Mechanics Topic D (Rotation) - 1 David Apsley

TOPIC D: ROTATION SPRING 2019 1. Angular kinematics 1.1 Angular velocity and angular acceleration 1.2 Constant-angular-acceleration formulae 1.3 Displacement, velocity and acceleration in circular motion 2. Angular dynamics 2.1 Torque 2.2 Angular momentum 2.3 The angular-momentum principle for motion in a circle 2.4 The angular-momentum principle for arbitrary motion 3. Rigid-body rotation 3.1 Moment of inertia 3.2 Second moments and the radius of gyration 3.3 The equations of rotational motion 3.4 Comparing translation and rotation 3.5 Examples 4. Calculation of moments of inertia 4.1 Methods of calculation 4.2 Fundamental shapes 4.3 Stretching parallel to an axis 4.4 Volumes of revolution 4.5 Change of axis 5. General motion of a rigid body (optional) 5.1 Rolling without slipping 5.2 Rolling with slipping Appendix 1: Moments of inertia of simple shapes Appendix 2: Second moment of area Mechanics Topic D (Rotation) - 1 David Apsley

1. ANGULAR KINEMATICS 1.1 Angular Velocity and Angular Acceleration P For a particle moving in a circular arc, or for a rigid body rotating about a fixed axis, the instantaneous position is defined by the angle between a radius vector and a fixed line. O r s If s is length of arc and r is radius then the angle θ in radians is defined such that (1) Angular velocity ω is the rate of change of angle: (2) Angular acceleration α is the rate of change of angular velocity: (3) It is common to use a dot to indicate differentiation w.r.t. time; e.g. means dθ/dt. 1.2 Constant-Angular-Acceleration Formulae There is a direct correspondence between linear and angular motion. Linear Angular Displacement s θ Velocity Acceleration Constant-acceleration formulae Mechanics Topic D (Rotation) - 2 David Apsley

For non-constant acceleration: distance is the area under a v t graph; angle is the area under an ω t graph; v acceleration is the gradient of a v t graph; angular acceleration is the gradient of a ω t graph. s t t Example 1. What is the angular velocity in radians per second of the minute hand of a clock? Example 2. A turbine starts from rest and has a constant angular acceleration of 0.1 rad s 2. How many revolutions does it make in reaching a rotation rate of 50 rpm? 1.3 Displacement, Velocity and Acceleration in Circular Motion Consider a particle moving at a fixed radius r. The following have already been derived in Topic A (Kinematics), as a special case of motion in general polar coordinates. v Velocity r Since s = rθ and r is constant, the velocity is tangential and its magnitude (speed) is (4) Acceleration dv dt Because it is not moving in a straight line, the particle has two components of acceleration: tangential, if its speed is changing: O r v 2 r or (5a) radially inward, because its direction is changing: or (5b) The latter is called the centripetal acceleration. A centripetal force is necessary to maintain this and keep the particle moving in a circular path. This force can be provided in many ways: for example, the tension in a cable, a normal reaction from an outer boundary or friction. Mechanics Topic D (Rotation) - 3 David Apsley

Example 3. Find the minimum coefficient of friction necessary to prevent slipping for a particle which is placed: (a) (b) 100 mm from the rotation axis on a turntable rotating at 78 rpm; on the inside of a cylindrical drum, radius 0.3 m, rotating about a vertical axis at 200 rpm. Example 4. (Exam 2017) A particle of mass 3 kg is whirled around in a horizontal circle by a light elastic string of unstretched length 1.5 m and stiffness 90 N m 1 attached to a fixed point O. At a particular speed, the cable makes an angle of 15º with the horizontal. Find: (a) the tension in the cable; (b) the extension of the cable; (c) the speed of the particle. O 15 o 3 kg Example 5. (Exam 2010) A building s roof consists of a smooth hemispherical dome with outside radius 20 m. A brief gust of wind dislodges a small object at the top of the dome and it slides down the roof. 20 m (a) (b) (c) (d) (e) Find an expression for the velocity v of the object when its position vector makes an angle θ with the vertical through the centre of the dome (see the figure). While it is in contact with the roof the object is undergoing motion in a circular arc. Write down an expression for its centripetal acceleration as a function of angle θ. Find an expression for the normal contact force as a function of angle θ and the mass m of the object. Hence determine the angle θ at which the object leaves the roof, as well as its height and speed at this point. Find the distance from the outside wall of the dome at which the object hits the ground. Mechanics Topic D (Rotation) - 4 David Apsley

