In the previous section, we saw the basics about angular momentum. In the section before that, we saw torque. In this section we will see some solved examples
Solved example 7.15
Find the torque of a force $\mathbf\small{\vec{F}=(-2\hat{i}+2\hat{j}+3\hat{k})}$ N about the origin O. The position vector of the 'point of application' of $\mathbf\small{\vec{F}}$ is $\mathbf\small{\vec{r}=(\hat{i}-2\hat{j}+\hat{k})}$ m
Solution:
1. The position vector $\mathbf\small{\vec{r}}$ is always drawn from the origin
• We have: $\mathbf\small{\vec{\tau}=\vec{r}\times \vec{F}}$
• For doing the cross multiplication, we form the table as shown below:
2. We get: $\mathbf\small{\vec{\tau}=(\vec{r}\times \vec{F})=(-2-6)\hat{i}+(-2-3)\hat{j}+(-4+2)\hat{k}}$
• Thus $\mathbf\small{\vec{\tau}=(-8\hat{i}-5\hat{j}-2\hat{k})}$ Nm
Solved example 7.16
Two forces act on a rigid laminar body shown in fig.7.90(a) below. The angles between the force vector and position vector are shown in fig.b
Find an expression for the net torque on the body about the pivot 'O'
Solution:
1. Torque $\mathbf\small{\vec{\tau}_1}$ created by $\mathbf\small{\vec{F}_1}$ is given by:
$\mathbf\small{\vec{\tau}_1=\vec{r}_1 \times \vec{F}_1}$
• Magnitude of this torque is given by: $\mathbf\small{|\vec{\tau}_1|=|\vec{r}_1| \times |\vec{F}_1| \times \sin \theta_1}$
2. To find the direction of this torque, we adopt the following steps:
(i) Assume that, the $\mathbf\small{\vec{F}_1}$ is shifted in such a way that, it's tail end coincides with the tail end of $\mathbf\small{\vec{r}_1}$
(ii) Imagine that, a right handed screw is placed perpendicular to the computer screen
(iii) Turn the screw from $\mathbf\small{\vec{r}_1}$ to $\mathbf\small{\vec{F}_1}$
(iv) We see that, the screw will move away from the screen, towards us
(v) So the direction of $\mathbf\small{\vec{\tau}_1}$ is perpendicular to the screen and towards us
3. Torque $\mathbf\small{\vec{\tau}_2}$ created by $\mathbf\small{\vec{F}_2}$ is given by:
$\mathbf\small{\vec{\tau}_2=\vec{r}_2 \times \vec{F}_2}$
• Magnitude of this torque is given by: $\mathbf\small{|\vec{\tau}_2|=|\vec{r}_2| \times |\vec{F}_2| \times \sin \theta_2}$
4. To find the direction of this torque, we adopt the following steps:
(i) Assume that, the $\mathbf\small{\vec{F}_2}$ is shifted in such a way that, it's tail end coincides with the tail end of $\mathbf\small{\vec{r}_2}$
(ii) Imagine that, a right handed screw is placed perpendicular to the computer screen
(iii) Turn the screw from $\mathbf\small{\vec{r}_2}$ to $\mathbf\small{\vec{F}_2}$
(iv) We see that, the screw will move into the screen, away from us
(v) So the direction of $\mathbf\small{\vec{\tau}_2}$ is perpendicular to the screen and away from us
5. The magnitude of the net torque is given by:
$\mathbf\small{|\vec{\tau}_1|-|\vec{\tau}_2|=\left [|\vec{r}_1| \times |\vec{F}_1| \times \sin \theta_1 \right ]-\left [|\vec{r}_2| \times |\vec{F}_2| \times \sin \theta_2 \right ]}$
6. Direction of the net torque will be perpendicular to the screen and towards us if:
$\mathbf\small{|\vec{\tau}_1|>|\vec{\tau}_2|}$
• Direction of the net torque will be perpendicular to the screen and away from us if:
