Systems Of Particles And Rotational MotionClass 11 Physics Notes

Introduction

So far, we have mostly studied the motion of single particles, which are idealized as point masses with no size. However, real-world objects have a finite size. To understand their motion, we need to think of them as a system of particles. This chapter explores the motion of these extended bodies.

A key concept is the rigid body, which is an ideal model for an object that has a perfectly definite and unchanging shape. In a rigid body, the distances between any two particles remain constant, even when forces are applied. While no real body is perfectly rigid (they all deform a little), objects like wheels, steel beams, and planets can often be treated as rigid bodies because their deformations are negligible.

What kind of motion can a rigid body have?

A rigid body can undergo several types of motion:

1. Pure Translational Motion: In this motion, every particle of the body has the same velocity at any given instant. Imagine a block sliding down a smooth inclined plane without tumbling. Every point on the block moves together.

2. Rotational Motion: If a body is constrained (e.g., fixed along a line), it can only rotate. The line it rotates around is called the axis of rotation.

   • Rotation about a Fixed Axis: In this case, every particle of the body moves in a circle. The center of each circle lies on the axis of rotation, and the plane of each circle is perpendicular to the axis. A ceiling fan is a perfect example.
   • Rotation about a Moving Axis: Sometimes, the axis of rotation itself moves. A spinning top is a great example. Its axis of rotation moves around a vertical line, a motion called precession. An oscillating fan also has a moving axis of rotation.

3. Combination of Translation and Rotation: The motion of most real objects is a mix of both. A cylinder rolling down an inclined plane is a classic example. Its center moves down the plane (translation), while the cylinder itself spins around its axis (rotation).

Centre of Mass

The centre of mass (CM) is a special point that represents the average position of all the mass in a system. Its motion describes the translational motion of the system as a whole.

Centre of Mass for a System of Particles

• For two particles with masses $m_1$ and $m_2$ at positions $x_1$ and $x_2$ on a line, the centre of mass $X$ is at: $X=\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}}$ This is the mass-weighted average of their positions. If the masses are equal ($m_1 = m_2$), the CM is exactly midway between them.

• For n particles along a line, the formula generalizes to: $X=\frac{\sum m_{i} x_{i}}{\sum m_{i}} = \frac{\sum m_{i} x_{i}}{M}$ where $M$ is the total mass of the system.

• For n particles in space, we find the coordinates $(X, Y, Z)$ of the centre of mass: $X=\frac{\sum m_{i} x_{i}}{M}$ $Y=\frac{\sum m_{i} y_{i}}{M}$ $Z=\frac{\sum m_{i} z_{i}}{M}$

In vector form, if $\mathbf{r}_i$ is the position vector of the $i^{th}$ particle, the position vector $\mathbf{R}$ of the centre of mass is: $\mathbf{R}=\frac{\sum m_{i} \mathbf{r}_{i}}{M}$

Centre of Mass for a Rigid Body

For a continuous body, we can't sum individual particles. Instead, we imagine the body is made of infinitesimally small mass elements, $dm$, and we replace the summation with integration. The position vector of the CM becomes: $\mathbf{R}=\frac{1}{M} \int \mathbf{r} \mathrm{d} m$

Centre of Mass of Homogeneous Bodies

For a homogeneous body (one with uniform mass distribution), the centre of mass is located at its geometric centre. This is due to symmetry. For every mass element on one side of the geometric centre, there is an identical mass element on the opposite side, and their effects cancel out.

• Sphere, Disc, Ring, Cube: CM is at the geometric centre.
• Thin Rod: CM is at its midpoint.

Given

• Mass 1, $m_1 = 100 \text{ g}$, at origin O with coordinates $(x_1, y_1) = (0, 0)$
• Mass 2, $m_2 = 150 \text{ g}$, at point A with coordinates $(x_2, y_2) = (0.5, 0)$
• Mass 3, $m_3 = 200 \text{ g}$, at point B with coordinates $(x_3, y_3) = (0.25, 0.25\sqrt{3})$
• Total mass, $M = m_1 + m_2 + m_3 = (100 + 150 + 200) \text{ g} = 450 \text{ g}$

To Find

The coordinates $(X, Y)$ of the centre of mass.

Formula

$X=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}}$ $Y=\frac{m_{1} y_{1}+m_{2} y_{2}+m_{3} y_{3}}{m_{1}+m_{2}+m_{3}}$

Solution

Substitute the given values into the formulas.

