Rotational Dynamics PDF

# Rotational Dynamics ## Rotational Motion of a Rigid Body A rigid body is one in which the distance between any two particles (points) does not change by any external force and motion of the body. ## Rotational motion: A rigid body is said to be in rotational motion, if its position particle...

# Rotational Dynamics ## Rotational Motion of a Rigid Body A rigid body is one in which the distance between any two particles (points) does not change by any external force and motion of the body. ## Rotational motion: A rigid body is said to be in rotational motion, if its position particles move in such a way that they took in different circles of different radii about the axis of rotation. ## Consider a rigid body rotating about an axis in the x-y plane. As the body passes from position P₁ to P₂, the point particles A and B move to A' and B'. <div align="center"><img src="https://i.imgur.com/6U4s7g8.png"></div> In this case, the angular displacements of A and B are the same, and they move through the same angle subtended by these arcs. Their angular velocities and angular accelerations are also the same. However, the linear displacements AA' and BB' are different. Hence, the linear velocities and linear accelerations of these particles are also different. So in a rotational motion of a rigid body, all constituent particles have the same angular displacement, angular velocities, and angular accelerations, but they have different linear displacements, linear velocities, and linear accelerations. ## Some Terms: **Angular Displacement (θ):** It is defined as the angle subtended by the arc traversed by a point particle in a small time. Then ΔS = r Δθ. Where ΔS is the linear displacement and r is the radius vector. In vector form: ΔS= r Δθ. **Angular velocity (ω):** It is defined as the rate of change of angular displacement. ω = Δθ/Δt. * Δθ = θ₂ - θ₁ is the angle subtended in a small time Δt. * Instantaneous angular velocity = ω = lim Δt→0 Δθ/Δt = dθ/dt. * Average angular velocity = ω = Δθ/Δt. ω is a vector quantity represented along the axis of rotation. Now, v = rω. Therefore, linear velocity = radius * angular velocity. In vector form v = r x ω . **Angular Acceleration (α):** Angular acceleration is defined as the rate of change of angular velocities. Then α = Δω/Δt = (ω₂ - ω₁)/(t₂ - t₁), where ω₁, ω₂ are the angular velocities at time t₁ and t₂. * Average angular acceleration α = Δω/Δt. * Instantaneous angular acceleration α = lim Δt→0 Δω/Δt = dω/dt. Therefore, angular acceleration is α = dω/dt. Linear acceleration = radius * angular acceleration. In vector form a = r x α. ## Where * a = r α (tangential acceleration) * a_r = ω²r (radial acceleration) Angular acceleration is also directed along the axis of rotation. ## Time Period of rotation (T) T is the time taken by a particle to complete one rotation. T = 2π/ω = v/2πr. ## Rotational Kinematics Where f = frequency of rotation, α = angular acceleration, ω₀ = initial angular velocity, ω = final angular velocity, t = time. Rotational kinematics are represented as: (1) ω = ω₀ + αt (2) θ = ω₀t + (1/2)αt² (3) ω² = ω₀² + 2αθ (4) θ = ω₀t + ½(ω - ω₀)t ## Terms in Rotational Dynamics * The branch of physics that deals with the study of rotational motion along with its cause of motion is called rotational dynamics. * Torque is the cause of rotational motion. It is a quantity corresponding to force in linear motion. * Torque is defined as the product of magnitude of force and the perpendicular of space from the axes of rotation to the line of action of the force. * It is the cross product of position vector and force. * Mathematically τ = r x F * τ = rF sin0 = F(r