2. ANGULAR DYNAMICS 2.1 Torque The torque 1 (or moment of force) T about an axis is given by: axis r F torque = force perpendicular distance from axis (6) Torque measures the turning effect of a force. When the force is not perpendicular to the radius vector then only the component perpendicular to the radius vector contributes torque. v 2.2 Angular Momentum Angular momentum (or moment of momentum) h is given by: axis r m angular momentum = momentum perpendicular distance from axis (7) For non-circular motion, v is the transverse component of velocity see Section 2.4. 2.3 The Angular-Momentum Principle For Motion in a Circle Force-momentum principle: If F is the tangential component of force and r is constant (i.e. circular motion) then torque = rate of change of angular momentum (8) In fact, (8) holds for non-circular motion, but the proof requires more effort; see Section 2.4. Equation (8) is the rotational analogue of the momentum principle for translational motion: force = rate of change of momentum For single particles the angular-momentum equation offers no advantage over the momentum equation. However, it is invaluable for rigid-body rotation, in which it is applied by summing over all masses in the system. The torque is then the sum of the moments of the external forces only, since internal forces between particles are equal and opposite and cancel in pairs. 1 Whilst one can have a moment of any physical quantity, torque is used almost exclusively for moment of force. Mechanics Topic D (Rotation) - 5 David Apsley

Example 6. (Ohanian) The original Ferris wheel built by George Ferris had radius 38 m and mass 1.910 6 kg. Assuming that all of its mass was uniformly distributed along the rim of the wheel, if the wheel was initially rotating at 0.05 rev min 1, what constant torque would stop it in 30 s? What force exerted on the rim of the wheel would have given such a torque? In the absence of an external torque, a direct corollary of the angular-momentum principle is: The Principle of Conservation of Angular Momentum The angular momentum of an isolated system remains constant. 2.4 The Angular-Momentum Principle For Arbitrary Motion For a particle of mass m, the angular momentum is O r sin r P v i.e. only the transverse component of velocity, v = v sin α, contributes to the angular momentum. The radial component, v r = v cos α, has no moment about the axis. A similar decomposition applies for the torque. O r v sin P v v cos Using a vector cross product (denoted by ), both angular momentum and torque may be represented by vectors oriented along the rotation axis (in the sense of a right-hand screw): angular momentum: or (9a) torque: or (9b) Differentiating the vector expression for angular momentum, using the product rule: Hence, which is, in vector form, the angular-momentum principle: rate of change of angular momentum = torque By summation this can be applied to the whole, with T the torque due to external forces only. (10) Mechanics Topic D (Rotation) - 6 David Apsley

3. RIGID-BODY ROTATION 3.1 Moment of Inertia Example. A bicycle wheel and a flat disk have the same mass, the same radius and are spinning at the same rate. Which has the greater angular momentum and kinetic energy? For rotating rigid bodies, different particles lie at different radii and hence have different speeds. Particles at greater radius move faster and contribute more to the body s angular momentum and kinetic energy. Thus, the angular momentum and kinetic energy depend on the distribution of mass relative to the axis of rotation. The total angular momentum and kinetic energy may be obtained by summing over individual particles of mass m at radius r. Most importantly, although particles at different radii have different speeds v, they all have the same angular velocity ω. Angular Momentum r m r Kinetic Energy The quantity (11) is the moment of inertia (or second moment of mass) of the body about the specified axis. angular momentum (12) kinetic energy (13) (c.f. momentum P = Mv and kinetic energy for translation). Mechanics Topic D (Rotation) - 7 David Apsley

Moment describes the distribution of mass relative to the selected axis. (It gives higher weighting to masses at greater radii.) Inertia refers to a resistance to a change in motion (acceleration). In this sense, the moment of inertia fulfils the same role for rotation as the mass of a body in translation. Example revisited. For a hoop (a close approximation to the bicycle wheel) all the mass is concentrated at the same radius R. Hence For a flat disk it turns out (see later) that the moment of inertia is ½MR 2. Other things being equal, the disk will have half the angular momentum and half the kinetic energy of the hoop. This is because some of its mass is at a smaller radius and is moving more slowly. 3.2 Second Moments and the Radius of Gyration The moment (strictly, the first moment) of any quantity is defined by first moment = quantity distance Similarly, second moment = quantity (distance) 2 For an extended body the distance varies, so we must sum over constituent parts; e.g. (14) (In Hydraulics and Structures courses a similar quantity second moment of area appears in connection with hydrostatic forces on and resistance to bending, respectively.) The centre of mass is where the same concentrated mass would have the same first moment: (15) The radius of gyration k is that at which the same mass would have the same second moment: (16) Distributed mass Concentrated mass R k Same total mass and moment of inertia Mechanics Topic D (Rotation) - 8 David Apsley

