Rotational Motion Notes PDF

# System of Particles and Rotational Motion - **Rigid Body**: Ideally a rigid body is a body with a perfectly definite and unchanging shape. The distance between all pairs of particles of such a body do not change. - **Motions that a Rigid Body Can Have**: - **Pure Translational Motion**: In p...

# System of Particles and Rotational Motion - **Rigid Body**: Ideally a rigid body is a body with a perfectly definite and unchanging shape. The distance between all pairs of particles of such a body do not change. - **Motions that a Rigid Body Can Have**: - **Pure Translational Motion**: In pure translational motion, every point in the body experiences the same velocity at any instant of time. - Ex: A rectangular block sliding down an inclined plane. - **Pure Rotational Motion**: In pure rotational motion, the rigid body rotates about a fixed axis that is parallel to a fixed plane. In other words, the axis is fixed and does not move or change its direction relative to an inertial frame of reference. - ex: ceiling fan. - In rotation of a rigid body about a fixed axis every particle of the body moves in a circle which lies in a plane perpendicular to the axis and has its center on the axis. - **Note**: The motion of a rigid body which is not pivoted or fixed in some way is either a pure translation or a combination of translation and rotation. The motion of a rigid body which is pivoted or fixed way is rotation. # Center of Mass - **COM**: COM is defined as the point where all of the system’s mass appears to be concentrated. - It may lie inside or outside the body. - Ex: - Solid sphere → inside the body. - Ring → outside the body. - **Definition**: COM is defined as the point at which the entire mass of the system is supposed to be concentrated. ## Center of Mass of the Two-Particle System - **Show**: Show that the center of mass of the two-particle system of equal masses lies at the center of the line joining them. - **Consider**: A system of two particles of masses $m_1$ & $m_2$. - **Assume**: At any instant of time, the position vector of masses $m_1$ & $m_2$ are $r_1$ and $r_2$ respectively. - **If $r_{cm}$**: If $r_{cm}$ be the position vector of the two-particle system, then it is given by: $$r_{cm} = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2}$$ ## Derivation - **Consider**: Consider the motion of the two particle system under a net external force $F_{net}$. - $F_{net} = F_1 + F_2$ - Where $F_{net}$ is the net force on particle $m_1$. - $F_{net} = F_{1x} + F_{1e} + F_{1a} + F_{1s}e$ - Where $F_{2}$ is the net force on particle $m_2$. - $F_{net} = F_e+ F_{re}$ - Where $F_{id} = - F_2$ - $ (m_1 + m_2) a_{cm} = m_1 a_1 + m_2 a_2$ - Applying Newton's third law we have: - $\frac{d}{dt}\left(\frac{d}{dt}r_{cm}\right) = \frac{d}{dt}\left(\frac{d}{dt}r_1\right) + \frac{d}{dt}\left(\frac{d}{dt}r_2\right)$ - We have: - $(m_1 + m_2)\frac{d^2}{dt^2}\left(r_{cm}\right) = \frac{d}{dt}\left(m_1\frac{d}{dt}r_1\right) + \frac{d}{dt}\left(m_2\frac{d}{dt}r_2\right)$ - $(m_1 + m_2) \frac{d^2}{dt^2}(r_{cm}) = \frac{d}{dt}(m_1\frac{d}{dt}r_1 + m_2\frac{d}{dt}r_2) = (m_1 + m_2)\frac{d^2}{dt^2}\left(\frac{m_1 r_1 + m_2r_2}{m_1 + m_2}\right)$. - Comparing both sides we have: - $\frac{d^2}{dt^2}(r_{cm}) = \frac{d^2}{dt^2}\left(\frac{m_1r_1 + m_2r_2}{m_1 + m_2}\right)$. - Thus we have: - $$r_{cm} = \frac{m_1 r_1 + m_2 r_2}{m_1 + m_2}$$ - **If**: If both the particles of the