$\mathbf\small{|\vec{\tau}_2|>|\vec{\tau}_1|}$
7. However, since the rigid body given in the problem is a lamina, this is a 2-D problem
■ We can write this about the direction of the net torque:
• Direction of the net torque will be anti-clockwise if:
$\mathbf\small{|\vec{\tau}_1|>|\vec{\tau}_2|}$
• Direction of the net torque will be clockwise if:
$\mathbf\small{|\vec{\tau}_2|>|\vec{\tau}_1|}$
Solved example 7.17
A car of mass 250 kg is travelling along a circular path of radius 75 m. It is moving with a constant speed of 25 ms-1. Find the angular momentum of the car
Solution:
1. We have: $\mathbf\small{\vec{L}=\vec{r}\times \vec{p}}$
So $\mathbf\small{|\vec{L}|=|\vec{r}| \times |\vec{p}| \times \sin \theta}$
2. Let the origin 'O' be at the center of the circular path
• Then at any instant that we consider, $\mathbf\small{\vec{r}}$ will be such that:
♦ The tail end of $\mathbf\small{\vec{r}}$ will be at 'O'
♦ The head end of $\mathbf\small{\vec{r}}$ will be at the 'point where the car is' at that instant
■ So the magnitude of $\mathbf\small{\vec{r}}$ at that instant = radius of the circular path = 75 m
3. At that instant, the $\mathbf\small{\vec{p}}$ will be tangential to the circular path and hence perpendicular to $\mathbf\small{\vec{r}}$
• So angle θ between $\mathbf\small{\vec{r}}$ and $\mathbf\small{\vec{p}}$ = 90o
■ The magnitude of $\mathbf\small{\vec{p}}$ at that instant
= mass of the car × velocity of the car at that instant
= 250 × 25 = 6250 kg ms-1
4. Substituting the values in (1), we get:
$\mathbf\small{|\vec{L}|}$ = 75 × 6250 × sin 90 = 468750 kg m2 s-1.
5. Direction of $\mathbf\small{\vec{L}}$ will be perpendicular to the ground
(i) Assume that, the car is moving in the anti-clockwise direction
(ii) Shift the $\mathbf\small{\vec{p}}$ in such a way that, it's tail end coincides with the tail end of $\mathbf\small{\vec{r}}$
(iii) Imagine that a right handed screw is placed perpendicular to the ground
(iv) Turn it in the direction from $\mathbf\small{\vec{r}}$ to $\mathbf\small{\vec{p}}$
(v) We will find that, the screw move upwards, away from the ground
(vi) So the direction of $\mathbf\small{\vec{L}}$ is perpendicular to the ground and upwards away from the ground
• If the car moves in the clockwise direction, the $\mathbf\small{\vec{L}}$ will be perpendicular to the ground and downwards into the ground
Solved example 7.18
Consider a single particle moving with constant velocity in any direction in space. Take the angular momentum $\mathbf\small{\vec{l}}$ of that particle about any point in space. Prove that, magnitude and direction of $\mathbf\small{\vec{l}}$ will always remain constant
Solution:
1. Given that:
• The particle move with a constant velocity. Let us denote this velocity as $\mathbf\small{\vec{v}}$
• The magnitude and direction of this $\mathbf\small{\vec{v}}$ does not change
2. Consider any point in space as the origin 'O'
• There will be a unique plane containing the three points:
(i) The tail end of $\mathbf\small{\vec{v}}$
(ii) The head end of $\mathbf\small{\vec{v}}$
(iii) The origin 'O'
3. This plane is shown in fig.7.91(a) below:
