91影视

• The disk is pivoted at a point on its rim

The physical pendulum exhibits SHM when the pivot point is at some distance from the center of mass. Its period of oscillation depends upon the position of the pivot point.

Formula:

The period of oscillations,$T=2蟿蟿\sqrt{\frac{l}{mgh}}$ (i)

Here, I is the moment of inertia of the object about an axis passing through the pivot point and h is the distance between the pivot point and centre of mass.

In the given situation, the pivot point is at the rim. Hence h = R and the moment of inertia can be determined using the parallel axes theorem. Hence, the moment of inertia is given by:

$$\begin{gathered} l=l_{COM}+M{K}^2 \\ =\frac{1}{2}M{R}^2+M{R}^2(鈭�l_{COM}ofphysicalpendulum=\frac{1}{2}M{R}^2) \\ =\frac{3}{2}M{R}^2 \end{gathered}$$

(Since, k = R )

Now, the period of oscillations using equation (i) can be given as:

$$\begin{gathered} T=2蟿蟿\frac{\sqrt{{\displaystyle \frac{3}{2}}M{R}^2}}{MgR} \\ =2蟿蟿\sqrt{\frac{3R}{2g}}.......................\left(a\right) \\ =0.873s \end{gathered}$$

Hence, the value of period is 0.873 s.

Let us assume the new pivot point is at distance, r < R.

Thus, we write, the moment of inertia is given by:

$$\begin{gathered} l=l_{COM}+M{k}^2 \\ =\frac{1}{2}M{R}^2+M{r}^2 \end{gathered}$$

And the expression for period using equation (i) will be given as:

$$\begin{gathered} T=2蟿蟿\frac{\sqrt{{\displaystyle \frac{3}{2}M{R}^2+M{r}^2}}}{MgR} \\ =2蟿蟿\sqrt{\frac{R^2+2r^2}{2gr}}(fromequation(a\left)\right) \\ 2蟿蟿\sqrt{\frac{R^2+2r^2}{2gr}}=2蟿蟿\sqrt{\frac{3R}{2g}} \end{gathered}$$

After simplifying it, we get the equation as:

$2r^2-3Rr+R^2=0$

Hence, the roots of this quadratic equation are given as:

$r=R\mathrm{or}\frac{R}{2}$

But r < R, hence the position is at,

$$\begin{gathered} r=\frac{R}{2} \\ =\frac{12.6}{2} \\ =6.3\mathrm{cm} \end{gathered}$$

Hence, the position at which the period is similar is 6.3 cm.