same mass $m_1 = m_2 = m$, then: - $r_{cm} = \frac{m_1 \vec{r}_1 + m_2\vec{r}_2}{m_1 + m_2} = \frac{ m (\vec{r}_1 + \vec{r}_2)}{2m}$. - That gives us: - $$r_{cm} = \frac{ \vec{r}_1 + \vec{r}_2}{2}$$ ## Center of Mass of n-Particle System - **Given**: Center of mass of n-particle system is given by - $$r_{cm} = \frac{m_1 r_1 + m_2 r_2 + m_3 r_3 + ... + m_n r_n}{m_1 + m_2 + M_3 + ... + m_n}$$ - That is equal to: - $$r_{cm} = \frac{\sum_{i=1}^{n} m_i r_i}{\sum_{i=1}^{n} m_i} = \frac{\sum_{i=1}^{n} m_i r_i}{\sum m_i}$$ - **For**: For two system describing motion in x-y plane, then we may write $r_{cm} = x_{cm} \hat{i} + y_{cm} \hat{j}$ - Where: - $$x_{cm} = \frac{m_1x_1 + m_2x_2}{m_1 + m_2}$$ - $$y_{cm} = \frac{m_1y_1 + m_2y_2}{m_1 + m_2}$$ - **General Expression**: The general expression for the COM for a two system getting in space - $$R= \sum_{i=1}^{n} m_ir_i $$ - $$X = \sum_{i=1}^{n} m_ix_i $$ - $$Y = \sum_{i=1}^{n} m_iy_i $$ - $$Z = \sum_{i=1}^{n} m_iz_i $$ - Where R be the position vector of the center of mass M - **For continuous distribution of mass**: For a continuous distribution of mass, we subdivide the body into n small elements of mass $Am_1$, $Am_2$, ..., $Am_n$, located at $(x_i, y_i, z_i)$. The coordinates of the center of mass are then approximately given by: - $X=\sum(Am_i)x_i$, $Y=\sum(Am_i)y_i$, $Z=\sum(Am_i)z_i$ - $X = \frac{\sum(\Delta m_i) x_i}{\sum\Delta m_i} = \frac{\int{x dm}}{M}$, $Y = \frac{\sum(\Delta m_i) y_i }{\sum\Delta m_i} = \frac{\int{y dm}}{M}$, $Z = \frac{\sum(\Delta m_i) z_i}{\sum\Delta m_i} = \frac{\int{z dm}}{M}$. - $R = \frac{\int{r dm}}{M}$. # Motion of Center of Mass - **Position vector of the center of mass of n-particles is**: - $$r_{cm} = \frac{m_1 r_1 + m_2 r_2 + ... + m_n r_n}{m_1 + m_2 + ... + m_n}$$ - **Velocity of the center of mass**: - $\frac{d}{dt}(r_{cm}) = \frac{d}{dt}\left(\frac{m_1 r_1 + m_2 r_2 + ... + m_n r_n}{m_1 + m_2 + ... + m_n}\right)$ - $$v_{cm} = \frac{m_1 \frac{d}{dt}r_1 + m_2 \frac{d}{dt}r_2 + ... + m_n \frac{d}{dt}r_n}{m_1 + m_2 + ... + m_n}$$ - $v_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$ - **Acceleration of the center of mass**: - $a_{cm} = \frac{d}{dt}(v_{cm})$ - $$a_{cm} = \frac{m_1 a_1 + m_2 a_2}{m_1 + m_2}$$ - $$F_{net} = m_1 a_1 + m_2 a_2 = \frac{d}{dt} (m_1 v_1) + \frac{d}{dt}(m_2 v_2) = \frac{d}{dt}(m_1 v_1 + m_2 v_2)$$ - $F_{net} = F_{cm} = \frac{d}{dt} (m_1 v_1) + \frac{d}{dt}(m_2 v_2) = \frac{d}{dt}(m_1 v_1 + m_2 v_2) = \frac{d}{dt} (m_1 v_1 + m_2 v_2) = \frac{d}{dt} (M v_{cm})$ - **Total linear momentum**: - $P = M v_{cm}$ - **The center of mass of a system of particles moves as if all the mass of the system was concentrated at the center of mass and all the external forces were applied at that point** # Conservation of Linear Momentum for System of Particles - **If**: If no external force is acting on the system of particles - $F_{net}=0$, then $\frac{dp}{dt}$= 0 or $P$ = constant - **That means**: That means the rate of change of momentum is zero. So momentum of system remains constant. - **So**: So $P_{initial} = P_{final}$ if $F_{net}=0$ - $P = MV_{cm} = constant$ - **That means**: That means if the mass of the system is constant then $V_{cm} = constant$. - **This shows that**: This shows that if net external force on the system is zero, then the velocity of COM of the system will remain constant. - **Note**: COM of uniformly distributed mass is found at their geometrical center. - Ex: COM of a