• The $\mathbf\small{\vec{v}}$ will be always along the line AB
• That means, the particle (shown as small red circle) will be always on the line AB
(Remember that, the directions of $\mathbf\small{\vec{v}}$ and $\mathbf\small{\vec{p}}$ are the same)
4. The $\mathbf\small{\vec{r}}$ at the instant when the particle is at 'P' is shown
• So now we can calculate $\mathbf\small{|\vec{l}|}$ at that instant
5. We have: $\mathbf\small{\vec{l}=\vec{r}\times \vec{p}}$
• So $\mathbf\small{|\vec{l}|=|\vec{r}| \times |\vec{p}| \times \sin \theta}$
• This can be rearranged as: $\mathbf\small{|\vec{l}|=\left (|\vec{r}| \times \sin \theta \right )\times |\vec{p}|}$
6. We see that, whatever be the position of the particle on the line AB, $\mathbf\small{\left (|\vec{r}| \times \sin \theta \right )}$ does not change
• It is the altitude of the right triangle OPQ
• Now, $\mathbf\small{|\vec{p}|}$ does not change because, mass and velocity are constants
• Thus we find that, every item in the right side of '$\mathbf\small{|\vec{l}|=\left (|\vec{r}| \times \sin \theta \right )\times |\vec{p}|}$' are constants
• So $\mathbf\small{|\vec{l}|}$ is constant
7. Now we check the direction of $\mathbf\small{\vec{l}}$
• The direction will be perpendicular to the plane shown in fig.7.91
(i) Shift the $\mathbf\small{\vec{p}}$ in such a way that, it's tail end coincides with the tail end of $\mathbf\small{\vec{r}}$
(iii) Imagine that a right handed screw is placed perpendicular to the plane
(iv) Turn it in the direction from $\mathbf\small{\vec{r}}$ to $\mathbf\small{\vec{p}}$
(v) We will find that, the screw moves into the plane
(vi) So the direction of $\mathbf\small{\vec{l}}$ is perpendicular to the plane and into the plane
• If the particle moves in the opposite direction, the $\mathbf\small{\vec{l}}$ will be perpendicular to the plane and away from the plane
• But such a situation does not arise here because, it is given that $\mathbf\small{\vec{v}}$ is a constant
• So both magnitude and direction of $\mathbf\small{\vec{l}}$ will always remain constant
8. It is worthy to note that, in this problem, as $\mathbf\small{|\vec{r}|}$ increases, θ decreases.
• We know that, when θ decreases, sin θ decreases
• That is the reason why we get a constant value for $\mathbf\small{\left (|\vec{r}| \times \sin \theta \right )}$
• This will become clear if we compare figs (a) and (b)
• In the fig.b, the particle is at P'
• The red dotted line at the bottom shows that, $\mathbf\small{|\vec{r}|\sin \theta=|\vec{r}'|\sin \theta'}$
Solved example 7.19
A particle of mass m is projected with velocity $\mathbf\small{\vec{v}}$ at an angle θ with the horizontal. Find the angular momentum about the point of projection when it is at the highest point of it's trajectory
Solution:
1. In fig.7.92(a) below, the point of projection is taken as the origin 'O'
• The horizontal through the point of projection is taken as the x-axis
• The trajectory is shown in yellow color
• 'P' is the highest point
2. We have: $\mathbf\small{\vec{l}_P=\vec{r}_P\times \vec{p}_P}$
Where:
• $\mathbf\small{\vec{l}_P}$ is the angular momentum at P
• $\mathbf\small{\vec{r}_P}$ the position vector at P