For the X-coordinate: $X=\frac{100(0)+150(0.5)+200(0.25)}{450} \text{ m}$ $X=\frac{0+75+50}{450} \text{ m} = \frac{125}{450} \text{ m} = \frac{5}{18} \text{ m}$

For the Y-coordinate: $Y=\frac{100(0)+150(0)+200(0.25 \sqrt{3})}{450} \text{ m}$ $Y=\frac{50 \sqrt{3}}{450} \text{ m} = \frac{\sqrt{3}}{9} \text{ m} = \frac{1}{3\sqrt{3}} \text{ m}$

Final Answer The centre of mass is located at $(\frac{5}{18} \text{ m}, \frac{1}{3\sqrt{3}} \text{ m})$. Note that because the masses are unequal, the CM is not at the geometric centre (centroid) of the triangle.

Solution

A lamina is a thin, flat plate. We can imagine the triangle is made of many narrow strips parallel to its base.

1. By symmetry, the centre of mass of each strip is at its midpoint.
2. If we connect the midpoints of all these strips, we get the median of the triangle (a line from a vertex to the midpoint of the opposite side).
3. Therefore, the centre of mass of the whole triangle must lie on this median.
4. We can repeat this argument for the other two sides of the triangle. The centre of mass must also lie on the other two medians.
5. The only point that lies on all three medians is their point of intersection, which is the centroid of the triangle.

Final Answer The centre of mass of a uniform triangular lamina is at its centroid.

Given

• Total mass of lamina = $3 \text{ kg}$.
• The lamina is uniform, so it can be divided into 3 squares, each with mass $1 \text{ kg}$.
• By symmetry, the CM of each square is at its geometric centre.
• Square 1: CM at $C_1 = (1/2, 1/2)$, mass $m_1 = 1 \text{ kg}$.
• Square 2: CM at $C_2 = (3/2, 1/2)$, mass $m_2 = 1 \text{ kg}$.
• Square 3: CM at $C_3 = (1/2, 3/2)$, mass $m_3 = 1 \text{ kg}$.

To Find

The coordinates $(X, Y)$ of the centre of mass of the L-shaped lamina.

Formula

$X=\frac{m_{1} x_{1}+m_{2} x_{2}+m_{3} x_{3}}{m_{1}+m_{2}+m_{3}}$ $Y=\frac{m_{1} y_{1}+m_{2} y_{2}+m_{3} y_{3}}{m_{1}+m_{2}+m_{3}}$

Solution

We can treat the problem as finding the CM of three point masses located at $C_1, C_2,$ and $C_3$.

For the X-coordinate: $X=\frac{[1(1 / 2)+1(3 / 2)+1(1 / 2)] \mathrm{kg} \mathrm{~m}}{(1+1+1) \mathrm{kg}} = \frac{1/2 + 3/2 + 1/2}{3} \text{ m} = \frac{5/2}{3} \text{ m} = \frac{5}{6} \text{ m}$

For the Y-coordinate: $Y=\frac{[1(1 / 2)+1(1 / 2)+1(3 / 2)] \mathrm{kg} \mathrm{~m}}{(1+1+1) \mathrm{kg}} = \frac{1/2 + 1/2 + 3/2}{3} \text{ m} = \frac{5/2}{3} \text{ m} = \frac{5}{6} \text{ m}$

Final Answer The centre of mass of the L-shape is at $(5/6 \text{ m}, 5/6 \text{ m})$.

Motion of Centre of Mass

The concept of the centre of mass is powerful because it simplifies the motion of complex systems. The position vector of the CM is given by $M \mathbf{R}=\sum m_{i} \mathbf{r}_{i}$.

• Velocity of CM: Differentiating with respect to time, we get the velocity of the centre of mass, $\mathbf{V}$: $M \mathbf{V}=m_{1} \mathbf{v}_{1}+m_{2} \mathbf{v}_{2}+\ldots+m_{n} \mathbf{v}_{n}$ where $\mathbf{v}_i$ is the velocity of the $i^{th}$ particle.

• Acceleration of CM: Differentiating again, we get the acceleration of the centre of mass, $\mathbf{A}$: $M \mathbf{A}=m_{1} \mathbf{a}_{1}+m_{2} \mathbf{a}_{2}+\ldots+m_{n} \mathbf{a}_{n}$ Since force $\mathbf{F}_i = m_i \mathbf{a}_i$ (Newton's Second Law), this becomes: $M \mathbf{A}=\mathbf{F}_{1}+\mathbf{F}_{2}+\ldots+\mathbf{F}_{n}$

The total force on the system is the sum of external forces (from outside the system) and internal forces (forces between particles within the system). By Newton's Third Law, internal forces always occur in equal and opposite pairs, so their vector sum is zero. This leaves us with a profound result: $M \mathbf{A}=\mathbf{F}_{ext}$

This equation means: The centre of mass of a system of particles moves as if all the mass of the system were concentrated at the centre of mass and all the external forces were applied at that point.

