In Figure, a 0.400kgball is shot directly upward at initial speed 40.0m/s.What is its angular momentum about P, 2.00m. horizontally from the launch point,(a) when the ball is at maximum height? (b) when the ball is halfway back to the ground? What is the torque on the ball about Pdue to the gravitational force (c) when the ball is at maximum height? (d) when the ball is halfway back to the ground?

1. The mass of the ball is m = 0.400kg
2. The initial velocity of the ball is v₀ = 40.0m/s
3. The horizontal distance of ball from launch point is x = 2.00m

The problem deals with the calculation of angular momentum. The angular momentum of a rigid object is product of the moment of inertia and the angular velocity. It is analogous to linear momentum. The angular momentum can be found using right-hand rule. The counterclockwise rotation of the particle’s position gives positive angular momentum and clockwise rotation of the particle’s position gives negative angular momentum.

Formulae:

$$\begin{gathered} l=r_{\perp }mv \\ \tau =rF\sin \theta \end{gathered}$$

When the ball reaches the maximum height, its speed is momentarily zero.

Hence, the angular momentum is zero.

$$\begin{gathered} l=r_{\perp }mv \\ \Rightarrow l=0kg\frac{m^2}{s} \end{gathered}$$

Find the velocity of the ball at halfway back to the ground by using kinematic equation. Treat it as afree fall.When, the ball goes at maximum height, its final velocity will be zero. Acceleration due to gravity will act on it throughout its motion. By using third kinematical equation.

$$\begin{gathered} v^2=v_0^2-2g{y}_{max} \\ {\left(0m/s\right)}^2=v_0^2-2g{y}_{max} \\ {y}_{max}=\frac{v_0^2}{2g} \\ {y}_{max}=\frac{{\left(40.0m/s\right)}^2}{2\times 9.8m/s^2} \\ \Rightarrow y_{max}=81.6m \end{gathered}$$

The half distance covered by the ball is

$$\begin{gathered} y=\frac{1}{2}{y}_{max} \\ y=\frac{1}{2}81.6m \\ y=40.8m \end{gathered}$$

We can find the velocity of the ball at the height y ; for that, we can again use third kinematic equation. The ball falls from maximum height; hence its initial velocity will be zero.

$\begin{array}{rcl}{v}_0& =& 0m/s\\ v^2& =& v_0^2+2gy\\ & =& 0+2gy\\ & =& \sqrt{2gy}\\ & =& \sqrt{2\times 9.8m/s^2\times 40.8m}\\ & =& 28.3m/s\end{array}$

At that point, the ball is going in the downward direction, hence for clockwise motion; the direction of angular momentum is into the page of paper. It is negative.

$\begin{array}{rcl}l& =& -r_{\perp }mv\\ & =& -2.00m\times 0.400kg\times 28.3m/s\\ & =& -22.6kg\frac{m^2}{s}\end{array}$

At maximum height, the force acting on the ball is gravitational force. It is acting in the downward direction. Hence, its direction is perpendicular to the horizontal distance when the ball is at point P.At this point, the ball is going in the downward direction; hence for clockwise motion, the direction of torque is into the page of paper. It is negative.

According to the expression of torque

$\begin{array}{rcl}\tau & =& -RF\sin \theta \\ & =& -rmg\sin 90\\ & =& -\left(2.00m\times 0.400kg\times 9.8m/s^2\right)\\ & =& -7.84N.m\end{array}$

The ball is halfway back to the ground, but its direction and perpendicular horizontal distance remain the same. Hence, the torque on the ball at point P remains the same as when ball is at maximum height.

$\begin{array}{rcl}\tau & =& -rmg\sin 90\\ & =& -\left(2.00m\times 0.400kg\times 9.8m/s^2\right)\\ & =& -7.84Nm\end{array}$