Question

A torsion pendulum consists of a metal disk with a wire running through its center and soldered in place. The wire is mounted vertically on clamps and pulled taut. 15-58a Figuregives the magnitude $\mathit{\tau }$of the torque needed to rotate the disk about its center (and thus twist the wire) versus the rotation angle $\mathit{\theta }$. The vertical axis scale is set by ${\mathit{\tau }}_{\mathbf{s}}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{N}\mathbf{.}\mathbf{m}\mathbf{.}$=.The disk is rotated to $\mathit{\theta }\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{200}$rad and then released. Figure 15-58bshows the resulting oscillation in terms of angular position $\mathit{\theta }$versus time t. The horizontal axis scale is set by ${\mathit{t}}_{\mathbf{s}}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{40}\mathbf{}\mathit{s}$. (a) What is the rotational inertia of the disk about its center? (b) What is the maximum angular speedof $\mathit{d}\mathbf{}\mathit{\theta }\mathbf{/}\mathit{d}\mathit{t}$the disk? (Caution: Do not confuse the (constant) angular frequency of the SHM with the (varying) angular speed of the rotating disk, even though they usually have the same symbol. Hint: The potential energy U of a torsion pendulum is equal to $\frac{\mathbf{1}}{\mathbf{2}}\mathit{k}{\mathit{\theta }}^{\mathbf{2}}$, analogous to $\mathit{U}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{k}{\mathit{x}}^{\mathbf{2}}$for a spring.)

Short answer

1. The disk’s rotational inertia at the center is $8.11\times {10}^{-5}\mathrm{kg}.{\mathrm{m}}^2/\mathrm{rad}$.
2. The maximum angular speed $\frac{d\theta }{dt}$of the disk is$3.14\mathrm{rad}/\mathrm{s}$

Step-by-step solution

The given data

• Graph forτvs θ and θ vs t.
• The vertical axis is set as:${\tau }_s=4.0\times {10}^{-3}\mathrm{N}.\mathrm{m}$
• The horizontal axis is set as:$t_s=0.40\mathrm{s}$
• The disk is rotated at an angle,$\theta =0.200\mathrm{rad}$
• The potential energy Uof a torsion pendulum,$U=\frac{1}{2}k{\theta }^2$

Understanding the concept of oscillations of torsion pendulum

A rotating pendulum in which the restoring force is torsion, is known as a torsion pendulum. The main use of this pendulum is for timekeeping. For example balance wheel in a wristwatch is a torsion pendulum.

Using the formula for the period of torsion pendulum, we can find the rotational inertia of the disk about its center. Then equating maximum kinetic energy and maximum potential energy as given in the problem we can find the maximum angular speed of the disk.

Formula:

The period of torsion pendulum, $T=2\pi \sqrt{\frac{l}{k}}$ (i)

Here, l is rotational inertia and k is force constant.

a) Calculation of rotational inertia of the disk

We can interpret from the graph in figure 15-58(b), that the period of the torsion pendulum is T = 0.40 s

The torsion constant is given as:

$$\begin{gathered} k=\frac{t}{0} \\ =\frac{4.0\times {10}^{-3}\mathrm{N}.\mathrm{m}}{0.2\mathrm{rad}} \\ =0.02\mathrm{N}.\mathrm{m}/\mathrm{rad} \end{gathered}$$

So, using equation (i), the rotational inertia of the disk can be given as:

$l=\frac{k{T}^2}{4{\pi }^2}$

Substitute the values of given terms, we get,

$$\begin{gathered} l=\frac{\left[\left(0.02\mathrm{N}.\mathrm{m}/\mathrm{rad}\right){\left(0.4\mathrm{s}\right)}^2\right]}{4{\left(3.142\right)}^2} \\ =8.11\times {10}^{-5}\mathrm{kg}.{\mathrm{m}}^2/\mathrm{rad} \end{gathered}$$

Therefore, the rotational inertia of the disk about its center is$8.11\times {10}^{-5}\mathrm{kg}.{\mathrm{m}}^2/\mathrm{rad}$

b) Calculation of maximum angular speed of the disk

For torsional pendulum, maximum rotational kinetic energy is

$K{E}_{rotation}=\frac{1}{2}l{\omega }_{max}^2$

${\omega }_{max}$is maximum angular velocity.

Maximum potential energy can be written using the given hint in the problem,

$P{E}_{max}=\frac{1}{2}k{\theta }_{max}^2$

Using the given hint, we can write,

$$\begin{gathered} \frac{1}{2}l{\omega }_{max}^2=\frac{1}{2}k{\theta }_{max}^2 \\ {\omega }_{max}={\theta }_{max}\sqrt{k//} \end{gathered}$$

From graph we interpret that:
${\theta }_{max}=0.20\mathrm{rad}$

Hence, the maximum angular speed of the disk can be given as:

$$\begin{gathered} {\omega }_{max}=\left(0.20\mathrm{rad}\right)\sqrt{\frac{0.02\mathrm{N}.\mathrm{m}/\mathrm{rad}}{8.11\times {10}^{-5}\mathrm{kg}.{\mathrm{m}}^2/\mathrm{rad}}} \\ =3.14\mathrm{rad}/\mathrm{s} \end{gathered}$$

Therefore, the maximum angular speed$\frac{d\theta }{dt}$ of the disk is 3.14 rad/s