Examples. Hoop of mass M and radius R (axis through centre, perpendicular to plane) Moment of inertia, radius of gyration, k = R Here, radius of gyration is geometric radius as all mass is concentrated at the circumference. Circular disc of mass M and radius R (axis through centre, perpendicular to plane) Moment of inertia radius of gyration The radius of gyration is less than the geometric radius because mass is distributed over a range of distances from the axis. 3.3 The Equations of Rotational Motion (Angular) Momentum For rigid-body rotation the equation of motion is the angular momentum equation: torque = rate of change of angular momentum (17) This is the rotational equivalent of Newton s Second Law: force = rate of change of momentum For solid bodies, where I and M are constant we can write these in terms of acceleration: (rotation) (translation) (18) (Angular) Impulse If we integrate (17) with respect to time we obtain an impulse equation: torque time = change in angular momentum The LHS is called the angular impulse. (19) Mechanics Topic D (Rotation) - 9 David Apsley

Energy Alternatively, integrate (17) wrt angle to obtain an energy equation. First rewrite it as Integrating with respect to angle θ gives the Mechanical Energy Principle: work done (torque angle) = change in kinetic energy (20) 3.4 Comparing Translation and Rotation Translation Rotation Displacement x θ Velocity v ω Acceleration a α Inertia m I Effective location of mass centre of mass radius of gyration Cause of motion force torque Translation Rotation Momentum Kinetic energy Power Fv Tω Equation of motion rate form force = rate of change of momentum torque = rate of change of angular momentum Equation of motion impulse form impulse (force time) = change of momentum angular impulse (torque time) = change of angular momentum Equation of motion energy form work done (force distance) = change of kinetic energy work done (torque angle) = change of kinetic energy Mechanics Topic D (Rotation) - 10 David Apsley

3.5 Examples Example 7. A bucket of mass M is fastened to one end of a light inextensible rope. The rope is coiled round a windlass in the form of a circular cylinder (radius r, moment of inertia I) which is left free to rotate about its axis. Prove that the bucket descends with acceleration r M Mg Example 8. A flywheel whose axial moment of inertia is 1000 kg m 2 rotates with an angular velocity of 300 rpm. Find the angular impulse which would be required to bring the flywheel to rest. Hence, find the frictional torque at the bearings if the flywheel comes to rest in 10 min under friction alone. Example 9. A flywheel of radius 500 mm is attached to a shaft of radius 100 mm, the combined assembly having a moment of inertia of 500 kg m 2. Long cables are wrapped around flywheel and shaft in opposite directions and are attached to masses of 10 kg and 20 kg respectively, which are initially at rest as shown. Calculate: (a) how far the 10 kg mass must drop in order to raise the 20 kg mass by 1 m; (b) the angular velocity of the shaft at this point. 500 mm 100 mm 10 kg 20 kg Mechanics Topic D (Rotation) - 11 David Apsley

Example 10. A 15 kg mass hangs in the loop of a light inextensible cable, one end of the cable being fixed and the other wound round a wheel of radius 0.3 m and moment of inertia 0.9 kg m 2 so that the lengths of cable are vertical (see the figure). The mass is released from rest and falls, turning the wheel. Neglecting friction between the mass and the loop of cable and between the wheel and its bearings, find: (a) a relationship between the downward velocity of the mass, v, and the angular velocity of the wheel, ω; (b) the downward acceleration of the mass; (c) the speed of the mass when it has fallen a distance 2 m; (d) the number of turns of the wheel before it reaches a rotation rate of 300 rpm. 15 kg Example 11. A square plate of mass 6 kg and side 0.2 m is suspended vertically from a frictionless hinge along one side. It is struck dead centre by a lump of clay of mass 1 kg which is moving at 10 m s 1 horizontally and remains stuck (totally inelastic collision). To what height will the bottom of the plate rise after impact? (The moment of inertia of a square lamina, side a and mass M, about one side, is ) Mechanics Topic D (Rotation) - 12 David Apsley

