Angular Momentum of Pendulum Bob Before Liftoff

To find the magnitude of the angular momentum $J$ of the pendulum about the point of suspension (pivot) just before the bob lifts off the floor, we proceed as follows:

1. Coordinate System and Geometry: Let the initial position of the bob on the floor be the origin $(0,0)$.

   • The pivot point is located at $(0, H) = (0, 0.9\text{ m})$.
   • The horizontal impulse $P = 0.2\text{ kg}\cdot\text{m/s}$ imparts a horizontal velocity to the bob along the $+x$-direction: $v_0 = \frac{P}{m} = \frac{0.2\text{ kg}\cdot\text{m/s}}{0.1\text{ kg}} = 2\text{ m/s}$

2. Motion Before Liftoff: Since the floor is frictionless and the string remains slack as the bob slides along the floor, no horizontal forces act on the bob. Thus, its linear momentum remains constant: $\vec{p} = P\hat{i} = 0.2\hat{i}\text{ kg}\cdot\text{m/s}$

   The position vector of the bob relative to the pivot point at any instant prior to liftoff is given by: $\vec{r} = x\hat{i} - H\hat{j}$ where $x$ is the horizontal displacement of the bob from the vertical line passing through the pivot.

3. Calculation of Angular Momentum $J$: The angular momentum $\vec{J}$ of the bob about the pivot point is: $\vec{J} = \vec{r} \times \vec{p} = (x\hat{i} - H\hat{j}) \times (P\hat{i})$

   Using vector cross-product properties ($\hat{i} \times \hat{i} = \vec{0}$ and $\hat{j} \times \hat{i} = -\hat{k}$): $\vec{J} = -H P (\hat{j} \times \hat{i}) = H P \hat{k}$

   The magnitude of the angular momentum $J$ is: $J = |\vec{J}| = H \cdot P$

4. Numerical Evaluation: Substituting the given values $H = 0.9\text{ m}$ and $P = 0.2\text{ kg}\cdot\text{m/s}$: $J = 0.9 \times 0.2 = 0.18\text{ kg}\cdot\text{m}^2/\text{s}$

The value of $J$ is $0.18$.