uniform rod of length $l$ with uniformly distributed mass, then COM = $(\frac{l}{2}, 0)$. # Moment of Force (Torque) - The turning effect of force is called torque $(\tau)$. - **Definition of Torque**: The turning effect of force about the axis of rotation is called the moment of force or torque due to the force. It is measured as the product of the magnitude of the force and the perpendicular distance (also known as the lever arm) of the line of action of the force from the axis of rotation. - **Mathematical Expression**: - $\tau = Force \times tar \ distance \ from \ axis \ of \ rotation$ - That is: - $\tau = F \times r_t$ - $\tau = F \times r$ - **SI Unit of Torque**: The SI unit of torque is newton-meter (Nm). - **Torque is a vector quantity**: Torque is a vector quantity. - **Dimensional Formula of Torque**: The dimensional formula is $[\tau] = [M^1 L^2 T^{-2}]$. - **Couple**: Two forces of an equal magnitude but opposite directions acting at two points of a body constitute a "couple". A couple only produces only rotational motion ie. a couple does not exert a net force on an object but exerts only a torque. - **Moment of Couple**: The moment of couple is the product of either of two forces and the perpendicular distance between their lines of action. - $\tau = F \times AB $ # Derivation of Torque in Cartesian Coordinate System - **Consider**: Consider a rigid body is rotating about an axis of rotation in x-y plane. Torque acting on the body is given by: - $\tau = r \times F$ - **In 2-D Cartesian coordinate system (x-y plane)**, $r$ and $F$ are given by: - $r = x\hat{i} + y\hat{j}$ - $F = F_x\hat{i} + F_y\hat{j}$. - Expanding the vector product we have: - $r \times F = (x\hat{i} + y\hat{j}) \times (F_x \hat{i} + F_y \hat{j})$ - Evaluating the cross products we have: - $r \times F = (x\hat{i} \times F_x\hat{i}) + (x\hat{i} \times F_y\hat{j}) + (y\hat{j} \times F_x\hat{i}) + (y\hat{j} \times F_y\hat{j})$. - We know that cross product of the same unit vectors are 0 and that $i \times j = k$ and $j \times i = -k$. That gives us: - $r \times F = \hat{k}(0) - \hat{j}(0) + \hat{k} (x F_y - y F_x) - \hat{k} (0)$ - $ \tau = (x F_y - y F_x)\hat{k} $ - **Required Expression in Cartesian form**: The required expression in cartesian form is: - $\tau = (x F_y - y F_x)\hat{k}$ - In polar form we know that x = rcos($\theta$) and y = rsin($\theta$). - That gives us: $F_x = Fcos(\theta)$ and $F_y = Fsin(\theta)$. - **Required Expression in Polar form**: The required expression in polar form is: - $\tau = rcos(\theta) \times Fsin(\theta) - rsin(\theta) \times Fcos(\theta) = rFsin(\theta - \theta)$ - Thus we have: - $\tau = rFsin(\phi)$, where $\theta - \theta = \phi$ - **Further**: We have: - $sin\phi = \frac{d}{r}$ - $d = rsin(\phi)$ - $\tau = F \times rsin(\phi)$ - $\tau = Fx d$ - $\tau = F \times lar \ distance$. ## Applications of Torque - **Opening or Closing a Door**: We use torques while opening or closing a door by applying force at the door handle, tightening a screw using a screw driver, using a wrench, wing a home grinder by applying force at the handle etc.. - **When**: When force is being applied at a point farthest from the axis of rotation and in a direction perpendicular to the axis of rotation. ## Applications of Torque - **Wrench**: We prefer to use a wrench with a long arm. For a wrench with a long arm, the turning effect is more, i.e. torque is more. So, it is easier to tighten or