Linear Momentum of a System of Particles

The total linear momentum of a system of particles, $\mathbf{P}$, is the vector sum of the individual momenta: $\mathbf{P}=\mathbf{p}_{1}+\mathbf{p}_{2}+\ldots+\mathbf{p}_{n} = m_{1} \mathbf{v}_{1}+m_{2} \mathbf{v}_{2}+\ldots+m_{n} \mathbf{v}_{n}$

Comparing this with the equation for the velocity of the centre of mass, we find: $\mathbf{P}=M \mathbf{V}$

This means the total linear momentum of a system is equal to the product of its total mass and the velocity of its centre of mass.

Newton's Second Law for a System of Particles

By differentiating the momentum equation with respect to time, we get: $\frac{\mathrm{d} \mathbf{P}}{\mathrm{~d} t}=M \frac{\mathrm{d} \mathbf{V}}{\mathrm{~d} t}=M \mathbf{A}$

Since we already know $M\mathbf{A} = \mathbf{F}_{ext}$, we arrive at Newton's Second Law for a system of particles: $\frac{\mathrm{d} \mathbf{P}}{\mathrm{~d} t}=\mathbf{F}_{ext}$ The time rate of change of the total linear momentum of a system is equal to the sum of all external forces acting on it.

Conservation of Linear Momentum

If the total external force on a system is zero ($\mathbf{F}_{ext} = 0$), then: $\frac{\mathrm{d} \mathbf{P}}{\mathrm{~d} t}=0 \quad \text{ or } \quad \mathbf{P}=\text{Constant}$

This is the law of conservation of total linear momentum. When the net external force on a system is zero, its total linear momentum remains constant. This also implies that the velocity of the centre of mass, $\mathbf{V}$, remains constant.

Vector Product of Two Vectors

Besides the scalar (dot) product, there is another way to multiply vectors called the vector product or cross product. The result of a cross product is a new vector.

The vector product of two vectors $\mathbf{a}$ and $\mathbf{b}$ is a vector $\mathbf{c} = \mathbf{a} \times \mathbf{b}$ with the following properties:

• Magnitude: The magnitude of $\mathbf{c}$ is $c = ab \sin\theta$, where $\theta$ is the smaller angle between $\mathbf{a}$ and $\mathbf{b}$.
• Direction: The vector $\mathbf{c}$ is perpendicular to the plane formed by $\mathbf{a}$ and $\mathbf{b}$. Its specific direction is given by the right-hand screw rule: If you turn a right-handed screw from vector $\mathbf{a}$ to vector $\mathbf{b}$, the direction the screw advances is the direction of $\mathbf{c}$.

Properties of the Vector Product

• Not Commutative: The order of multiplication matters. $\mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})$. The resulting vectors have the same magnitude but point in opposite directions.
• Distributive: $\mathbf{a} \times (\mathbf{b} + \mathbf{c}) = (\mathbf{a} \times \mathbf{b}) + (\mathbf{a} \times \mathbf{c})$.
• The cross product of any vector with itself is the null vector: $\mathbf{a} \times \mathbf{a} = \mathbf{0}$.

Vector Product of Unit Vectors

For the standard unit vectors $\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}$:

• $\hat{\mathbf{i}} \times \hat{\mathbf{i}} = \hat{\mathbf{j}} \times \hat{\mathbf{j}} = \hat{\mathbf{k}} \times \hat{\mathbf{k}} = \mathbf{0}$
• $\hat{\mathbf{i}} \times \hat{\mathbf{j}} = \hat{\mathbf{k}}$
• $\hat{\mathbf{j}} \times \hat{\mathbf{k}} = \hat{\mathbf{i}}$
• $\hat{\mathbf{k}} \times \hat{\mathbf{i}} = \hat{\mathbf{j}}$ (Note the cyclic order i-j-k gives a positive result). Reversing the order gives a negative result (e.g., $\hat{\mathbf{j}} \times \hat{\mathbf{i}} = -\hat{\mathbf{k}}$).

Component Form

The cross product can be calculated using a determinant:

Angular Velocity and its Relation with Linear Velocity

When a rigid body rotates about a fixed axis, every particle moves in a circle.

• Angular Displacement ($\theta$): The angle through which a particle moves.
• Angular Velocity ($\omega$): The rate of change of angular displacement, $\omega = d\theta/dt$.

For a rigid body rotating about a fixed axis, every particle has the same angular velocity at any instant.

The magnitude of the linear velocity $v$ of a particle is related to its angular velocity $\omega$ and its perpendicular distance (radius) $r$ from the axis of rotation by: $v = \omega r$

Angular velocity is a vector ($\boldsymbol{\omega}$). Its direction is along the axis of rotation, determined by the right-hand rule: if you curl the fingers of your right hand in the direction of rotation, your thumb points in the direction of $\boldsymbol{\omega}$.