• $\mathbf\small{\vec{p}_P}$ the linear momentum at P
3. So $\mathbf\small{|\vec{l}_P|=|\vec{r}_P| \times |\vec{p}_P| \times \sin \phi}$
• Where φ is the angle between $\mathbf\small{\vec{r}_P}$ and $\mathbf\small{\vec{p}_P}$
• This can be rearranged as: $\mathbf\small{|\vec{l}_P|=\left (|\vec{r}_P| \times \sin \phi \right )\times |\vec{p}_P|}$
4. So we want two items:
(i) $\mathbf\small{\left (|\vec{r}_P| \times \sin \phi \right )}$
(ii) $\mathbf\small{|\vec{p}_P|}$
5. First we will find item (i):
• In fig,b, a perpendicular is dropped from P onto the x-axis
• Q is the foot of the perpendicular
• From the right triangle, we have:
Altitude PQ = $\mathbf\small{\left (|\vec{r}| \times \sin \phi \right )}$
• But PQ is the maximum height 'H' of the projectile
It is given by: $\mathbf\small{H=\frac{(|\vec{v}|\sin \theta)^2}{2g}}$ (Details here)
6. Now we will find item (ii):
• $\mathbf\small{|\vec{p}_P|=m|\vec{v}_P|}$
• $\mathbf\small{\vec{v}_P}$ is the velocity at P
♦ It will be horizontal
♦ It's magnitude is given by: $\mathbf\small{|\vec{v}_P|=|\vec{v}|\times \cos \theta}$
• Thus we get: $\mathbf\small{|\vec{p}_P|=m|\vec{v}|\times \cos \theta}$
7. So the result in (3) becomes:
• $\mathbf\small{|\vec{l}_P|=\left (|\vec{r}_P| \times \sin \phi \right )\times |\vec{p}_P|}$
= $\mathbf\small{|\vec{l}_P|=\left (\frac{(|\vec{v}|\sin \theta)^2}{2g} \right )\times m|\vec{v}|\times \cos \theta}$
= $\mathbf\small{|\vec{l}_P|=\frac{m|\vec{v}|^3\sin^2 \theta \cos \theta}{2g}}$
Solved example 7.20
Find the components along x, y and z axes of the angular momentum $\mathbf\small{\vec{l}}$ of a particle, whose position vector is $\mathbf\small{\vec{r}}$ with components x, y, z and momentum $\mathbf\small{\vec{p}}$ with components px, py and pz. Show that, if the particle moves only in the xy-plane, the angular momentum has only a z-component
Solution:
Part (i):
1. Given that the position vector $\mathbf\small{\vec{r}}$ has components x, y, z
• So we get: $\mathbf\small{\vec{r}=x\;\hat{i}+y\;\hat{j}+z\;\hat{k}}$
2. Given that the linear momentum vector $\mathbf\small{\vec{p}}$ has components px, py and pz
• So we get: $\mathbf\small{\vec{p}=p_x\;\hat{i}+p_y\;\hat{j}+p_z\;\hat{k}}$
3. We have: $\mathbf\small{\vec{l}=\vec{r}\times \vec{p}}$
• We can set up the table as shown below:
4. From the table, we get:
$\mathbf\small{\vec{l}=\vec{r}\times \vec{p}}$
= $\mathbf\small{(yp_z-zp_y)\;\hat{i}+(zp_x-xp_z)\;\hat{j}+(xp_y-yp_x)\;\hat{k}}$
5. So the components of the angular momentum can be written as follows:
• $\mathbf\small{\vec{l}_x=(yp_z-zp_y)\;\hat{i}}$
• $\mathbf\small{\vec{l}_y=(zp_x-xp_z)\;\hat{j}}$
• $\mathbf\small{\vec{l}_z=(xp_y-yp_x)\;\hat{k}}$
Part (ii):
1. If the particle moves only in the xy-plane, then:
• The position vector will have only x and y components
♦ So we can write: $\mathbf\small{\vec{r}=x\;\hat{i}+y\;\hat{j}}$
• The linear momentum vector will have only x and y components
♦ So we can write: $\mathbf\small{\vec{p}=p_x\;\hat{i}+p_y\;\hat{j}}$
2. We have: $\mathbf\small{\vec{l}=\vec{r}\times \vec{p}}$
We can set up the table as shown below:
3. From the table, we get:
$\mathbf\small{\vec{l}=\vec{r}\times \vec{p}}$
= $\mathbf\small{(0)\;\hat{i}+(0)\;\hat{j}+(xp_y-yp_x)\;\hat{k}}$
= $\mathbf\small{(xp_y-yp_x)\;\hat{k}}$
• Thus we prove that, the angular momentum in this case has only a z-component
Solved example 7.15
Find the torque of a force $\mathbf\small{\vec{F}=(-2\hat{i}+2\hat{j}+3\hat{k})}$ N about the origin O. The position vector of the 'point of application' of $\mathbf\small{\vec{F}}$ is $\mathbf\small{\vec{r}=(\hat{i}-2\hat{j}+\hat{k})}$ m
Solution:
1. The position vector $\mathbf\small{\vec{r}}$ is always drawn from the origin
• We have: $\mathbf\small{\vec{\tau}=\vec{r}\times \vec{F}}$
• For doing the cross multiplication, we form the table as shown below:
• Thus $\mathbf\small{\vec{\tau}=(-8\hat{i}-5\hat{j}-2\hat{k})}$ Nm
Solved example 7.16
Two forces act on a rigid laminar body shown in fig.7.90(a) below. The angles between the force vector and position vector are shown in fig.b
 |
| Fig.7.90 |
Solution:
1. Torque $\mathbf\small{\vec{\tau}_1}$ created by $\mathbf\small{\vec{F}_1}$ is given by:
$\mathbf\small{\vec{\tau}_1=\vec{r}_1 \times \vec{F}_1}$
• Magnitude of this torque is given by: $\mathbf\small{|\vec{\tau}_1|=|\vec{r}_1| \times |\vec{F}_1| \times \sin \theta_1}$
2. To find the direction of this torque, we adopt the following steps:
(i) Assume that, the $\mathbf\small{\vec{F}_1}$ is shifted in such a way that, it's tail end coincides with the tail end of $\mathbf\small{\vec{r}_1}$
(ii) Imagine that, a right handed screw is placed perpendicular to the computer screen
(iii) Turn the screw from $\mathbf\small{\vec{r}_1}$ to $\mathbf\small{\vec{F}_1}$
(iv) We see that, the screw will move away from the screen, towards us
(v) So the direction of $\mathbf\small{\vec{\tau}_1}$ is perpendicular to the screen and towards us
3. Torque $\mathbf\small{\vec{\tau}_2}$ created by $\mathbf\small{\vec{F}_2}$ is given by:
$\mathbf\small{\vec{\tau}_2=\vec{r}_2 \times \vec{F}_2}$
• Magnitude of this torque is given by: $\mathbf\small{|\vec{\tau}_2|=|\vec{r}_2| \times |\vec{F}_2| \times \sin \theta_2}$
4. To find the direction of this torque, we adopt the following steps:
(i) Assume that, the $\mathbf\small{\vec{F}_2}$ is shifted in such a way that, it's tail end coincides with the tail end of $\mathbf\small{\vec{r}_2}$
(ii) Imagine that, a right handed screw is placed perpendicular to the computer screen
(iii) Turn the screw from $\mathbf\small{\vec{r}_2}$ to $\mathbf\small{\vec{F}_2}$
(iv) We see that, the screw will move into the screen, away from us
(v) So the direction of $\mathbf\small{\vec{\tau}_2}$ is perpendicular to the screen and away from us
5. The magnitude of the net torque is given by:
$\mathbf\small{|\vec{\tau}_1|-|\vec{\tau}_2|=\left [|\vec{r}_1| \times |\vec{F}_1| \times \sin \theta_1 \right ]-\left [|\vec{r}_2| \times |\vec{F}_2| \times \sin \theta_2 \right ]}$
6. Direction of the net torque will be perpendicular to the screen and towards us if:
$\mathbf\small{|\vec{\tau}_1|>|\vec{\tau}_2|}$
• Direction of the net torque will be perpendicular to the screen and away from us if:
$\mathbf\small{|\vec{\tau}_2|>|\vec{\tau}_1|}$