sin0) = Fd * τ = magnitude of force * moment arm. In case of three dimensional motion, in terms of rectangular components, r = xi + yj + zk and F = Fxi + Fyj + Fzk. Then τ = (Fy*z - Fz*y)i + (Fz*x - Fx*z)j + (Fx*y - Fy*x)k ## Torque on a Rigid Body ## For rotation of a rigid body m₁, m₂, m₃ be the masses of particles situated at distances r₁, r₂, r₃ from the axis. The moment of force f₁ = mir₁α (: a = rα). So, the moment of force f₁: m₁r₁α x r₁ = m₁r₁²α. Similarly, the moment of force f₂ = m₂r₂²α and so on. By definition, torque τ = (m₁r₁²α + m₂r₂²α + m₃r₃²α ...) = (m₁r₁² + m₂r₂² + m₃r₃² ...) α = Iα. Where I = moment of inertia = (m₁r₁² + m₂r₂² + ...). So Torque τ = Iα (vector form) ## Relation Between Torque and Angular Momentum ## (Iω) = τ Torque is defined as the rate of change of angular momentum. τ = d(Iω)/dt = I(dω/dt) = Iα ## Vector form: τ = dL/dt τ = d/dt (Emr²ω) = (Emr²) dω/dt = Iα [:: d/dt(Emr²) = 0 and α = dω/dt] τ = Iα → Equation is analogue to Newton's second law ## Conservation of Angular Momentum States that the net angular momentum of an isolated system remains constant when the external torque on the system is zero (i.e. external torque on the system is zero). For a single particle system: Consider a particle moving with an angular momentum L. In an isolated system, there is no external torque on the particle. So dL/dt = 0 ⇒ L = constant. Torque can be zero in the following cases: * When the external torque is zero. * When external forces are zero. * When the line of action of the external forces passes through the axis of rotation. * Like a spinning ball or a planet around the sun (this is a category of central force). * The line of action of the force passes through the centre of motion. In the case of the central force, the angular momentum remains constant, and the body continues to rotate in the same plane. Constancy of L means angular velocity of such a body will not change in magnitude, but a change in direction is allowed according to the law of conservation of angular momentum. ## Conservation of Angular Momentum for a many-particle system Consider a system consisting of many particles. The angular momentum of the system may change due to the torque exerted on the particles by external forces or due to internal forces between the particles. Newton's third law vanishes and hence internal forces vanish. Now, let Pi be the momentum of the ith particle of the system, having a position vector r_i with origin O. The angular momentum Li = r_i x P_i. The total angular momentum of the system is L = ΣL_i = Σ(r_i x P_i) = Σ(r_i x m_i v_i) = Σ(m_ir_i x v_i) But r_i x v_i = r_i x (r_iω) = 0 *Net force F_i on the ith particle is due to external forces as well as internal forces.* ΣF_i = ΣF_i (ext) + ΣF_i (int) where ΣF_i (ext) is the total external forces due to all particles which vanishes. So ΣF_i becomes ΣF_i = ΣF_i (ext) = ΣF_i (ext) = (Σm_i) a = Ma. In an isolated system, Στ = 0 So dL/dt = 0 ⇒ L is constant. Thus, the angular momentum of the system of particles remains (const) conserved. ## Applications of Law of Conservation of Momentum: ### Planetary motion: The angular velocity of a planet increases when it is closer to the sun in an elliptical path during its revolution. As L = Iω is constant. When the planet is closer to the sun, I = mr² decreases and hence ω increases. ### Suppose a person standing on a rotating platform in his hands stretched out heavy weights is rotating on a platform. When the person folds his arms, this moment of inertia I decreases, hence angular velocity ω increases. This is the same as a rotating body. ### The destructive effect of a tornado is greater near the center of