loosen the nuts. - **Door**: Handle in a door is provided at the free edge of a door (i.e. at a maximum distance from hinges) so that turning moment of applied force is more and the door can be opened easily. - **Potter's Wheel**: A potter puts a wooden handle normally near the circumference of his wheel and tries to rotate the wheel rapidly # Derive Expression for Vector Form of Torque - **Torque**: $\tau = force \times far \ distance$ - Expanding the equation we have: - $\tau = F\sin{\theta} \times r$ - Simplifying we have: - $\tau = r F\sin{\theta}$ - $\tau = r \times F$. - **Direction of Force**: The direction of force is given by the right-hand thumb rule. # Angular Momentum $(L\rightarrow)$ - **Definition**: The angular momentum of a particle rotating about an axis gives a measure of turning motion of the body. - **What is it**: It is the product of the linear momentum and the normal distance of the line of action from the axis of rotation. - **Angular Momentum is a vector quantity**: It is denoted by $L$. - **Mathematical Expression**: Mathematically, angular momentum, $L = r \times p = rp\sin{\theta}$. - **SI Unit of Angular Momentum**: SI unit of angular momentum is $kg \ m^2 \ s^{-1}$. - **Dimensional Formula of Angular Momentum**: And its dimensional formula is $[M^1 L^2 T^{-1}]$. # Relation between Torque and Angular Momentum - Let: Let $\frac{d\vec{L}}{dt} = r \times (r \times P)$ - Expanding the equation we get: - $\frac{d\vec{L}}{dt} = r \times \frac{d}{dt}(\vec{r} \times P)= r \times \left(\frac{d}{dt} \vec{r} \times P + \vec{r} \times \frac{d}{dt}P\right)$. - We know that $ \frac{d}{dt}\vec{r} = \vec{v}$ and $\frac{d}{dt}\vec{P} = \frac{d}{dt}(mv) = m\frac{d}{dt}\vec{v} = m\vec{a}$. - $\frac{d\vec{L}}{dt} = r \times (\vec{v} \times P + \vec{r} \times m\vec{a})$ - We know that $\vec{v} \times P = \vec{v} \times (m\vec{v}) = m(\vec{v} \times \vec{v})=0$ - $\frac{d\vec{L}}{dt} = r \times (\vec{v} \times P + \vec{r} \times m\vec{a}) = r \times m\vec{a} + 0$ - $\frac{d\vec{L}}{dt} = r \times m\vec{a}$ - $\frac{d\vec{L}}{dt} = r\times F$ - $\frac{d\vec{L}}{dt} = \tau$ - **Conclusion**: The rate of change of angular momentum is torque. # Law of Conservation of Angular Momentum - **If**: If no net external torque acts on a system, then the total angular momentum of the system remains conserved - Mathematically if $\tau_{ext} = 0$, then $\frac{d\vec{L}}{dt}$ = 0 → $\vec{L}$ = constant. - **This law is known as**: This law is known as the "conservation law of angular momentum". # Applications of Law of Conservation of Angular Momentum: - **Planetary Motion**: The angular velocity of a planet revolving in an elliptical orbit around the sun increases, when it comes closer the sun because its moment of inertia about the axis through the sun decreases. - When it goes far away from the sun, its moment of inertia increases and hence angular velocity decreases so as to conserve angular momentum. - $L = constant$ - $L = I\omega$ - $I \uparrow$ , $ \omega \downarrow$. - **Ice Skater**: An ice skater or ballet dancer can increase their angular velocity by folding their arms and bringing the stretched leg close to the other leg. - **Diver**: A diver jumping from a spring board exhibits somersaults in air before touching the water. - After leaving the spring board, a diver curls his body by pulling his arms and legs towards the center of his body. - This decreases his moment of inertia and he