7. However, since the rigid body given in the problem is a lamina, this is a 2-D problem
■ We can write this about the direction of the net torque:
• Direction of the net torque will be anti-clockwise if:
$\mathbf\small{|\vec{\tau}_1|>|\vec{\tau}_2|}$
• Direction of the net torque will be clockwise if:
$\mathbf\small{|\vec{\tau}_2|>|\vec{\tau}_1|}$
Solved example 7.17
A car of mass 250 kg is travelling along a circular path of radius 75 m. It is moving with a constant speed of 25 ms-1. Find the angular momentum of the car
Solution:
1. We have: $\mathbf\small{\vec{L}=\vec{r}\times \vec{p}}$
So $\mathbf\small{|\vec{L}|=|\vec{r}| \times |\vec{p}| \times \sin \theta}$
2. Let the origin 'O' be at the center of the circular path
• Then at any instant that we consider, $\mathbf\small{\vec{r}}$ will be such that:
♦ The tail end of $\mathbf\small{\vec{r}}$ will be at 'O'
♦ The head end of $\mathbf\small{\vec{r}}$ will be at the 'point where the car is' at that instant
■ So the magnitude of $\mathbf\small{\vec{r}}$ at that instant = radius of the circular path = 75 m
3. At that instant, the $\mathbf\small{\vec{p}}$ will be tangential to the circular path and hence perpendicular to $\mathbf\small{\vec{r}}$
• So angle θ between $\mathbf\small{\vec{r}}$ and $\mathbf\small{\vec{p}}$ = 90o
■ The magnitude of $\mathbf\small{\vec{p}}$ at that instant
= mass of the car × velocity of the car at that instant
= 250 × 25 = 6250 kg ms-1
4. Substituting the values in (1), we get:
$\mathbf\small{|\vec{L}|}$ = 75 × 6250 × sin 90 = 468750 kg m2 s-1.
5. Direction of $\mathbf\small{\vec{L}}$ will be perpendicular to the ground
(i) Assume that, the car is moving in the anti-clockwise direction
(ii) Shift the $\mathbf\small{\vec{p}}$ in such a way that, it's tail end coincides with the tail end of $\mathbf\small{\vec{r}}$
(iii) Imagine that a right handed screw is placed perpendicular to the ground
(iv) Turn it in the direction from $\mathbf\small{\vec{r}}$ to $\mathbf\small{\vec{p}}$
(v) We will find that, the screw move upwards, away from the ground
(vi) So the direction of $\mathbf\small{\vec{L}}$ is perpendicular to the ground and upwards away from the ground
• If the car moves in the clockwise direction, the $\mathbf\small{\vec{L}}$ will be perpendicular to the ground and downwards into the ground
Solved example 7.18
Consider a single particle moving with constant velocity in any direction in space. Take the angular momentum $\mathbf\small{\vec{l}}$ of that particle about any point in space. Prove that, magnitude and direction of $\mathbf\small{\vec{l}}$ will always remain constant
Solution:
1. Given that:
• The particle move with a constant velocity. Let us denote this velocity as $\mathbf\small{\vec{v}}$
• The magnitude and direction of this $\mathbf\small{\vec{v}}$ does not change
2. Consider any point in space as the origin 'O'
• There will be a unique plane containing the three points:
(i) The tail end of $\mathbf\small{\vec{v}}$
(ii) The head end of $\mathbf\small{\vec{v}}$
(iii) The origin 'O'
3. This plane is shown in fig.7.91(a) below:
 |
| Fig.7.91 |
• That means, the particle (shown as small red circle) will be always on the line AB