4. CALCULATION OF MOMENTS OF INERTIA 4.1 Methods of Calculation The moment of inertia I depends on: the distribution of mass; the axis of rotation. Some common methods of calculating I are as follows. Method 1. First Principles For isolated particles this can be done by direct summation. For continuous bodies integration is necessary. Method 2. Combination of Fundamental Elements (Hoop, Disk, Rod) hoop surface of revolution First principles disc solid of revolution rod rectangular lamina Method 3. Stretching Parallel to the Axis If a shape is simply stretched parallel to an axis then the moment of inertia is unchanged since the relative disposition of mass about the axis is not changed. e.g. hoop cylindrical shell disc solid cylinder rod rectangular lamina Method 4. Change of Axis hoop/disc rod cylinder rectangle Calculations may be performed first about some convenient (typically symmetry) axis; the moment of inertia about the actual axis is then determined by one of two techniques for changing axes: the parallel-axis theorem and the perpendicular-axes theorem. Mechanics Topic D (Rotation) - 13 David Apsley

4.2 Fundamental Shapes 4.2.1 Hoop For a hoop (an infinitesimally-thin circular arc) of mass M and radius R, rotating about its symmetry axis, all the mass is concentrated at the single distance R from the axis. Hence, R For a hoop of mass M and radius R, about the symmetry axis perpendicular to its plane: (21) 4.2.2 Disc Consider the moment of inertia of a uniform circular disc (an infinitesimally-thin, circular plane lamina) of mass M and radius R, about the axis of symmetry perpendicular to its plane. The disc can be broken down into sub-elements which are hoops of radius r and thickness r. Let ρ be the mass per unit area. Sum over all elements: r r R For a hoop of mass M and radius R, about the symmetry axis perpendicular to its plane: (22) 4.2.3 Rod Consider the moment of inertia of a rod (an infinitesimally-thin line segment) of mass M and length L, about its axis of symmetry. The rod can be broken down into sub-elements of length δx, distance x from the axis. Let ρ be the mass per unit length. L x x Summing: Mechanics Topic D (Rotation) - 14 David Apsley

For a rod of mass M and length L, about its axis of symmetry: (23) 4.3 Stretching Parallel to an Axis The distribution of mass about the axis and hence the moment of inertia is not changed by stretching parallel to the axis of rotation without change of mass. Hence, for the axes shown: hoop cylindrical shell: R disc solid cylinder: rod rectangular lamina: hoop/disc cylinder b (In the last case the axis is in the plane of the lamina.) a The only dependence on the dimension parallel to the axis of rotation is via its effect on the total mass M. rod rectangle Mechanics Topic D (Rotation) - 15 David Apsley

Example 12. outer steel rim (part (b)) 40 cm 6 cm flywheel shaft 30 cm (a) (b) (c) 10 cm A flywheel consists of an aluminium disc of diameter 40 cm and thickness 6 cm, mounted on an aluminium shaft of diameter 10 cm and length 30 cm as shown. Calculate the moment of inertia of flywheel + shaft. To increase the moment of inertia a steel rim is fixed to the outside of the flywheel. Calculate the outer radius of the steel rim required to double the moment of inertia of the assembly. If the flywheel is initially rotating at 100 rpm, calculate the constant frictional braking force which needs to be applied to the outside of the steel rim in part (b) if the flywheel is to be brought to rest in 30 seconds. For this question you may require the following information. Density of aluminium: 2650 kg m 3 ; steel: 7850 kg m 3. Moment of inertia of a solid cylinder of radius R and mass M about its axis:. Mechanics Topic D (Rotation) - 16 David Apsley

4.4 Volumes of Revolution Moments of inertia for volumes of revolution may be deduced by summing over infinitesimal discs (or very thin cylinders) of radius y and length δx. Let ρ be the mass per unit volume. Then the elemental mass and moment of inertia are, respectively: mass: moment of inertia: y x x Summing over all elemental masses and moments of inertia: (24) (25) Example. Find the moment of inertia of a solid sphere, mass M and radius R about an axis of symmetry. For a solid sphere,. Hence, between. Thus, Hence, R axis whence Mechanics Topic D (Rotation) - 17 David Apsley

4.5 Change of Axis 4.5.1 Parallel-Axis Rule If the moment of inertia of a body of mass M about an axis through its centre of mass is I G, then the moment of inertia about a parallel axis A is given by where M is the mass of the body and d is the distance between axes. (26) Proof. Take (x,y,z) coordinates relative to the centre of mass, with the z direction parallel to the axes of rotation. By Pythagoras, G (0,0) d P(x,y) A(x A, y A) Expanding the second of these: The last two terms vanish because there are no moments about the centre of mass. Corollary 1. The corresponding radii of gyration are related by (27) Corollary 2. For a set of parallel axes, the smallest moment of inertia is about an axis through the centre of mass. Example. The moment of inertia of a rod of mass M and length L about an axis through its centre and normal to the rod is the end of the rod is. Hence the moment of inertia about a parallel axis through A G 1 2 L 1 2 L Mechanics Topic D (Rotation) - 18 David Apsley