spins fast in midair. - Just before hitting the water surface, he stretches out his arms. This decreases his moment of inertia and the diver enters water at a gentle speed. # Moment of Inertia $(I)$ - **Rotational Inertia**: Rotational inertia or moment of inertia of a rotating body is its inherent tendency to continue rotate with a given angular velocity in a given sense of direction unless some external torque acts on it. - **What is Moment of Inertia**: It is the physical quantity which opposes any change in rotational motion of a body about an axis of rotation. It is the rotational analog of mass. ## Moment of Inertia of a Rigid Body in terms of Constituent Particles: - **Consider**: Consider a rigid body of mass M is rotating about an axis of rotation with constant angular velocity ω. Assume, it is composed of “n” particles of masses $m_1$, $m_2$,..., $m_n$ located at a perpendicular distance $r_1$, $r_2$,... $r_n$ from the axis of rotation at any instant t. - **Moment of Inertia**: The moment of inertia of the body about the axis of rotation in terms of constituent particles is given by: - $I = m_1 r_1^2 + m_2 r_2^2 + m_3 r_3^2 + ... + m_n r_n^2 $ - $I = \sum_{i=1}^{n} m_i r_i^2$. - **Note**: Moment of Inertia about a particular axis of rotation is a scalar positive quantity. - **SI Unit of Moment of Inertia**: The S.I. unit of Moment of inertia is $ kg. m^2$. - **Dimensional Formula of Moment of Inertia**: And its dimensional formula is $[M^1 L^2 T^0]$. - **Moment of inertia depends on**: Moment of inertia depends on: - a) Mass of the rigid body - b) Shape and size of the rigid body - c) Location and orientation of the axis of rotation. - d) Distribution of mass of particles about the axis of rotation # Radius of Gyration - **Definition**: Radius of gyration is defined as the distance from the axis of rotation at which if whole mass of the body is supposed to be concentrated then its moment of inertia would be same as with the actual distribution of mass of the body. - **Mathematical Expression**: Mathematically, the radius of gyration K of a body about a given axis is root mean square distance of the particles of given body from the given axis of rotation. - **Moment of Inertia**: The moment of inertia of a body is given as: - $I = m_1 r_1^2 + m_2 r_2^2 + ...+ m_n r_n^2$ - If all particles are identical, then $m_1=m_2=m_3=...=m_n = m$ (say). - $I = m r_1^2 + m r_2^2 + ... + m r_n^2$ - $I = m(r_1^2 + r_2^2 + ... + r_n^2)$ - **Derive Expression**: Multiplying both the numerator and denominator by n, we have: - $I = \frac{mn}{n} (r_1^2 + r_2^2 + ... + r_n^2)$ - $I = M k^2$, where: - $M=mn = mass \ of \ rigid \ body$ - $k^2 = \frac{r_1^2 + r_2^2 + ... + r_n^2}{n}$, where k = radius of gyration - $$k = \sqrt\frac{r_1^2 + r_2^2 + ... + r_n^2}{n}$$ - **Radius of Gyration as a Scalar Quantity**: Radius of gyration is a scalar quantity. - **SI Unit of Radius of Gyration**: Its S.I unit is meter (m) and its dimensional formula is [L]. - **In terms of radius of gyration**: In terms of radius of gyration, the moment of inertia of a body is may be expressed as the product of mass of body and square of its radius of gyration about the given axis of rotation. Thus, I = M.k^2. # Theorem of Perpendicular Axes - **Application**: This Theorem is valid only for 2-D objects. - **Definition**: According to the theorem of perpendicular axes, the moment of inertia of a plane laminar body about an axis