The relationship between the linear velocity vector $\mathbf{v}$ and the angular velocity vector $\boldsymbol{\omega}$ is given by the cross product: $\mathbf{v}=\boldsymbol{\omega} \times \mathbf{r}$ where $\mathbf{r}$ is the position vector of the particle from an origin on the axis of rotation. The vector $\mathbf{v}$ is tangent to the circular path of the particle.

Angular Acceleration

Angular acceleration ($\boldsymbol{\alpha}$) is the rotational analogue of linear acceleration. It is defined as the time rate of change of angular velocity: $\boldsymbol{\alpha}=\frac{\mathrm{d} \boldsymbol{\omega}}{\mathrm{d} t}$ For rotation about a fixed axis, the direction of $\boldsymbol{\alpha}$ is also fixed along the axis, and the equation can be treated as a scalar equation: $\alpha=\frac{\mathrm{d} \omega}{\mathrm{d} t}$

Torque and Angular Momentum

Just as force causes linear acceleration, a quantity called torque causes angular acceleration. Similarly, angular momentum is the rotational analogue of linear momentum.

Moment of Force (Torque)

Torque ($\boldsymbol{\tau}$), or moment of force, is the turning effect of a force. It depends not just on the magnitude of the force, but also on where it is applied.

If a force $\mathbf{F}$ acts on a particle at a position $\mathbf{r}$ from an origin O, the torque about that origin is defined as the vector product: $\boldsymbol{\tau}=\mathbf{r} \times \mathbf{F}$

• Magnitude of Torque: $\tau = rF \sin\theta$, where $\theta$ is the angle between $\mathbf{r}$ and $\mathbf{F}$.
• Direction of Torque: Perpendicular to the plane of $\mathbf{r}$ and $\mathbf{F}$, given by the right-hand rule.
• Units: The SI unit of torque is the newton-metre ($\text{N m}$).

Torque can also be expressed as $\tau = r_{\perp}F$ (where $r_{\perp}$ is the perpendicular distance from the origin to the line of action of the force) or $\tau = rF_{\perp}$ (where $F_{\perp}$ is the component of the force perpendicular to the position vector).

Angular Momentum of a Particle

Angular momentum ($\boldsymbol{l}$) is the rotational analogue of linear momentum ($\mathbf{p}$). For a particle with linear momentum $\mathbf{p}$ at position $\mathbf{r}$ from an origin O, its angular momentum is defined as: $\boldsymbol{l}=\mathbf{r} \times \mathbf{p}$ where $\mathbf{p} = m\mathbf{v}$.

• Magnitude of Angular Momentum: $l = rp \sin\theta$, where $\theta$ is the angle between $\mathbf{r}$ and $\mathbf{p}$.
• Units: The SI unit is joule-second ($\text{J s}$).

Relationship between Torque and Angular Momentum

By differentiating the definition of angular momentum with respect to time, we find a relationship that is the rotational analogue of $\mathbf{F} = d\mathbf{p}/dt$: $\frac{\mathrm{d} \boldsymbol{l}}{\mathrm{~d} t}=\boldsymbol{\tau}$ The time rate of change of the angular momentum of a particle is equal to the torque acting on it.

Torque and Angular Momentum for a System of Particles

For a system of particles, the total angular momentum $\mathbf{L}$ is the vector sum of the individual angular momenta: $\mathbf{L}=\sum_{i} \boldsymbol{l}_{i}=\sum_{i} \mathbf{r}_{i} \times \mathbf{p}_{i}$

The time rate of change of the total angular momentum is equal to the sum of all external torques acting on the system. The internal torques cancel out, just like internal forces. $\frac{\mathrm{d} \mathbf{L}}{\mathrm{~d} t}=\boldsymbol{\tau}_{ext}$

Conservation of Angular Momentum

If the total external torque on a system is zero ($\boldsymbol{\tau}_{ext} = \mathbf{0}$), then: $\frac{\mathrm{d} \mathbf{L}}{\mathrm{~d} t}=0 \quad \text{ or } \quad \mathbf{L}=\text{constant}$ This is the law of conservation of angular momentum: If the net external torque on a system is zero, its total angular momentum is conserved.

Given

• Position vector, $\mathbf{r}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$
• Force vector, $\mathbf{F}=7 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$

To Find

The torque, $\boldsymbol{\tau}$.

Formula

$\boldsymbol{\tau}=\mathbf{r} \times \mathbf{F}$

Solution

We use the determinant rule for the cross product:

Equilibrium of a Rigid Body

A rigid body is in mechanical equilibrium if its linear momentum and angular momentum do not change with time. This means the body has neither linear acceleration nor angular acceleration.