disturbance than the edges. In a tornado, the moment of inertia of the wind decreases as air moves towards its centre (I decreases), resulting in the angular velocity increasing (ω increases), and it is more destructive at its center. ## Moment of Inertia (M.I.) The moment of inertia of a body about an axis is defined as the property of a body by virtue of which it is unable to change its state of rest or uniform rotational motion without the help of an external torque. #### Physical significance of M.I. Moment of inertia plays the same role in rotational motion as mass does in linear motion. i.e. M.I is the analogue of mass in linear motion. It is the measure of opposition to the change in the state of rest or rotational motion. #### Mathematical Expression for M.I. The equation for M.I of a rigid body can be derived from its K.E of rotation. Consider a rigid body rotating about an axis xy with an angular velocity ω. Let m₁, m₂, m₃ be the masses of different various particles situated at a distance r₁, r₂, r₃ respectively from the axis of rotation. The equation for M.I. of a rigid body is K.E of particle of mass m₁ = ½m₁v₁² = ½m₁r₁²ω² (v = rω) K.E of particle of mass m₂ = ½m₂v₂²= ½m₂r₂²ω² and so on. Therefore, the total K.E of a rotating body = ½m₁r₁²ω² + ½m₂r₂²ω² + ½m₃r₃²ω²+ ..... Where I = Σmr² = (m₁r₁² + m₂r₂² + m₃r₃²) → is called moment of inertia of the body. ## From equation I = Σmr² * The moment of inertia about an axis is defined as the sum of the product of mass and the square of the distances of different particles from the axis of rotation. * From the same equation, it is clear that M.I depends upon the mass of the body and the distribution of mass about the axis of rotation. *Unit of M.I is kgm². ## Note * The dimensional formula is [ML²T⁰] * When ω = 1, I = 2E_rot . * So, M.I is twice to the rotational kinetic energy when angular velocity is unity. ## Radius of gyration ## K The radius of gyration of a body about an axis is that distance at which if the whole mass of the body were concentrated, it would have the same moment of inertia (or K.E of rotation) of the body originally. Consider a rigid body of mass m rotating about an axis xy with a moment of inertia I. Moment of inertia I = m₁r₁² + m₂r₂² + m₃r₃² + ..... Let m₁ = m₂ = m₃ = ... = m (say) I = m (r₁² + r₂² + r₃² + ..... ) I = mk². Where k = radius of gyration. So, mk² = m( r₁² + r₂² + r₃² + .....) k² = (r₁² + r₂² + r₃² + .....)/n Therefore, the radius of gyration of a body = √(mean of the square of various particles of the body from the axis of rotation). ## Moment of Inertia of a solid disc about an axis passing through its center and perpendicular to its plane: Consider a circular solid disc rotating about an axis xy passing through its centre and perpendicular to its plane. Let M be the mass of the disc, R be the radius of the disc. *Area = πR² *Mass per unit area = M/πR² <div align="center"><img src="https://i.imgur.com/N4m03jQ.png"></div> Now let us consider an element (in the form of a ring) bounded between two concentric circles of radii r and r + dr. Let dr be the thickness of the ring. Mass of the ring = mass per unit area * area of the element = M/πR² * 2πrdr = 2M/R² * rdr Moment of inertia of the element (dI): dI = mass of the element * (distance)² = 2M/R² * rdr * r² = 2M/R² * r³ dr. Therefore, the moment of inertia of the whole disc about xy is I = ∫dI = ∫₀^R 2M/R² * r³ dr = 2M/R² [r⁴/4]₀^R = 2M/R² *(R⁴/4) = ½MR² Thus, M.I of the disc = ½(Mass of the disc) x (radius)² ## Note: * Moment of inertia about its diameter = ½MR². * Moment of inertia of an annular disc about an axis passing through its centre and perpendicular to its plane = M(R² + r²)/2. ## Theorems of Moment of inertia ### Perpendicular Axis Theorem: It states that the moment of inertia of a plane lamina about an axis perpendicular to the plane of the lamina is equal to the sum of its moments of inertia of the lamina about two mutually perpendicular axes lying on the plane of the lamina, intersecting each other at the point where the perpendicular axis passes through the plane of the lamina. Mathematically, I_z = I_x + I_y Where I_x = M.I. of lamina about xx', I_y = M.I. of the lamina about yy'. Let xx' and yy' be two perpendicular axes lying on the plane of the lamina intersecting at O. Let zz' be an axis perpendicular to the plane of the lamina, and the axis passes through the point of intersection O. <div align="center"><img src="https://i.imgur.com/H7QXaP9.png"></div> Consider a point mass ‘m’ situated at P(x,y) at a distance r from O. As the body rotates about zz', the moment of inertia is given by I_z = Σmr². As the body rotates about xx', the moment of inertia is given by I_x = Σmy². Similarly, the moment of inertia about yy' is given by I_y = Σmx². So I_z = Σmr² = Σm(x² +y²) = Σmx² + Σmy² = I_x + I_y. Hence, I_z = I_x + I_y, proved. ### Parallel Axis Theorem: It states that the moment of inertia of a body about an axis is equal to the sum of its moment of inertia about a parallel axis through its centre of gravity (or C.G.) and the product of its mass and the square of the distance between the two axes. Let X'Y', XY be the axes of rotation. <div align="center"><img src="https://i.imgur.com/50K1kQ1.png"></div> Let X'Y' be the axis passing through C.G. and parallel to XY. Let ‘h’ be the distance between these two axes of rotation. Consider a point mass ‘m’ at a distance of ‘r’ from C.G. and at a distance of (h + r) from the axis XY. Moment of inertia of the body about X'Y' is given by I_G = Σmx². The moment of inertia of the body about XY is given by I = Σm(x + h)² I = Σm(x² + h² + 2xh) = Σmx² + 2mh² + 2Σmxh Since the body has to rotate about X'Y', the moment of inertia of the body would be I_G = Σmx². And mh² = h² Σm = Mh². Where M = total mass. So I = 2Σmx + I_G + Mh² I = I_G + Mh². Where Σ2mxh = 0, since the body can be balanced at the C.G. So, the equation becomes I = I_G + Mh² (proved) ## Moment of inertia of a solid cylinder about its own axis. Let us consider a solid cylinder of mass M, radius R, and length L. The axis of rotation passes through its center and is perpendicular to its length, axis yy'. *Volume of the cylinder = πR²L *Mass per unit volume of the cylinder = M/πR²L <div align="center"><img src="https://i.imgur.com/7g6Zv0e.png"></div> We can imagine the soild cylinder to be made up of a number of coaxial cylindrical shells. Let us consider an elementary shell with inner radius r and thickness dr., along the axis of rotation. *Cross sectional area of the elementary shell = 2πrdr. *Volume of the shell = 2πrdrL *Mass of the cylindrical shell = 2πrdrL * (M/πR²L) = 2M/R² * rdr. Moment of inertia of the elementary shell: dI = mass of the shell * (distance)² = 2M/R² * rdr * r² = 2M/R² * r³ dr. So total moment of inertia of the whole cylinder is I =∫dI =∫₀^R 2M/R² * r³ dr = 2M/R² * [r⁴/4]₀^R = 2M/R² *(R⁴/4) = ½MR² ## Moment of inertia of a hollow cylinder about its own axis Let us consider a hollow cylinder of mass M, inner radius r, and outer radius R, and length L. The axis of rotation passes through its centre and is perpendicular to its length, axis yy'. Mass per unit length = M/2πRL. <div align="center"><img src="https://i.imgur.com/C92M9wK.png"></div> We can imagine the hollow cylinder to be made up of a number of circular discs, each with its plane parallel to yy'. Consider one