(Remember that, the directions of $\mathbf\small{\vec{v}}$ and $\mathbf\small{\vec{p}}$ are the same)
4. The $\mathbf\small{\vec{r}}$ at the instant when the particle is at 'P' is shown
• So now we can calculate $\mathbf\small{|\vec{l}|}$ at that instant
5. We have: $\mathbf\small{\vec{l}=\vec{r}\times \vec{p}}$
• So $\mathbf\small{|\vec{l}|=|\vec{r}| \times |\vec{p}| \times \sin \theta}$
• This can be rearranged as: $\mathbf\small{|\vec{l}|=\left (|\vec{r}| \times \sin \theta \right )\times |\vec{p}|}$
6. We see that, whatever be the position of the particle on the line AB, $\mathbf\small{\left (|\vec{r}| \times \sin \theta \right )}$ does not change
• It is the altitude of the right triangle OPQ
• Now, $\mathbf\small{|\vec{p}|}$ does not change because, mass and velocity are constants
• Thus we find that, every item in the right side of '$\mathbf\small{|\vec{l}|=\left (|\vec{r}| \times \sin \theta \right )\times |\vec{p}|}$' are constants
• So $\mathbf\small{|\vec{l}|}$ is constant
7. Now we check the direction of $\mathbf\small{\vec{l}}$
• The direction will be perpendicular to the plane shown in fig.7.91
(iii) Imagine that a right handed screw is placed perpendicular to the plane
(iv) Turn it in the direction from $\mathbf\small{\vec{r}}$ to $\mathbf\small{\vec{p}}$
(v) We will find that, the screw moves into the plane
(vi) So the direction of $\mathbf\small{\vec{l}}$ is perpendicular to the plane and into the plane
• If the particle moves in the opposite direction, the $\mathbf\small{\vec{l}}$ will be perpendicular to the plane and away from the plane
• But such a situation does not arise here because, it is given that $\mathbf\small{\vec{v}}$ is a constant
• So both magnitude and direction of $\mathbf\small{\vec{l}}$ will always remain constant
8. It is worthy to note that, in this problem, as $\mathbf\small{|\vec{r}|}$ increases, θ decreases.
• We know that, when θ decreases, sin θ decreases
• That is the reason why we get a constant value for $\mathbf\small{\left (|\vec{r}| \times \sin \theta \right )}$
• This will become clear if we compare figs (a) and (b)
• In the fig.b, the particle is at P'
• The red dotted line at the bottom shows that, $\mathbf\small{|\vec{r}|\sin \theta=|\vec{r}'|\sin \theta'}$
Solved example 7.19
A particle of mass m is projected with velocity $\mathbf\small{\vec{v}}$ at an angle θ with the horizontal. Find the angular momentum about the point of projection when it is at the highest point of it's trajectory
Solution:
1. In fig.7.92(a) below, the point of projection is taken as the origin 'O'
 |
| Fig.7.92 |
• The trajectory is shown in yellow color
• 'P' is the highest point
2. We have: $\mathbf\small{\vec{l}_P=\vec{r}_P\times \vec{p}_P}$
Where:
• $\mathbf\small{\vec{l}_P}$ is the angular momentum at P
• $\mathbf\small{\vec{r}_P}$ the position vector at P
• $\mathbf\small{\vec{p}_P}$ the linear momentum at P
3. So $\mathbf\small{|\vec{l}_P|=|\vec{r}_P| \times |\vec{p}_P| \times \sin \phi}$
• Where φ is the angle between $\mathbf\small{\vec{r}_P}$ and $\mathbf\small{\vec{p}_P}$
• This can be rearranged as: $\mathbf\small{|\vec{l}_P|=\left (|\vec{r}_P| \times \sin \phi \right )\times |\vec{p}_P|}$
