Question

In the figure, a 0.400 kg ball is shot directly upward at initial speed 31.1 m/s. What is its angular momentum about P 8.65 m horizontally from the launch point, when the ball is (a) at maximum height and (b) halfway back to the ground? What is the torque on the ball about P due to the gravitational force when the ball is (c) at maximum height and (d) halfway back to the ground? Ball (a) Number O (b) Numbe (o) Numbe nts um Unit Units (d) Numbe Units TN-m

Answer 1

Before we begin let us first calculate few needed variables

max height

$H=\frac{v^2}{2g} \Rightarrow \frac{31.1^2}{2\times 9.8}=49.35 \ m$

velocity at top is 0 but at half distance i.e H=24.675 m

$v_{h_{1/2}}=\sqrt{gH} \Rightarrow \sqrt{9.8\times 49.35}=21.99 \ m/s$

Also, the angles made at at point P when the particle is at top

$\theta_{top}=\tan^{-1}\left (\frac{49.35}{8.65} \right ) = 80.06^0$

at half distance

$\theta_{H_{1/2}}=\tan^{-1}\left (\frac{24.675}{8.65} \right ) = 70.68^0$

Radial distance at top

$R_{top}=\sqrt{49.35^2+8.65^2} = 50.1 \ m$

at half distance

$R_{H_{1/2}}=\sqrt{24.675^2+8.65^2} = 26.15 \ m$

Now we can start with answering the main question.

(a) as velocity at top is zero angular momenutm at top is zero. unit for angluar momentum is Kg-m/s

(b) At half distance.

$L=m(v\times r) = mvrsin(\theta)$

note this theta is the angle between velocity vector and the radial distance vector joining point H half and P.

$\\ L=m(v\times r) = mvR_{H_{1/2}}sin(90-\theta_{H_{1/2}}) \Rightarrow mvR_{H_{1/2}}cos(\theta_{H_{1/2}}) \\ L= 0.4\times 21.99\times 26.15cos(70.68) \\L=76.1 \ \text{Kg-m/s}$

Torque

$\\ \tau=F\times R \Rightarrow MgRsin(\theta)$

(c) at top

$\\ \tau=MgR_{top}sin(90-\theta_{top}) \\ \tau=0.4\times 9.8\times 50.1cos(80.06) \\ \tau=33.901 \ \text{N-m}$

(d) at half distance

$\\ \tau=MgR_{H_{1/2}}sin(90-\theta_{H_{1/2}}) \\ \tau=0.4\times 9.8\times 26.15cos(70.68) \\ \tau=33.914 \ \text{N-m}$