There are two conditions for mechanical equilibrium:

1. Translational Equilibrium: The vector sum of all external forces acting on the body must be zero. $\sum \mathbf{F}_{i}=\mathbf{0}$
2. Rotational Equilibrium: The vector sum of all external torques acting on the body about any point must be zero. $\sum \boldsymbol{\tau}_{i}=\mathbf{0}$

For problems where all forces are in a single plane (coplanar), this simplifies to three scalar conditions:

• Sum of x-components of forces is zero: $\sum F_x = 0$.
• Sum of y-components of forces is zero: $\sum F_y = 0$.
• Sum of torques about any axis perpendicular to the plane is zero: $\sum \tau_z = 0$.

A couple is a pair of equal and opposite forces that do not act along the same line. A couple produces rotation without translation. The net force of a couple is zero, so it satisfies translational equilibrium, but it produces a net torque.

Solution

Consider two forces, $\mathbf{F}$ and $-\mathbf{F}$, acting at points with position vectors $\mathbf{r}_2$ and $\mathbf{r}_1$ respectively, relative to an origin O. The total torque about O is the sum of the individual torques: $\boldsymbol{\tau} = (\mathbf{r}_2 \times \mathbf{F}) + (\mathbf{r}_1 \times (-\mathbf{F}))$ $\boldsymbol{\tau} = \mathbf{r}_2 \times \mathbf{F} - \mathbf{r}_1 \times \mathbf{F}$ $\boldsymbol{\tau} = (\mathbf{r}_2 - \mathbf{r}_1) \times \mathbf{F}$ The vector $(\mathbf{r}_2 - \mathbf{r}_1)$ is the vector pointing from the point of application of $-\mathbf{F}$ to the point of application of $\mathbf{F}$. Let's call this vector $\mathbf{AB}$. So, $\boldsymbol{\tau} = \mathbf{AB} \times \mathbf{F}$. This final expression depends only on the separation vector between the forces and the force itself. It does not contain any reference to the origin O.

Final Answer The moment of a couple is independent of the origin chosen to calculate it.

Principle of Moments

A lever is a simple machine in mechanical equilibrium. It pivots on a point called a fulcrum. If a force $F_1$ (the load) is at a distance $d_1$ (the load arm) from the fulcrum, and a force $F_2$ (the effort) is at a distance $d_2$ (the effort arm), the condition for rotational equilibrium is that the clockwise moments must equal the anticlockwise moments.

This gives the principle of moments: $d_{1} F_{1}=d_{2} F_{2}$ $\text{load arm} \times \text{load} = \text{effort arm} \times \text{effort}$

The Mechanical Advantage (M.A.) of a lever is the ratio of the load to the effort: $\text { M.A. }=\frac{F_{1}}{F_{2}}=\frac{d_{2}}{d_{1}}$ If the effort arm is longer than the load arm ($d_2 > d_1$), the M.A. is greater than one, allowing a small effort to lift a large load.

Centre of Gravity

The centre of gravity (CG) of a body is the point where the total gravitational torque on the body is zero. It is the point where the entire weight of the body can be considered to act. $\boldsymbol{\tau}_{g}=\sum \mathbf{r}_{i} \times m_{i} \mathbf{g}=\mathbf{0}$ If the acceleration due to gravity, $\mathbf{g}$, is uniform over the entire body, it can be taken out of the summation: $(\sum m_{i} \mathbf{r}_{i}) \times \mathbf{g}=\mathbf{0}$ This implies that $\sum m_{i} \mathbf{r}_{i} = \mathbf{0}$. This is the same condition that defines the centre of mass.

Given

• Length of bar AB = $70 \text{ cm} = 0.7 \text{ m}$
• Mass of bar, $M = 4.00 \text{ kg}$. Its weight is $W = 4g$.
• The bar is uniform, so its centre of gravity G is at the midpoint, $35 \text{ cm}$ from either end.
• Suspended mass, $m_1 = 6.00 \text{ kg}$. Its weight is $W_1 = 6g$.
• Position of knife-edge $K_1$ is at $10 \text{ cm}$ from end A.
• Position of knife-edge $K_2$ is at $10 \text{ cm}$ from end B (or $60 \text{ cm}$ from A).
• Position of load P is at $30 \text{ cm}$ from end A.
• Let $R_1$ and $R_2$ be the upward reaction forces at $K_1$ and $K_2$.
• Distances from G: $K_1G = 35 - 10 = 25 \text{ cm}$; $K_2G = 60 - 35 = 25 \text{ cm}$; $PG = 35 - 30 = 5 \text{ cm}$.

To Find

The reaction forces $R_1$ and $R_2$.

Solution

The bar is in equilibrium. We apply the two conditions.