such disc of thickness dx situated at a distance x from O. *Mass of the elementary disc = π(R² - r²)dx * (M/2πRL) = M(R² - r²)dx/2RL Moment of inertia of the elementary disc about a diameter AB (parallel to yy): dI = mass of the disc * (radius)²= M(R² - r²)dx/2RL * (R²/4) = M(R² - r²)R²dx/8RL Therefore, the moment of inertia of the disc about yy' at distance x from O is dI = I_G + Mh² = ½M(R² - r²)dx/RL + M(R² - r²)dx/ (2RL) * x² Hence, M.I. of the whole cylinder about yy' is I = ∫dI = ∫₀^L ½M(R² - r²)dx/RL + ∫₀^L M(R² - r²)dx/ (2RL) * x² = M(R² - r²)/2RL * [x]₀^L + M(R² - r²)/2RL * [x³/3]₀^L = M(R² - r²)/2RL * L + M(R² - r²)/2RL * L³/3 = ½M(R² - r²) + (1/6)ML²(R² - r²) = M(R² + r²)/2 + (1/6)ML²(R² - r²) = M(R² + r²)/2 (for thin cylinder) So, moment of inertia of the hllow cylinder about yy': I = ½M(R² + r²). ## Moment of inertia of a solid sphere about its diameter. Consider a solid sphere of mass, M, and radius R. Let xx' be the axis of rotation passing along the diameter. *Volume of the sphere = (4/3)πR³ *Mass per unit volume = M/((4/3)πR³). <div align="center"><img src="https://i.imgur.com/pVxG28t.png"></div> Now let us consider the sphere to be made up of a number of thin discs placed perpendicular to the axis xx' one after another. Consider one such disc with a thickness dx at a distance x from O. Distance from O to the edge of the disc = √(R²-x²). *Radius of the disc: y =√(R² - x²) *Area of the disc: π(R²-x²)dx *Mass of the element = π(R²-x²)dx * (3M/4πR³) = 3M/4R³ * (R²-x²)dx Moment of inertia of the elementary disc about xx': dI = ½(mass of the element) * (radius)² = ½(3M/4R³ * (R²-x²)dx) * (R²-x²) = 3M/8R³(R²-x²)²dx = 3M/8R³(R⁴ + x⁴ − 2R²x²). So total moment of inertia of the whole sphere is I =∫dI =∫_-R^R 3M/8R³(R⁴ + x⁴ − 2R²x²) dx = 3M/8R³[R⁴x + (x⁵/5) − 2R²(x³/3)]_-R^R = 3M/8R³[(R⁵ + (R⁵/5) − 2R²(R³/3)) − ((-R⁵) + (-R⁵/5) − 2R²(-R³/3)] = 3M/8R³[(R⁵ + (R⁵/5) − 2R²(R³/3)) + (R⁵ + (R⁵/5) − 2R²(R³/3)] = 3M/8R³[2(R⁵ + (R⁵/5) − 2R²(R³/3)] = 3M/4R³[R⁵ + (R⁵/5) − 2R²(R³/3)] = 3M/4R³[(16R⁵/15)] = (2/5)MR² Therefore, the moment of inertia of a solid sphere = (2/5)MR². ### Applying parallel axes theorem: To get the M.I. of the sphere about the tangent. I = I_G + Mh² Where I_G = (2/5)MR² and h = R. I = (2/5)MR² + MR² = (7/5)MR² ## Note: * M.I of a spherical shell = (2/3)MR². * M.I of a thick shell (hollow sphere) about diameter = (2/3)M[R⁵-r⁵]/(R³-r³). ## This rule is usually used to find the moment of inertia of symmetrical bodies. ### Routh’s rule: Routh’s rule states that the moment of inertia of a body about any of the three mutually perpendicular axes of symmetry passing through the center of mass is equal to the mass of the body * the sum of the squares of the other two perpendicular semi-axes. * I = MS/n * I = moment of inertia about the required axis. * M = mass of the body. * S = sum of the squares of the two semi-axes except the axis about which the moment of inertia is to be found. * n = positive integer. * n = 4 for a square, rectangle or parallelogram. * n = 4 for a circle or ellipse. * n = 5 for a sphere or cylinder. This rule can be used to find the moment of inertia of symmetrical bodies. It is a very useful method for finding the moment of inertia of many bodies, particularly for finding moments of inertia of symmetrical bodies. ## Examples: ### Moment of inertia of a rectangular lamina about an axis passing through its center and perpendicular to its plane. From Routh’s rule I = MS/n For a rectangle, n = 2 S = sum of squares of two perpendicular semi-axes = (b/2)² + (l/2)² = 1/4(b²+l²) So I = MS/n = M(1/4 (b²+l²) )/2 = (1/8)M(b² + l²) Therefore, the moment of inertia of a rectangular lamina about an axis