perpendicular to its plane is equal to the sum of the moments of inertia of the laminar body about any two mutually perpendicular axes in its own plane intersecting each other at the point through which the perpendicular axis passes. Thus, for a plane lamina lying in x-y plane, $ I_z = I_x + I_y $. ## Applications of the Theorem: - **Ring**: Find MOI about the axis shown (in plane) - $I_z = I_x + I_y$ - $MR^2 = I + I$ - $I = \frac{MR^2}{2}$ - **Disc**: Disc (in the plane) passing through COM. - $I_z = I_x + I_y$ - $I = \frac{MR^2}{4}$ # Theorem of Parallel Axes - **Definition**: According to the theorem of parallel axes, the moment of inertia of a body about any axis is equal to its moment of intertia about a parallel axis passing through its center of mass plus the product of the mass of the body and the square of the perpendicular distance between the two parallel axes. - **Mathematical Expression**: Thus $I = I_{cm} + Md^2$ where: - d = far distance bl/w parallel axis - M = mass of the body ## Applications of the Theorem - **Rod**: $I_{xx'} = I_{COM}+Ma^2$ = $\frac{ML^2}{12} + M\left(\frac{L}{2}\right)^2 = \frac{ML^2}{12} + \frac{ML^2}{4} = \frac{4ML^2}{12} = \frac{ML^2}{3}$ - **Disc**: $I_{xx'} = I_{COM} + Ma^2$. - $I_{total} = 4 \times I_{xx'} = 4 \times \frac{ML^2}{3} = \frac{4ML^2}{3}$. - $I_{rod \ end} = \frac{ML^2}{3}$. - $I_{total} = 2 \times \frac{M}{2}\left(\frac{L}{2}\right)^2 = \frac{ML^2}{12}$. - **Moment of Inertia of a Disc**: $I_{ring + plane} = MR^2$ - $I_{disc + plane} = MR^2$ - $I_{xx'} = I_{disc \ end \ circumference} = I_{centre} + MR^2 = \frac{MR^2}{2} + MR^2 = \frac{3MR^2}{2}$ # Moment of Inertia of Some Standard Bodies - **Rod**: For a rod of length l and mass m. - $I_{COM} = \frac{Ml^2}{12}$ - $I_{end} = \frac{Ml^2}{3}$ - **Ring**: For a ring of radius r and mass m. - $I_{COM} = MR^2$ - $I_{plane} = MR^2$ - **Disc**: For a disc of radius r and mass m. - $I_{COM} = \frac{MR^2}{2}$ - $I_{plane} = \frac{MR^2}{2}$ - **Rectangle**: For a rectangle of length l and width b and mass m. - $I_{yy'} = \frac{Mb^2}{12}$ - $I_{xx'} = \frac{Ml^2}{12}$ # Moment of Inertia - **Rectangle**: For a rectangle of length l and width b and mass m with the axis at its center of mass. - $I_{COM} = I_x + I_y = \frac{Mb^2}{12} + \frac{Ml^2}{12}$ - $I_{plane} = \frac{Mb^2}{12} + \frac{Ml^2}{12}$ - **Rectangle Diagonal**: For a rectangle of length l and width b and mass m where the axis is along its diagonal. - $I_{diagonal} = \frac{Mb^2}{6(l^2 + b^2)}$ - **Triangle**: For a triangle of base b and height h and mass m: - $I_{base} = \frac{Mb^2}{6}$ - **Square**: For a square of side a and mass m: - $I_{COM} = \frac{Ma^2}{12}$ (in plane) - $I_{diagonal} = I + I = \frac{Ma^2}{6}$ (in plane) - **Hollow Sphere**: For a hollow sphere of radius r and mass m: - $I_{COM} = \frac{2}{3}MR^2$ - **Solid Sphere**: For a solid sphere of radius r and mass m, - $I_{COM} = \frac{2}{5}MR^2$ - **Hollow Cone**: For a hollow cone of radius r, height h, and mass m: - $I_{hollow \ cone} = \frac{MR^2}{2y^2}$ - **Solid Cone**: For a solid cone of radius r, height h, and mass m: - $I_{solid \ cone} = \frac{3MR^2}{10y^2}$ - **Hollow Cylinder**: For a hollow cylinder of radius r, height h, and mass m: - $I_{yy'} = MR^2$ - $I_{bisector} = \frac{MR^2}{2} + \frac{ML^2}{12}$ - **Solid Cylinder**: For a solid cylinder of radius r, height h, and mass m: - $I_{yy'}=\frac{MR^2}{2}$ - $I_{bisector} = \frac{MR^2}{4} + \frac{ML^2}{12}$ # Ques: Find the moment of Inertia of a solid sphere about a tangent to the sphere where M is mass & R is radius. - **Answer**: - $I_{xx'} = I_{COM} + Ma^2 = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2 $ # Ques: 3 identical shells. Find $I_{xx'}$. - **Answer:** - $I_{xx'} = I_1 + I_2 + I_3$ - $I_{xx'} = 2 \times (I_{COM} + Ma^2) + I_{COM}$ - $I_{xx'} = 2 \times (\frac{2}{3}MR^2 + MR^2 ) + \frac{2}{3}MR^2$ - $I_{xx'} = 2 \times \frac{5}{3}MR^2 + \frac{2}{3}MR^2 = \frac{10}{3}MR^2 + \frac{2}{3}MR^2 = \frac{12}{3}MR^2$ - $I_{xx'} = 4MR^2$ # Equilibrium of a Rigid Body - **Definition**: A rigid body is said to be in equilibrium, if both of its linear momentum and angular momentum are not changing with time. - **This means**: Thus, equilibrium body does not possess linear acceleration or angular acceleration. - **Equilibrium Conditions**: This means: - i) The total force, i.e. the vector sum of all forces acting on the rigid body, is zero. - $\sum_{i=1}^n F_i = 0$. - ii) The total torque , i.e. the vector sum of all torques acting on the body must be zero. - $\sum_{i=1}^{n} \tau_i=0$. - iii) The sum of the components of the torques along any axis perpendicular to the plane of the forces must be zero. - **Note**: A body may remain in partial equilibrium as well. It means that body may remain only in translational equilibrium or only in rotational equilibrium. # Rotational Equilibrium Only- - **Definition**: If the net torque acting on the rigid body is zero but net force is non-zero, then the rigid is in rotational equilibrium only. - **Example**: Two equal forces are acting at the ends of a rod. In this case, the net torque produced by two forces will be zero, but net force is non-zero. It means that the rigid body will have zero angular acceleration but non-zero linear acceleration. # Translational Equilibrium Only - **Definition**: If the net force acting on the rigid body is zero but net torque is non-zero, then the rigid body is in translational equilibrium only. - **Example**: Two equal and opposite forces are acting at the ends of a rod. In this case, the net force is zero, but both the forces will produce a non-zero torque. It means that rigid body will have zero linear acceleration, but non-zero angular acceleration. - **Center of Gravity**: A body is supported on a point such that total gravitational torque about this point is zero, then this point is called the center of gravity of the body. # Laws of Rotational Motion - **Corresponding to**: Corresponding to Newton’s three laws of translatory motion, we have three laws of rotational motion. ## Laws of Rotational Motion - **First Law**: Every body continues in its state of rest or of uniform rotational motion unless an external torque is applied on it to change that state. - **Second Law**: The rate of change of angular momentum of a body is directly proportional to the torque applied and the change occurs in the direction of the torque applied. - **Third Law**: To every external torque applied, there is an equal and opposite reaction torque. # Kinematics of Rotational Motion about a Fixed Axis - **Angular Displacement**: The angle made by the body from its point of rest at any point in the rotational motion is the angular displacement. - $ \theta = \frac{s}{r}$. - where $\theta$ is the angular displacement - s is the distancer travelled by the body - r is the radius of the circle along which it is moving. - **Angular