4.3.2 Perpendicular-Axis Rule Important note. This is applicable to plane laminae only. However, it can often be combined with stretching parallel to the axis to give 3-d shapes. If a plane body has moments of inertia I x and I y about perpendicular axes Ox and Oy in the plane of the body, then its moment of inertia about an axis Oz, perpendicular to the plane, is: (28) Proof. By Pythagoras, z y Hence r x y x Example. Find the moment of inertia of a rectangular lamina, mass M and sides a and b, about an axis through the centre, perpendicular to the lamina. Solution. From the earlier examples, the moments of inertia about axes in the plane of the lamina are b z y x a Example. Find the moment of inertia of a circular disc, radius R, about a diameter. Solution. In this case we use the perpendicular-axis theorem in reverse, because we already know the moment of inertia about an axis through the centre, perpendicular to the plane of the disc:. By rotational symmetry the unknown moment of inertia I about a diameter is the same for both x and y axes. Hence, 1 I MR 2 z 2 I I Mechanics Topic D (Rotation) - 19 David Apsley

Example 13. Find the radius of gyration of the square-frame lamina shown about an axis along one side. axis axis 0.1 m 0.5 m 0.1 m 0.5 m Example 14. A rigid framework consists of four rods, each of length L and mass M, connected in the form of a square ABCD as shown. Find expressions, in terms of L and M, for the moment of inertia of the framework about: (a) the axis of symmetry SS; (b) the side AB; (c) an axis perpendicular to the framework and passing through centre O; (d) an axis perpendicular to the framework and passing through vertex A; (e) the diagonal AC. Data: the moment of inertia of a rod, length L and mass M, about an axis through its centre and perpendicular to the rod is. B S C O A S D Mechanics Topic D (Rotation) - 20 David Apsley

5. GENERAL MOTION OF A RIGID BODY (Optional) The motion of a rigid body which is allowed to rotate as well as translate (e.g. a rolling body) can be decomposed into: It may be shown (optional exercise) that, for a system of particles (e.g. a rigid body): (1) The centre of mass moves like a single particle of mass M under the resultant of the external forces: where (29) (2) The relationship torque = rate of change of angular momentum : holds for the torque of all external forces about a point which is either: fixed, or moving with the centre of mass. (3) The total kinetic energy can be written as the sum: kinetic energy of the centre of mass ( ) + kinetic energy of motion relative to the centre of mass For a rigid body, motion relative to the centre of mass must be rotation and hence: (30) 5.1 Rolling Without Slipping Consider a body with circular cross-section rolling along a plane surface. If the body rolls without slipping then the distance moved by the point of contact must equal the length of arc swept out: v r r Hence the linear and angular velocities are related by: Mechanics Topic D (Rotation) - 21 David Apsley

The instantaneous point of contact with the plane has zero velocity; hence the friction force does no work but it is responsible for rotating the body! The total kinetic energy is given by Example 15. (Synge and Griffiths) A wheel consists of a thin rim of mass M and n spokes each of mass m, which may be considered as thin rods terminating at the centre of the wheel. If the wheel is rolling with linear velocity v, express its kinetic energy in terms of M, m, n, v. A common example is of a spherical or cylindrical body rolling down an inclined plane. The forces on the body are its weight Mg, the normal reaction force R and the friction force F. R v mg F Consider the linear motion of the centre of mass and the rotational motion about it. force = mass acceleration for translation of the centre of mass: (along slope) (normal to slope) torque = rate of change of angular momentum for rotation about the centre of mass: v and ω are related, if not slipping, by v = rω Mechanics Topic D (Rotation) - 22 David Apsley

Example. (Ohanian) A piece of steel pipe, mass 360 kg, rolls down a ramp inclined at 30 to the horizontal. What is the acceleration if the pipe rolls without slipping? What is the magnitude of the friction force that acts at the point of contact between the pipe and ramp? Solution. Linear motion: Rotation about centre of mass: (i) (ii) Eliminate F by (i) r + (ii), noting that : Hence, But for a hoop, and hence (by stretching parallel to the axis) a pipe,. Thus, This is the linear acceleration. For the friction force use either linear or rotational equation of motion; e.g. from (i): Answer: 2.45 m s 2 ; 883 N. Mechanics Topic D (Rotation) - 23 David Apsley