passing through its center and perpendicular to its plane = (1/8)M(b² + l²). ### Moment of inertia of a circular disc about an axis passing through its center and perpendicular to its plane For a circular disc, n = 4 Here, S = R² + R² = 2R² So I = MS/n = M(2R²)/4 = ½MR² Therefore, the moment of inertia of a circular disc about an axis passing through its center and perpendicular to its plane = ½MR². ### Moment of inertia of a solid sphere about its diameter. For a sphere, n= 5 Here, S = R² + R² = 2R² So I = MS/n = M(2R²)/5 = (2/5)MR² Therefore, the moment of inertia of a solid sphere about its diameter = (2/5)MR² ### Moment of inertia of an elliptical disc about an axis passing through its center and perpendicular to its plane For an elliptical body, n = 4 S = a² + b² So I = MS/n = M(a² + b²)/4 Therefore, the moment of inertia of an elliptical disc about an axis passing through its center and perpendicular to its plane = M(a² + b²)/4 ## Moment of inertia of a circular body: I = AR²/4 where A = area of circular section. For I about zz. A = ππR² I = π(R²)*(1/4) *2R² = πR⁴ I = area of a circular body = πR² =π(D/2)² I = (πD⁴) = πD⁴/16. ## Calculate the moment of inertia of a rectangular lamina in terms of the area of the body. Let the dimensions of the lamina be length (l) and breadth (b). About zz: Mass = σlb <div align="center"><img src="https://i.imgur.com/bP8P5a1.png"></div> Area = lb S = (l/2)² + (b/2)² = 1/4(l² + b²) I = σlb * (1/4(l² + b²))/3 = σlb/12 * (l² + b²) I = (σlb/12) * (l²+ b²) I = (mass/12) * (l²+ b²) ## Euler's Equation of Motion The fundamental equation of rotational motion of a rigid body is: τ = Iω where τ = the net torque applied, I = the moment of inertia and ω = the angular momentum. This equation represents Newton's second law for rotation of a rigid body in an inertial frame. However, the equation needs modification for a fixed coordinate system and for the motion of a body in a rotating coordinate system. From the theory of rotating coordinate system, the equation is transformed as: d/dt(Iω)_fixed = (τ + Iω x ω)_rot where * (Iω)_fixed is the total angular momentum of the rigid body in the fixed coordinate system. * (τ + Iω x ω)_rot is the sum of the torque and the Coriolis torque acting on the body in the rotating coordinate system. If the Cartesian axes in the rotating frame lie along the fixed axes 1, 2, 3 (say), then the equation becomes: R = I₁ω₁i + I₂ω₂j + I₃ω₃k where ω₁, ω₂, ω₃ are the components of ω along the principal axes. So, the equation becomes: d/dt(I₁ω₁i + I₂ω₂j + I₃ω₃k) = (τ + (I₁ω₁i + I₂ω₂j + I₃ω₃k) x (ω₁i + ω₂j + ω₃k) Now, let us consider the component along axis 1 (i). The equation becomes: d/dt(I₁ω₁) = [τ_x + (I₃ω₃ω₂ − I₂ω₂ω₃)] And similarly for the component along axes 2 and 3. ## Which are called Euler’s equation of motion of rigid body rotation: d/dt(I₁ω₁) = τ_x + (I₃ω₃ω₂ − I₂ω₂ω₃) d/dt(I₂ω₂) = τ_y + (I₁ω₁ω₃ − I₃ω₃ω₁) d/dt(I₃ω₃) = τ_z + (I₂ω₂ω₁ − I₁ω₁ω₂) These equations are very important in studying the motion of rigid bodies, especially in situations where the axis of rotation is not fixed. ## Kinetic Energy of Rotation For translational motion, KE depends upon mass and velocity and is given by KE = ½mv². But for rotational motion, KE not only depends upon the mass (m) but also depends upon the moment of inertia (I) and the angular velocity (ω). Consider a rigid body rotating about the xy axis with angular velocity ω. Let m₁, m₂, m₃... be the masses of particles situated at distances r₁, r₂, r₃ from the axis of rotation and v₁, v₂, v₃... be their respective linear velocities. So, the total KE = ½m₁v₁² + ½m₂v₂² + ½m₃v₃² + ... But v = rω So total KE of rotation is = ½m₁r<sub>1