4. So we want two items:
(i) $\mathbf\small{\left (|\vec{r}_P| \times \sin \phi \right )}$
(ii) $\mathbf\small{|\vec{p}_P|}$
5. First we will find item (i):
• In fig,b, a perpendicular is dropped from P onto the x-axis
• Q is the foot of the perpendicular
• From the right triangle, we have:
Altitude PQ = $\mathbf\small{\left (|\vec{r}| \times \sin \phi \right )}$
• But PQ is the maximum height 'H' of the projectile
It is given by: $\mathbf\small{H=\frac{(|\vec{v}|\sin \theta)^2}{2g}}$ (Details here)
6. Now we will find item (ii):
• $\mathbf\small{|\vec{p}_P|=m|\vec{v}_P|}$
• $\mathbf\small{\vec{v}_P}$ is the velocity at P
♦ It will be horizontal
♦ It's magnitude is given by: $\mathbf\small{|\vec{v}_P|=|\vec{v}|\times \cos \theta}$
• Thus we get: $\mathbf\small{|\vec{p}_P|=m|\vec{v}|\times \cos \theta}$
7. So the result in (3) becomes:
• $\mathbf\small{|\vec{l}_P|=\left (|\vec{r}_P| \times \sin \phi \right )\times |\vec{p}_P|}$
= $\mathbf\small{|\vec{l}_P|=\left (\frac{(|\vec{v}|\sin \theta)^2}{2g} \right )\times m|\vec{v}|\times \cos \theta}$
= $\mathbf\small{|\vec{l}_P|=\frac{m|\vec{v}|^3\sin^2 \theta \cos \theta}{2g}}$
Solved example 7.20
Find the components along x, y and z axes of the angular momentum $\mathbf\small{\vec{l}}$ of a particle, whose position vector is $\mathbf\small{\vec{r}}$ with components x, y, z and momentum $\mathbf\small{\vec{p}}$ with components px, py and pz. Show that, if the particle moves only in the xy-plane, the angular momentum has only a z-component
Solution:
Part (i):
1. Given that the position vector $\mathbf\small{\vec{r}}$ has components x, y, z
• So we get: $\mathbf\small{\vec{r}=x\;\hat{i}+y\;\hat{j}+z\;\hat{k}}$
2. Given that the linear momentum vector $\mathbf\small{\vec{p}}$ has components px, py and pz
• So we get: $\mathbf\small{\vec{p}=p_x\;\hat{i}+p_y\;\hat{j}+p_z\;\hat{k}}$
3. We have: $\mathbf\small{\vec{l}=\vec{r}\times \vec{p}}$
• We can set up the table as shown below:
4. From the table, we get:
$\mathbf\small{\vec{l}=\vec{r}\times \vec{p}}$
= $\mathbf\small{(yp_z-zp_y)\;\hat{i}+(zp_x-xp_z)\;\hat{j}+(xp_y-yp_x)\;\hat{k}}$
5. So the components of the angular momentum can be written as follows:
• $\mathbf\small{\vec{l}_x=(yp_z-zp_y)\;\hat{i}}$
• $\mathbf\small{\vec{l}_y=(zp_x-xp_z)\;\hat{j}}$
• $\mathbf\small{\vec{l}_z=(xp_y-yp_x)\;\hat{k}}$
Part (ii):
1. If the particle moves only in the xy-plane, then:
• The position vector will have only x and y components
♦ So we can write: $\mathbf\small{\vec{r}=x\;\hat{i}+y\;\hat{j}}$
• The linear momentum vector will have only x and y components
♦ So we can write: $\mathbf\small{\vec{p}=p_x\;\hat{i}+p_y\;\hat{j}}$
2. We have: $\mathbf\small{\vec{l}=\vec{r}\times \vec{p}}$
We can set up the table as shown below:
3. From the table, we get:
$\mathbf\small{\vec{l}=\vec{r}\times \vec{p}}$
= $\mathbf\small{(0)\;\hat{i}+(0)\;\hat{j}+(xp_y-yp_x)\;\hat{k}}$
= $\mathbf\small{(xp_y-yp_x)\;\hat{k}}$
• Thus we prove that, the angular momentum in this case has only a z-component
In the next section, we will see equilibrium of rigid bodies
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