5.2 Rolling With Slipping For a body which is rolling along a surface the condition for no slipping is that the instantaneous point of contact is not moving; that is, the linear velocity of the centre of mass (v) must be equal and opposite to that of the relative velocity of a point on the rim which is rotating (rω). Hence, slipping occurs whilst v rω (31) v r r R In this case, friction will act to oppose slipping. If a spinning body is placed on a surface then it is the friction force which initiates its forward motion. Note that, while slipping occurs, there is relative motion and so friction is maximal: F mg Example. (Synge and Griffiths) A hollow spherical ball of radius 5 cm is set spinning about a horizontal axis with an angular velocity of 10 rad s 1. It is then gently placed on a horizontal plane and released. If the coefficient of friction between the ball and the plane is 0.34, find the distance traversed by the ball before slipping ceases. [The moment of inertia of a spherical shell of mass m and radius r is ]. Solution. Initially slipping must occur, because the ball is not moving forward but it is rotating. Whilst slipping it is friction which (a) accelerates the translational motion from rest; (b) decelerates the rotation. Slipping ceases when v = rω, but until this point friction is maximal and given by. Linear motion Whilst slipping,. Hence, with v = 0 at t = 0. (i) Mechanics Topic D (Rotation) - 24 David Apsley

Rotational motion Whilst slipping,. Also,. Hence, with ω = ω 0 = 10 rad s 1 at t = 0. (ii) Slipping stops when v = rω. From (i) and (ii), this occurs when This distance travelled may be determined from the linear constant-acceleration formula, with Hence, u = 0, a = μg, Using consistent length units of metres: Answer: 6.0 mm. Mechanics Topic D (Rotation) - 25 David Apsley

Appendix 1: Moments of Inertia for Simple Shapes Many formulae are given in the textbooks of Meriam and Kraige or Gere and Timoshenko. Only some of the more common ones are summarised here. Geometric figures are assumed to have a uniform density and have a total mass M. Geometry Axis I Rod, length L (1) Through centre (2) End Rectangular lamina, sides L (perpendicular to axis) and W (1) In-plane; symmetry (2) Side (3) Perpendicular to plane; symmetry ) Triangular lamina, base B, altitude H Base Circular ring, radius R (1) Perpendicular to plane; symmetry (2) Diameter Circular disc, radius R (1) Perpendicular to plane; symmetry (2) Diameter Circular cylinder, radius R, height H. Symmetry axis Solid sphere (or hemisphere), radius R Any diameter Spherical (or hemispherical) shell, radius R Any diameter Moments of inertia for many different shapes or axes can be constructed from these by: use of the parallel-axis or perpendicular-axes rules; stretching parallel to an axis (without change of mass distribution); combination of fundamental elements. Mechanics Topic D (Rotation) - 26 David Apsley

Appendix 2: Second Moment of Area Second moment of area rather than second moment of mass appears in structural engineering (resistance to beam bending) and hydrostatics (pressure force). The formulae for second moments of area of plane figures are exactly the same as those in the table above except that mass M is replaced by area A. The same symbol (I) is used. Dependence on a length dimension parallel to the axis is often hidden inside M or A; e.g. second moment of area of a rectangular lamina about an in-plane symmetry axis: (since ) second moment of area of a triangular lamina about a side of length B: (since ) You will meet second moments of area a great deal in your Structures courses. Mechanics Topic D (Rotation) - 27 David Apsley

Numerical Answers to Examples in the Text Full worked answers are given in a separate document online. Example 1. 1.7510 3 rad s 1 Example 2. 21.8 Example 3. (a) 0.680; (b) 0.0746 Example 4. (a) 114 N; (b) 1.26 m; (c) 9.88 m s 1 Example 5. (a) ; (b) (c) ; (d) 48.2 ; 13.3 m; 11.4 m s 1 ; (e) 2.49 m Example 6. 4.7910 5 N m; 12600 N Example 7. No numerical answer Example 8. 3.1410 4 N m s; 52.4 N m Example 9. (a) 5 m; (b) 1.08 rad s 1 Example 10. (a) ; (b) 2.68 m s 2 ; (c) 3.27 m s 1 ; (d) 4.41 rev Example 11. 0.162 m Example 12. (a) 0.407 kg m 2 ; (b) 215 mm; (c) 1.32 N Example 13. 0.301 m Example 14. (a) ; (b) ; (c) ; (d) ; (b) Example 15. Mechanics Topic D (Rotation) - 28 David Apsley