Two circular discs of radius \(10\) cm each are joined at their centres by a rod, as shown in the figure. The length of the rod is \(30\) cm and its mass is \(600\) g. The mass of each disc is also \(600\) g. If the applied torque between the two discs is \(43\times10^{-7}\) dyne·cm, then the angular acceleration of the system about the given axis \(AB\) is ________ rad s\(^{-2}\).
Show Hint
Always include the moment of inertia of connecting rods and use the parallel axis theorem when rotation is not about the centre of mass.
Concept:
Angular acceleration \( \alpha \) is given by:
\[
\alpha=\frac{\tau}{I}
\]
where \( \tau \) is the applied torque and \( I \) is the moment of inertia of the system about the given axis.
Step 1: Convert given quantities into SI units
\[
\text{Mass of each disc } = 600\text{ g} = 0.6\text{ kg}
\]
\[
\text{Mass of rod } = 600\text{ g} = 0.6\text{ kg}
\]
\[
\text{Radius of each disc } r = 10\text{ cm} = 0.1\text{ m}
\]
Given torque:
\[
43\times10^{-7}\text{ dyne·cm}
\]
Since:
\[
1\text{ dyne·cm} = 10^{-7}\text{ N·m}
\]
\[
\tau = 43\times10^{-14}\text{ N·m}
\]
Step 2: Moment of inertia of each disc about axis \(AB\)
The axis \(AB\) passes through the midpoint of the rod.
Each disc’s centre is at a distance of \(0.2\) m from the axis.
Moment of inertia of a disc about its own central axis:
\[
I_{\text{disc,cm}}=\frac{1}{2}mr^2
=\frac{1}{2}(0.6)(0.1)^2=0.003\text{ kg m}^2
\]
Using parallel axis theorem:
\[
I_{\text{disc}}=I_{\text{disc,cm}}+md^2
=0.003+0.6(0.2)^2
=0.003+0.024=0.027
\]
For two discs:
\[
I_{\text{discs}}=2\times0.027=0.054
\]
Step 3: Moment of inertia of the rod
The rod rotates about an axis through its centre and perpendicular to its length:
\[
I_{\text{rod}}=\frac{1}{12}ML^2
=\frac{1}{12}(0.6)(0.3)^2
=0.0045
\]
Step 4: Total moment of inertia
\[
I_{\text{total}}=0.054+0.0045=0.0585\text{ kg m}^2
\]
Step 5: Calculate angular acceleration
\[
\alpha=\frac{43\times10^{-14}}{0.0585}
\approx 11\text{ rad s}^{-2}
\]
