Question

A uniform thin rod of length 0.766 m is hung from a horizontal nail passing through a small hole in the rod located 0.044 m from the rod's end. When the rod is set swinging about the nail at small amplitude, what is the period T of oscillation? T=______ s

Answer 1

We given with data Length of the rod L = 0.766 m and pivoted at a point 0.044 m from end.

Figure 1

Considering the figure Let ' h ' be the distance from pivot point to the cetre of the rod

Then h =( 0.766/2)- 0.044 = 0.339 m

Time period of a physical prndulam is T=2π√(Imgh) Equation (1)  Where, I - Moment of inertia ,

g - acceleration due gravity

Hence by parellel axes theoram,

Here I = I _cm + mh² Equation (2)  ; I _cm - rotational inertia about the centre of mass.

And for a thin rod I _cm = 1 /12 mL²

ie, I = 1 /12 mL² + mh² = m ( L²/12 + h² )

So Equation 1 become, T= 2π √ ( (m( L²/12 + h² )) / mgh ) = 2π √( (L²/12 + h² ) / gh)

T = 2π √ (L² + 12 h² ) / ( 12gh)

Substituting values for L= 0.766m, h = 0.339m, g = 9.8 ms^-2

T=  2π √ ((0.766)² + 12 x( 0.339)² ) / (12 x 9.8 x 0.339)

=  2π √ 39.8664

Time period of oscillation,T = 39.67 s.

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