Question

The balance wheel of an old-fashioned watch oscillates with angular amplitude$\mathbf{\pi }\mathbf{}\mathbf{rad}$and periodrole="math" localid="1655102340305" $\mathbf{0}\mathbf{.}\mathbf{500}\mathbf{}\mathbf{s}$.

1. Find the maximum angular speed of the wheel.
2. Find the angular speed at displacemeradrole="math" localid="1655102543053" $\mathbf{\pi }\mathbf{}\mathbf{/}\mathbf{2}\mathbf{}\mathbf{rad}$.
3. Find the magnitude of the angular acceleration at displacement$\mathit{\pi }\mathbf{}\mathbf{/}\mathbf{}\mathbf{4}$rad.

Short answer

1. Maximum angular speed is$39.5\mathrm{rad}/\sec$.
2. Angular speed at displacement is$\mathrm{\pi }/2\mathrm{rad}\mathrm{is}-34.2\mathrm{rad}/\sec$.
3. Magnitude of angular acceleration at displacement is$\mathrm{\pi }/4\mathrm{rad}\mathrm{is}124\mathrm{rad}/{\sec }^2$.

Step-by-step solution

The given data

1. Amplitude of watch,${\mathrm{\theta }}_{\mathrm{m}}=\mathrm{\pi }\mathrm{rad}$.
2. Time period,$T=0.500\sec$.

Understanding the concept of simple harmonic motion

Using the given information and the standard equation of SHM, we can find the required values.

Formula:

Angular frequency of a body, $\omega =\frac{2\pi }{T}$….. (i)

Angular displacement of a body,$\Omega \left(t\right)={\theta }_m\cos \left(\omega t\right)$……(ii)

Angular velocity of a body,$\Omega =-\left(\frac{2\pi }{T}\right){\theta }_m\sin \left(\frac{2\pi t}{T}\right)$……….(iii)

Angular acceleration of a body,$\alpha =-\Omega {\left(\frac{2\pi }{T}\right)}^2{\theta }_m\cos \left(\frac{2\pi t}{T}\right)$………(iv)

The maximum angular speed of a body, ${\Omega }_m=\frac{2\pi {\theta }_m}{T}$……..(v)

(a) Calculation of maximum angular speed

Using equation (v) and the given values, we get the maximum angular speed as:

$$\begin{gathered} {\Omega }_m=\frac{2\tau \tau \left(\tau \tau \right)}{0.50} \\ =39.5\mathrm{rad}/\mathrm{s} \end{gathered}$$

Hence, the angular speed value is found to be$39.5rad/s$.

(b) Calculation of angular speed at π/4 rad

Using equation (i) and substituting the values, we get the angular speed at required displacement as:

$$\begin{gathered} \Omega \left(t\right)=\pi \cos \left(\omega t\right) \\ \frac{\pi }{2}=\pi \cos \left(\omega t\right) \\ \cos \left(\omega t\right)=\frac{1}{2} \\ \left(\omega t\right)=\frac{\pi }{3} \end{gathered}$$

Calculate the value of angular velocity as below.

$$\begin{gathered} \mathrm{\omega }=\frac{2\mathrm{\pi }}{T} \\ =\frac{2\left(3.14\right)}{0.50} \\ =12.56\mathrm{rad}/\mathrm{s} \end{gathered}$$

Now, the angular velocity using equation (iii) and the given values is given as:

$$\begin{gathered} \mathrm{\Omega }=-\left(\mathrm{\omega \tau \tau }\right)\sin \left(\mathrm{\omega t}\right) \\ =-\left(12.56\right)\left(3.14\right)\left(\frac{\sqrt{3}}{2}\right) \\ =-34.11\mathrm{rad}/\mathrm{s} \\ \approx -34.11\mathrm{rad}/\mathrm{s} \end{gathered}$$

Hence, the angular speed value is found to be .

(c) Calculation of angular acceleration at π/4 rad

Using equation (iii) and the given values, the angular acceleration is given as:

$$\begin{gathered} \alpha \left(t\right)=-{\left(\omega \right)}^2\theta \\ =-{\left(12.56\right)}^2\left(\frac{\tau \tau }{4}\right) \\ =-123.8\mathrm{rad}/{\mathrm{s}}^2 \\ \approx -124\mathrm{rad}/{\mathrm{s}}^2 \\ \left|\mathrm{\alpha }\left(\mathrm{t}\right)\right|=124\mathrm{rad}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the value of angular acceleration is found to be .