1. Translational Equilibrium (sum of vertical forces = 0): $R_1 + R_2 - W - W_1 = 0$ $R_1 + R_2 = W + W_1 = 4g + 6g = 10g$ Using $g = 9.8 \text{ m/s}^2$, $R_1 + R_2 = 10(9.8) = 98.0 \text{ N}$. (Equation i)

2. Rotational Equilibrium (sum of torques = 0): Let's take moments about the centre of gravity, G. Clockwise moments are negative, anticlockwise are positive.

• Torque from $R_1$: $-R_1 \times (K_1G) = -R_1(0.25 \text{ m})$
• Torque from $W_1$: $+W_1 \times (PG) = +6g(0.05 \text{ m})$
• Torque from $R_2$: $+R_2 \times (K_2G) = +R_2(0.25 \text{ m})$

Sum of torques = 0: $-0.25 R_1 + 0.05(6g) + 0.25 R_2 = 0$ $0.25(R_2 - R_1) = -0.3g$ $R_1 - R_2 = \frac{0.3g}{0.25} = 1.2g = 1.2(9.8) = 11.76 \text{ N}$. (Equation ii)

Now we solve the two simultaneous equations: (i) $R_1 + R_2 = 98.00 \text{ N}$ (ii) $R_1 - R_2 = 11.76 \text{ N}$

Adding (i) and (ii): $2R_1 = 109.76 \text{ N} \implies R_1 = 54.88 \text{ N}$

Substituting $R_1$ back into (i): $54.88 + R_2 = 98.00 \implies R_2 = 43.12 \text{ N}$

Final Answer The reaction at the first knife-edge ($K_1$) is $54.88 \text{ N}$ and at the second ($K_2$) is $43.12 \text{ N}$.

Given

• Length of ladder AB = $3 \text{ m}$.
• Weight of ladder, $W = 20g = 20 \times 9.8 = 196.0 \text{ N}$. It acts at the centre of gravity D (midpoint).
• Distance AC = $1 \text{ m}$.
• Using Pythagoras theorem, height BC = $\sqrt{AB^2 - AC^2} = \sqrt{3^2 - 1^2} = \sqrt{8} = 2\sqrt{2} \text{ m}$.
• The wall is frictionless, so its reaction force $F_1$ is purely horizontal.
• The floor exerts a normal reaction $N$ (vertical) and a friction force $F$ (horizontal, towards the wall). The total floor reaction is $F_2$.

To Find

The reaction forces of the wall ($F_1$) and the floor ($F_2$).

Solution

The ladder is in equilibrium.

1. Translational Equilibrium:

• Sum of vertical forces = 0: $N - W = 0 \implies N = W = 196.0 \text{ N}$.
• Sum of horizontal forces = 0: $F - F_1 = 0 \implies F = F_1$.

2. Rotational Equilibrium: Let's take moments about point A (the base of the ladder). This is convenient as the forces $N$ and $F$ pass through A and thus have zero torque.

• Torque from wall reaction $F_1$ (anticlockwise, positive): $+F_1 \times (\text{vertical distance BC}) = F_1 \times 2\sqrt{2}$.
• Torque from weight $W$ (clockwise, negative): $-W \times (\text{horizontal distance from A to D})$. Since D is the midpoint, this distance is half of AC, which is $1/2 \text{ m}$. So, the torque is $-W \times (1/2)$.

Sum of torques = 0: $F_1 (2\sqrt{2}) - W(1/2) = 0$ $F_1 = \frac{W}{4\sqrt{2}} = \frac{196.0}{4\sqrt{2}} = 34.6 \text{ N}$

Now we can find the other forces:

• From horizontal equilibrium, friction force $F = F_1 = 34.6 \text{ N}$.
• The total floor reaction $F_2$ is the vector sum of $N$ and $F$: $F_2 = \sqrt{F^2 + N^2} = \sqrt{(34.6)^2 + (196.0)^2} \approx 199.0 \text{ N}$

Final Answer The reaction force of the wall is $F_1 = 34.6 \text{ N}$. The reaction force of the floor is $F_2 = 199.0 \text{ N}$.

Moment of Inertia

What is the rotational analogue of mass? In linear motion, mass is a measure of inertia (resistance to change in motion). In rotational motion, that role is played by the moment of inertia ($I$).

