Question

The 3.00 kgcube in Figure 15-47 has edge lengths d=6.00 cmand is mounted on an axle through its center. A spring (k=1200 N/m)connects the cube’s upper corner to a rigid wall. Initially the spring is at its rest length. If the cube is rotated 3⁰ and released, what is the period of the resulting SHM?

Short answer

The period of the resulting SHM is 0.18 s.

Step-by-step solution

The given data:

• The spring constant of the spring, k=1200 N/m
• The edge length of the cube, $\mathrm{d}=6.00\mathrm{cm}=0.06\mathrm{m}$
• The rotational angle of the plane,$\mathrm{\theta }=3^{°}=0.05\mathrm{rad}$
• The mass of the cube, $\mathrm{M}=3.00\mathrm{kg}$

Understanding the concept of energy

For a given system, when a body undergoes both the translational and the rotational motions, you can solve for the required kinematics considering that energy is neither created nor destroyed, which means it s always conserved.

Here, the cube as it is attached to the spring when displaced horizontally undergoes a translational change preserving the potential energy in its motion, while due to the spring attachment; 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,

$\mathrm{KE}=\frac{1}{2}{\mathrm{I\omega }}_{\mathrm{m}}^2$ ….. (ii)

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

The displacement of the body in transverse motion,

$\mathrm{x}=\mathrm{r\theta }$ ….. (iii)

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

The maximum angular frequency of the oscillation,

${\mathrm{\omega }}_{\mathrm{m}}=\mathrm{\omega \theta }=\left(\frac{2\mathrm{\pi }}{\mathrm{T}}\right)\mathrm{\theta }$ ….. (iv)

Where, $\mathrm{\omega }$is the angular frequency of an oscillation, T is the period of oscillation, $\mathrm{\theta }$is the angular displacement of the body.

Calculate the potential energy of the spring:

Let us consider the given figure

From this figure, you can infer that if the cube is rotated through some angle, its arc length L will increase, and this increase in length will be the stretching length of the spring.

Now, the spring is rotated through angle $\mathrm{\theta }=3^{°}$, so the arc length or the stretching length of the spring is given using equation (iii) as follows:

$\begin{array}{rcl}\mathrm{x}& =& \mathrm{y\theta }\\ & =& \sqrt{{\left(\frac{\mathrm{d}}{2}\right)}^2+{\left(\frac{\mathrm{d}}{2}\right)}^2}\mathrm{\theta }\\ & =& \left(\frac{\mathrm{d}}{\sqrt{2}}\right)\mathrm{\theta }\end{array}$

Substitute known values in the above equation.

$\begin{array}{rcl}\mathrm{x}& =& \left(\frac{0.06\mathrm{m}}{\sqrt{2}}\right)0.05\mathrm{rad}\\ & =& 0.0022\mathrm{m}\end{array}$

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

$\begin{array}{rcl}\mathrm{UE}& =& \frac{1}{2}\times {\left(1200\mathrm{N}/\mathrm{m}\right)}^2\times {\left(0.0022\mathrm{m}\right)}^2\\ & =& 0.0029\mathrm{J}\end{array}$

Calculate the angular speed:

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

So, the maximum angular speed of the body can be using the above data in equation (ii) as follows:

$\begin{array}{rcl}\mathrm{UE}& =& \mathrm{KE}\\ 0.0029\mathrm{J}& =& \frac{1}{2}{\mathrm{I\omega }}_{\mathrm{m}}^2\end{array}$

Rearranging this equation for angular speed ${\mathrm{\omega }}_{\mathrm{m}}$, you get

${\mathrm{\omega }}_{\mathrm{m}}=\sqrt{\frac{2\times \left(0.0029\mathrm{J}\right)}{\mathrm{I}}}$ ….. (v)

Here, I is the moment of inertia of cube and it is given as follows.

$\begin{array}{rcl}\mathrm{I}& =& \frac{1}{6}{\mathrm{Md}}^2\\ & =& \frac{1}{6}\times \left(3.00\mathrm{kg}\right)\times {\left(0.06\mathrm{m}\right)}^2\\ & =& 0.0018\mathrm{kg}.{\mathrm{m}}^2\end{array}$

So, the angular speed from equation (v) becomes:

$\begin{array}{rcl}{\mathrm{\omega }}_{\mathrm{m}}& =& \sqrt{\frac{2\times \left(0.0029\mathrm{J}\right)}{\left(0.0018\mathrm{kg}.{\mathrm{m}}^2\right)}}\\ & =& 1.80\mathrm{rad}/\mathrm{s}\end{array}$

Calculate the time period:

The period of oscillation of the body in rotational motion can be given using the data in equation (iv) as follows:

$\begin{array}{rcl}\mathrm{T}& =& \left(\frac{2\mathrm{\pi }}{{\mathrm{\omega }}_{\mathrm{m}}}\right)\mathrm{\theta }\\ & =& \left(\frac{2\times 3.14}{1.80\mathrm{rad}/\mathrm{s}}\right)\times 0.05\mathrm{rad}\\ & =& 0.18\mathrm{s}\end{array}$

Hence, the time period is 0.18 s.