91Ó°ÊÓ

The spring constant of the spring is k=2000 N/m

The distance of the point attached to the spring from the center of mass of the plate, $\mathrm{r}=2.5\mathrm{cm}=0.025\mathrm{m}$

The plate is rotated by,$\mathrm{θ}=7^{°}=0.122\mathrm{rad}$

The time scale, ${\mathrm{t}}_{\mathrm{s}}=20\mathrm{ms}=20×{10}^{-3}\mathrm{s}$

For a given system, when a body undergoes both the translational and the rotational motions, we solve for the required kinematics considering that energy is neither created nor destroyed, which means its always conserved. Here, the metal plate as it is mounted on an axle when displaced horizontally undergoes a translational change preserving the potential energy in its motion, while due to the rotation about the mount; it also undergoes the kinetic change due to the release of the body afterward. Thus, applying the principle of energy conservation, you can relate the body's translational motion to the rotational motion.

Formulae:

The potential energy of a body in spring oscillations,

$\mathrm{UE}=\frac{1}{2}{\mathrm{kx}}^2$ ….. (i)

Where, k is the spring constant, x is the displacement of the body.

The kinetic energy of a body in rotational motion,

$\mathrm{KE}=\frac{1}{2}{\mathrm{\pm õÓ\neg }}_{\mathrm{m}}^2$ ….. (ii)

Where, I is the moment of inertia of a body about its central axis, ${\mathrm{Ó\neg }}_{\mathrm{m}}$is the maximum angular speed of the body.

The displacement of the body in transverse motion,

$\mathrm{x}=\mathrm{°ùθ}$ ….. (iii)

Where, r is the radial distance about which the rotational motion occurs, $\mathrm{θ}$is the angular displacement of the body,

The maximum angular frequency of the oscillation,

${\mathrm{Ó\neg }}_{\mathrm{m}}=\mathrm{Ó\neg θ}=\left(\frac{2\mathrm{Ï€}}{\mathrm{T}}\right)\mathrm{θ}$ ….. (iv)

Where, $\mathrm{Ó\neg }$is the angular frequency of an oscillation, T is the period of oscillation, $\mathrm{θ}$is the angular displacement of the body.

We are given that the spring is rotated through angle$\mathrm{θ}=7^{°}$so the arc length or the stretching length of the spring is given using equation (iii) as follows:

$\begin{array}{rcl}\mathrm{x}& =& 0.025\mathrm{m}×0.122\\ & =& 0.0031\mathrm{m}\end{array}$

So, the potential energy of the spring is given using the above data in equation (i) as follows:

$\begin{array}{rcl}\mathrm{UE}& =& \frac{1}{2}×\left(2000\mathrm{N}/\mathrm{m}\right)×{\left(0.0031\mathrm{m}\right)}^2\\ & =& 0.0096\mathrm{J}\end{array}$

Now, as per the conservation of energy principle, the potential energy of the spring should be equal to the rotational kinetic energy of the plate.

So,

$\begin{array}{rcl}\mathrm{UE}& =& \mathrm{KE}\\ 0.0096\mathrm{J}& =& \frac{1}{2}{\mathrm{\pm õÓ\neg }}_{\mathrm{m}}^2\end{array}$

Rearranging this equation for moment of inertia I, you get the value of moment of inertia of the metal plate as follows:

$\mathrm{I}=\frac{2×0.0096\mathrm{J}}{{\mathrm{Ó\neg }}_{\mathrm{m}}^2}$ ….. (vi)

Here, ${\mathrm{Ó\neg }}_{\mathrm{m}}$is the maximum angular speed and it is given by using equation (iv) as follows:

$\begin{array}{rcl}{\mathrm{Ó\neg }}_{\mathrm{m}}& =& \frac{2\mathrm{Ï€}}{\left(2×{10}^{-3}\mathrm{s}\right)}×0.122\mathrm{rad}\\ & =& 38.33\mathrm{rad}/\mathrm{s}\end{array}$

By substituting this value in the equation (iv), we get the value of the moment of inertia as follows:

$\begin{array}{rcl}\mathrm{I}& =& \frac{2×0.0096\mathrm{J}}{{\mathrm{Ó\neg }}_{\mathrm{m}}^2}\\ & =& 1.3×{10}^{-5}\mathrm{kg}.{\mathrm{m}}^2\end{array}$

Hence, the rotational inertia of the plate about its center of mass is $1.3×{10}^{-5}\mathrm{kg}.{\mathrm{m}}^2$.