The kinetic energy of a single particle $i$ in a rotating body is $k_i = \frac{1}{2} m_i v_i^2$. Since $v_i = r_i \omega$, this becomes $k_i = \frac{1}{2} m_i r_i^2 \omega^2$. The total kinetic energy $K$ of the rotating body is the sum of the kinetic energies of all its particles: $K = \sum k_i = \sum \left(\frac{1}{2} m_i r_i^2 \omega^2\right)$ Since angular velocity $\omega$ is the same for all particles, we can factor it out: $K=\frac{1}{2} \omega^{2}\left(\sum_{i=1}^{n} m_{i} r_{i}^{2}\right)$ We define the term in the parenthesis as the moment of inertia, $I$. $I=\sum_{i=1}^{n} m_{i} r_{i}^{2}$ With this definition, the rotational kinetic energy is: $K=\frac{1}{2} I \omega^{2}$ This is perfectly analogous to the translational kinetic energy, $K = \frac{1}{2}mv^2$.

The moment of inertia depends on:

• The total mass of the body.
• The shape and size of the body.
• The axis of rotation (how the mass is distributed around the axis).

The SI unit for moment of inertia is $\text{kg m}^2$.

Radius of Gyration

The radius of gyration ($k$) is the distance from the axis of rotation at which all the mass of the body could be concentrated to give the same moment of inertia. It is defined by the relation: $I = M k^2$ where $M$ is the total mass of the body.

Moments of Inertia for Common Shapes

Kinematics of Rotational Motion about a Fixed Axis

The equations of rotational motion for constant angular acceleration are directly analogous to the linear kinematic equations.

Here, $\omega_0$ and $\theta_0$ are the initial angular velocity and angular displacement at $t=0$.

Given

• Initial angular speed, $\omega_0 = 1200 \text{ rpm}$ (revolutions per minute)
• Final angular speed, $\omega = 3120 \text{ rpm}$
• Time, $t = 16 \text{ s}$

To Find

(i) Angular acceleration, $\alpha$ (ii) Number of revolutions

Formula

$\omega = \omega_0 + \alpha t$ $\theta = \omega_0 t + \frac{1}{2}\alpha t^2$

Solution

First, we must convert the angular speeds from rpm to rad/s. $1 \text{ rpm} = \frac{2\pi \text{ rad}}{60 \text{ s}}$

Initial angular speed: $\omega_0 = \frac{1200 \times 2\pi}{60} \text{ rad/s} = 40\pi \text{ rad/s}$

Final angular speed: $\omega = \frac{3120 \times 2\pi}{60} \text{ rad/s} = 104\pi \text{ rad/s}$

(i) Calculate the angular acceleration

Using $\omega = \omega_0 + \alpha t$, we can solve for $\alpha$: $\alpha = \frac{\omega - \omega_0}{t} = \frac{104\pi - 40\pi}{16} \text{ rad/s}^2$ $\alpha = \frac{64\pi}{16} \text{ rad/s}^2 = 4\pi \text{ rad/s}^2$

Answer for part (i) = $4\pi \text{ rad/s}^2$

(ii) Calculate the number of revolutions

First, find the total angular displacement $\theta$: $\theta = \omega_0 t + \frac{1}{2}\alpha t^2$ $\theta = (40\pi)(16) + \frac{1}{2}(4\pi)(16)^2$ $\theta = 640\pi + 2\pi(256) = 640\pi + 512\pi = 1152\pi \text{ rad}$

Since one revolution is $2\pi$ radians, the number of revolutions is: $\text{Number of revolutions} = \frac{\theta}{2\pi} = \frac{1152\pi}{2\pi} = 576$

Answer for part (ii) = $576$ revolutions

Dynamics of Rotational Motion about a Fixed Axis

We can now establish the dynamic relationship between torque, moment of inertia, and angular acceleration.

Work Done by a Torque

Just as work done by a force in linear motion is $dW = F ds$, the work done by a torque in rotational motion is: $dW = \tau d\theta$

The power (rate of doing work) is: $P = \frac{dW}{dt} = \tau \frac{d\theta}{dt} = \tau \omega$

Newton's Second Law for Rotation

The work done on a rigid body increases its kinetic energy. Therefore, the power delivered by the torque must equal the rate of change of rotational kinetic energy. $P = \frac{dK}{dt} = \frac{d}{dt}\left(\frac{1}{2}I\omega^2\right)$ Assuming $I$ is constant: $\frac{dK}{dt} = \frac{1}{2}I(2\omega)\frac{d\omega}{dt} = I\omega\alpha$

Equating the two expressions for power, $P = \tau\omega$ and $P = I\omega\alpha$: $\tau\omega = I\omega\alpha$ $\tau = I\alpha$

This is Newton's Second Law for rotational motion about a fixed axis. It is the rotational analogue of $F=ma$. Torque causes angular acceleration, and the moment of inertia is the measure of resistance to this angular acceleration.

