Question

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1-)A physical pendulum consists of a meter stick that is pivoted at a small hole drilled through the stick a distance d from the 50 cm mark. The period of oscillation is 2.24 s. Find d (in cm's).

Answer 1

The time period of a simple pendulum is given by the formula,

Where,

T = Time period

g = acceleration due to gravity = 9.8 m/s²

= Moment of inertia

d = distance from the pivot point to the center of mass

We have to use the parallel axis theorem to find out the moment of inertia of the pendulum.

Where,

= Moment of inertia about the axis passing through the point of pivot

= Moment of inertia about the axis passing through the center of mass

= Mass of the rod

= distance of the rod from the center of mass

Here the center of mass of the stick is at the 50 cm point

We know that a thin rod has a moment of inertia about the center of mass is given by the formula,

Where,

L = Full length of the stick = 100 cm = 1 m

Thus, by substitution,

canceling out the mass

multiplying and dividing all terms in the root by 12

and we can write

Squaring the equation,

We have a quadratic type of equation in variable ' d ' as below

this is similar to the equation in the form

whose solution is given by the formula

Here,

  

Thus

Now T = 2.24 s

and L = 1 m

we know, g = 9.8 m/s²

Thus we get

So either

which is impossible

or

which is possible

Thus the value of ' d ' = 7.1 cm

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