

(a) Time period of compound pendulum \[\left( T \right)\] is given by ,

\[T = 2\sqrt{\frac{I}{mgl}} = 2\sqrt{\frac{I}{mgr}}\]

\[\left( \because l = r \right)\] ...(1)  
Moment of inertia about the point of suspension  is calculated as,
I = mr2 + mr2 = 2mr2
On substituting the value of moment of inertia I in equation (1), we get:

\[T = 2\pi\sqrt{\frac{2m r^2}{mgr}} = 2\pi\sqrt{\frac{2r}{g}}\] 

\[ \Rightarrow \frac{2}{2\pi} = \sqrt{\frac{2r}{g}}\] 

\[ \Rightarrow \frac{2r}{g} = \frac{1}{\pi^2}\] 

\[ \Rightarrow r = \frac{g}{2 \pi^2}\] 

\[           = 0 . 5  m = 50  cm\]

(b) From the energy equation, we have:

\[\frac{1}{2}I \omega^2  - 0 = mgr  \left( 1 - \cos  \theta \right)\] 

\[\frac{1}{2}I \omega^2  - 0 = mgr  \left( 1 - \cos  2^\circ \right)\]

\[\Rightarrow \left( \frac{1}{2} \right)2m r^2  \cdot  \omega^2  = mgr\left( 1 - \cos  2^\circ\right)                    \left( \because I = 2m r^2 \right)\] 

\[ \Rightarrow  \omega^2  = \frac{g}{r}\left( 1 - \cos  2^\circ\right)\] 

\[\text { On  substituing  the  value  of  g  and  r  in  the  above  equation,   we  get:}\] \[\omega = 0 . 11  rad/s\] 

\[ \Rightarrow v = \omega \times 2r = 11   {\text { cms }}^{- 1}\]

(c) The acceleration is found to be centripetal at the extreme position.
    Centripetal acceleration at the extreme position \[\left( a_n \right)\] is given by,
    an = ω2(2r) = (0.11) × 100 = 12 cm/s2
    The direction of an is towards the point of suspension.

(d) The particle has zero centripetal acceleration at the extreme position. However, the particle will still have acceleration due to the S.H.M.
 Angular frequency \[\left( \omega \right)\] is given by ,

\[\omega = \frac{2\pi}{T}\] 

\[   = \frac{2\pi}{2} = 3 . 14\]

\[\therefore\] Angular Accelaration \[\left( a \right)\] at the extrame position is given as ,

\[\alpha =  \omega^2 \theta\] 

\[\alpha =  \omega^2 2^\circ=  \pi^2  \times \frac{2\pi}{180}\] 

\[   = \frac{2 \pi^3}{180}  \left[ 1^\circ= \frac{\pi}{180}\text { radian } \right]\]

Thus , tangential acceleration

\[= \alpha\left( 2r \right) = \left( \frac{2 \pi^3}{180} \right) \times 100\]

  = 34 cm/s2

  Is there an error in this question or solution?
Advertisement Remove all ads


HC Verma Class 11, 12 Concepts of Physics 1
Chapter 12 Simple Harmonics Motion
Q 52 | Page 255
Advertisement Remove all ads

Video TutorialsVIEW ALL [1]

Advertisement Remove all ads

View all notifications

      Forgot password?
View in app×