91Ó°ÊÓ

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

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, $T=2\mathrm{Ï€}\sqrt{\frac{l}{k}}$ (i)

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

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:

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

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

$l=\frac{k{T}^2}{4{\mathrm{Ï€}}^2}$

Substitute the values of given terms, we get,

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

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

For torsional pendulum, maximum rotational kinetic energy is

$K{E}_{rotation}=\frac{1}{2}l{\mathrm{Ó\neg }}_{max}^2$

${\mathrm{Ó\neg }}_{max}$is maximum angular velocity.

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

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

Using the given hint, we can write,

$$\begin{gathered} \frac{1}{2}l{\mathrm{Ó\neg }}_{max}^2=\frac{1}{2}k{\mathrm{θ}}_{max}^2 \\ {\mathrm{Ó\neg }}_{max}={\mathrm{θ}}_{max}\sqrt{k//} \end{gathered}$$

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

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

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

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