© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: State the condition for angular momentum to be conserved
Angular momentum about the origin remains constant (i.e., conserved) if the net torque about the origin is zero. In mathematical form:
$ \vec{\tau} = \frac{d\vec{L}}{dt} = 0 \quad \Longrightarrow \quad \vec{\tau} = 0 $
Step 2: Express torque in terms of position and force
Torque $ \vec{\tau} $ is given by the cross product of the position vector $ \vec{r} $ and the force vector $ \vec{F} $:
$ \vec{\tau} = \vec{r} \times \vec{F} $
For angular momentum to be conserved, we require
$ \vec{r} \times \vec{F} = 0. $
Step 3: Write down the given vectors
Position vector: $ \vec{r} = 2\,\hat{i} - 6\,\hat{j} - 12\,\hat{k} $
Force vector: $ \vec{F} = \alpha\,\hat{i} + 3\,\hat{j} + 6\,\hat{k} $
Step 4: Calculate the cross product
We set up the cross product for $ \vec{r} \times \vec{F} $:
$$
\vec{r} \times \vec{F} \;=\;
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & -6 & -12 \\
\alpha & 3 & 6
\end{vmatrix}.
$$
For the vector to be zero, each component of this determinant must be zero. The determinant expands as:
$ \hat{i} $ component: $(-6)(6) - (-12)(3) = -36 + 36 = 0 $
$ \hat{j} $ component: $ (2)(6) - (-12)(\alpha) = 12 + 12\alpha $
$ \hat{k} $ component: $ (2)(3) - (-6)(\alpha) = 6 + 6\alpha $
Step 5: Impose the zero-torque condition
From the $ \hat{i} $ component, the expression automatically becomes $ 0 $. Now for the cross product to be the zero vector, both the $ \hat{j} $ and $ \hat{k} $ components must also be zero, so we solve these:
1) $ 12 + 12\alpha = 0 $
2) $ 6 + 6\alpha = 0 $
Both equations are essentially the same condition. Solving $ 12 + 12\alpha = 0 $ gives:
$ 12\alpha = -12 \quad \Longrightarrow \quad \alpha = -1.
$
Step 6: Conclude the value of $ \alpha $
The value of $ \alpha $ that ensures the torque is zero (hence the angular momentum remains conserved) is $ \alpha = -1 $.
