91影视

$T=2蟺\sqrt{\frac{I}{mgl}}$

Where, I is the moment of inertia, m is the mass of the object, g is the acceleration due to gravity and l is the distance between pivot and center of mass.

a)The rod's length is given as

L is the length of the pivot, and the distance between the pivot and the center of mass is $\mathit{x}$.

The expression for an object's moment of inertia about any other point other than the moment of inertia is:

localid="1651404148929" $I_{\mathrm{rod}}=I_{\mathrm{cm}}+M{x}^2$

Where, localid="1651404155522" $I_{\mathrm{cm}}$is the moment of inertia at center of mass, M is the mass of rod and x is the distance between pivot and center of mass.

As a result, the moment of inertia of the rod about the pivot is:

localid="1651404158982" $I_{\mathrm{rod}}=\frac{1}{2}M{L}^2+M{x}^2$

Substitute localid="1651404163198" $I_{\text{rod}}$for I in the equation for time period,

localid="1651404167466" $T=\sqrt{\frac{I}{Mgx}}$

localid="1651404172342" $=2蟺\sqrt{\frac{\frac{1}{12}M{L}^2+M{x}^2}{Mgx}}$

localid="1651404177706" $=2蟺{\left(\frac{\frac{1}{12}{L}^2+x^2}{gx}\right)}^{1/2}$

Take the derivate of T with respect to time and equate it to zero to find the length of x for which the time period is shortest. The time period derivate with respect to x is:

localid="1651404184908" $\frac{dT}{dx}=蟺{\left(\frac{\frac{1}{12}{L}^2+x^2}{gx}\right)}^{鈭�1/2}\left(\frac{2x^2鈭�\left(\frac{1}{12}{L}^2+x^2\right)}{g{x}^2}\right)$

Equatelocalid="1651404192448" $\frac{dT}{dx}=0$, but only term with subtracting values will be possibly equal to zero Thus,

localid="1651404198632" $2x^2鈭�\left(\frac{1}{12}{L}^2+x^2\right)=0$

localid="1651404236683" $2x^2=\frac{1}{12}{L}^2+x^2$

localid="1651404243754" $x=\frac{1}{\sqrt{12}}L$

As a result, the value of the distance between the pivot and the center of mass is localid="1651404251141" $\frac{1}{\sqrt{12}}L$.

b)A physical pendulum is a solid object that swings back and forth on a pivot as a result of gravity. A physical pendulum's time period is expressed as:

$T=2蟺\sqrt{\frac{I}{mgl}}$

Where I denotes the moment of inertia, m the mass of the object, g the gravitational acceleration, and l the distance between the pivot and the center of mass.

b)The mass of the rod is$15\mathrm{kg}$and the length of the rod is $1.0\mathrm{m}$.

The time period for value $\frac{1}{\sqrt{12}}L$For x, the value in the expression for time period is substituted.

$T=2蟺\sqrt{\frac{I_{\text{rod}}}{Mgx}}$

Substitute $\frac{1}{\sqrt{12}}L$for $x,\frac{1}{2}M{L}^2+M{x}^2$for $I_{\text{rod}},1.0\mathrm{m}$for L and $9.8\mathrm{m}/{\mathrm{s}}^2$for g in the equation,

$T=2蟺\sqrt{\frac{\frac{1}{12}{L}^2+x^2}{gx}}$

$=2蟺\sqrt{\frac{\frac{1}{12}{L}^2+\frac{1}{12}{L}^2}{g\left(\frac{1}{\sqrt{12}}L\right)}}$

$=1.5\mathrm{s}$

As a result, the time period is $1.5\mathrm{s}$