Given

• Mass of flywheel, $M = 20 \text{ kg}$
• Radius of flywheel, $R = 20 \text{ cm} = 0.20 \text{ m}$
• Force applied, $F = 25 \text{ N}$
• Initial angular velocity, $\omega_0 = 0$
• Length of unwound cord, $l = 2 \text{ m}$

To Find

(a) Angular acceleration, $\alpha$ (b) Work done by the pull, $W$ (c) Final kinetic energy, K.E. (d) Compare W and K.E.

Formula

$\tau = I\alpha$ $I_{flywheel} = \frac{1}{2}MR^2$ $W = F \times l$ $K.E. = \frac{1}{2}I\omega^2$ $\omega^2 = \omega_0^2 + 2\alpha\theta$ $\theta = l/R$

Solution

(a) Compute the angular acceleration

First, calculate the torque $\tau$ and moment of inertia $I$. $\tau = F \times R = 25 \text{ N} \times 0.20 \text{ m} = 5.0 \text{ Nm}$ $I = \frac{1}{2}MR^2 = \frac{1}{2}(20.0 \text{ kg})(0.20 \text{ m})^2 = 0.4 \text{ kg m}^2$ Now use Newton's second law for rotation: $\alpha = \frac{\tau}{I} = \frac{5.0 \text{ Nm}}{0.4 \text{ kg m}^2} = 12.5 \text{ s}^{-2}$

Answer for part (a) = $12.5 \text{ s}^{-2}$

(b) Find the work done

Work done is force times distance: $W = 25 \text{ N} \times 2 \text{ m} = 50 \text{ J}$

Answer for part (b) = $50 \text{ J}$

(c) Find the final kinetic energy

First, find the angular displacement $\theta$ and the final angular velocity $\omega$. $\theta = \frac{\text{length of unwound string}}{\text{radius}} = \frac{2 \text{ m}}{0.2 \text{ m}} = 10 \text{ rad}$ $\omega^2 = \omega_0^2 + 2\alpha\theta = 0 + 2(12.5)(10) = 250 \text{ (rad/s)}^2$ Now calculate the final kinetic energy: $K.E. = \frac{1}{2}I\omega^2 = \frac{1}{2}(0.4)(250) = 50 \text{ J}$

Answer for part (c) = $50 \text{ J}$

(d) Compare the answers

The work done by the pull (50 J) is equal to the kinetic energy gained by the wheel (50 J). This is consistent with the work-energy theorem, as there is no energy loss due to friction.

Angular Momentum in Case of Rotation about a Fixed Axis

For a rigid body rotating about a fixed axis (say, the z-axis), the angular momentum of a particle in the body is $\boldsymbol{l} = \mathbf{r} \times \mathbf{p}$. This vector does not necessarily point along the axis of rotation.

However, when we sum the angular momenta of all particles to get the total angular momentum $\mathbf{L}$, an important simplification occurs for symmetric bodies. A symmetric body is one where the axis of rotation is also an axis of symmetry (like a cylinder, sphere, or disc). For such bodies, the components of angular momentum perpendicular to the axis of rotation cancel out. This leaves only the component along the axis of rotation, $\mathbf{L}_z$.

For a symmetric body rotating about a fixed axis, the total angular momentum is: $\mathbf{L} = \mathbf{L}_z = I\omega\hat{\mathbf{k}}$ where $\hat{\mathbf{k}}$ is the unit vector along the axis of rotation. The angular momentum vector $\mathbf{L}$ is parallel to the angular velocity vector $\boldsymbol{\omega}$.

We know that $\frac{d\mathbf{L}}{dt} = \boldsymbol{\tau}_{ext}$. Applying this to the equation above (and assuming $I$ is constant): $\frac{d}{dt}(I\omega\hat{\mathbf{k}}) = I \frac{d\omega}{dt} \hat{\mathbf{k}} = I\alpha\hat{\mathbf{k}}$ This gives us another derivation of Newton's second law for rotation: $\boldsymbol{\tau} = I\boldsymbol{\alpha}$

Conservation of Angular Momentum

Revisiting the principle of conservation of angular momentum for a body rotating about a fixed axis: if the net external torque is zero, then the angular momentum is constant. For a symmetric body, this means: $L_z = I\omega = \text{constant}$

This principle explains many fascinating phenomena.

1. The person is spun with their arms stretched out. They have a moment of inertia $I_1$ and an angular speed $\omega_1$. Their angular momentum is $L = I_1\omega_1$.
2. They then pull their arms in close to their body. This reduces their moment of inertia to a new value $I_2$ (since mass is now closer to the axis of rotation, so $I_2 < I_1$).
3. Because angular momentum must be conserved, $I_1\omega_1 = I_2\omega_2$. Since $I_2$ is smaller than $I_1$, the final angular speed $\omega_2$ must be larger than $\omega_1$. The person spins faster!

This is the same principle used by ice skaters, divers, and acrobats to control the speed of their spins.