Dynamics of Rotation Dynamics of Rotation Torque (τ) : The tendency or ability of a force to rotate an object about an axis, fulcrum, or pivot. Just as a force is a push or a pull, a torque can be thought of as a twist. F Axis τ=Fxr θ Moment Arm (r) Torque (τ) τ is the Cross Product of the Force vector & Moment Arm Line of Force Consider only the magnitude : τ = Fr sinθ , τ is in N-m (MKS) & Dyne-cm (CGS) Where F – applied force r – moment arm θ – angle between F & r Dynamics of Rotation Torque (τ) : Could go in either a clockwise (CW) or counter-clockwise (CCW) in direction F Axis CW Torque (τCW) θ Moment Arm (r) Line of Force Line of Force Axis CCW Torque (τCCW) θ Moment Arm (r) F Torque (τ) : F1 r1 F3 r2 F4 r1 F2 r2 θ1 θ2 τ2 = (F2)(r2) sinθ2 F1 τ1 = (F1)(r1) sinθ1 but θ2 = 90° τ2 = (F2)(r2) F2 Torque (Γ) : F1 r1 F3 r2 F2 F4 r2 F3 θ3 τ4 = (F4)(0) sinθ4 τ3 = (F3)(r2) sinθ3 but θ3 = 0° τ3 = 0 τ4 = 0 F4 Dynamics of Rotation MOMENT OF INERTIA (I) – It is the measure of the object’s resistance to changes in rotational speed about an axis. The greater the value of the mass moment of inertia, the smaller the angular acceleration about an axis (slower rotation) I of a point mass General form of I I = ∫r2 dm Axis m Moment Arm (r) I = mr2 Axis Moment Arm (r) I is in kg-m2 (MKS) & g-cm2 (CGS) Dynamics of Rotation Common Moments of Inertia I = cMR2 c is a fraction http://hyperphysics.phy-astr.gsu.edu Dynamics of Rotation Parallel Axes Theorem -The moment of inertia can be calculated if moved to another axis as long as that (new) axis is parallel to the (original) axis that passes through the center of mass and the distance between them is known. New Axis Old Axis d IP = ICM + Md2 Where: IP : Moment of Inertia about another axis that is parallel to the axis that goes through the center of mass. ICM : Moment of Inertia about an axis at the center of mass. M : Mass of the object d - distance between the center of mass and the point of the parallel axis. Newton’s Second Law in Rotational Motion a = Rα τ = FR sinθ for a point mass moving in a circle, F is always perpendicular to the Radius thus R τ = FR Fnet If F is the net force (Fnet) Applying NSLM Fnet = Ma , where a is the tangential acceleration , a = Rα τnet = (Ma)R = M(Rα)R τnet = MR2α But I = M MR2 (for a point mass) where I is really I = cMR2 τnet = Iα Rotation about a FIXED Axis A bucket of water with a mass of 20kg is suspended by a rope wrapped around a windlass in the form of a solid cylinder 0.4m in radius, also with a mass of 20kg. The cylinder is pivoted on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 20m to the water. N a) What is the tension in the rope while the bucket is falling? a = Rα b) With what speed does the bucket strike the water? c) What is the time of fall? R R M = 20 kg FBD : R = 0.4 m T Mg m = 20 kg T h = 20 m a mg Rotation about a FIXED Axis A bucket of water with a mass of 20kg is suspended by a rope wrapped around a windlass in the form of a solid cylinder 0.4m in radius, also with a mass of 20kg. The cylinder is pivoted on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 20m to the water. a) What is the tension in the rope while the bucket is falling? b) With what speed does the bucket strike the water? c) What is the time of fall? T NSLM ΣFy = ma ↑+ T – mg =−ma T =mg −ma >eq-1 a mg Rotation about a FIXED Axis A bucket of water with a mass of 20kg is suspended by a rope wrapped around a windlass in the form of a solid cylinder 0.4m in diameter, also with a mass of 20kg. The cylinder is pivoted on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 20m to the water. a) What is the tension in the rope while the bucket is falling? b) With what speed does the bucket strike the water? c) What is the time of fall? N a = Rα τnet + τ net = τT + τ N + τ Mg τ net = − TR + 0 + 0 τ net = − TR >eq-2 τ net = Iα + I = ½ MR2 (for solid cylinder) α = a/R τ net = − ½ MR2 (a/R) T τ net = − ½ MR a >eq-3 eq. 2 = ea. 3 R Mg R Rotation about a FIXED Axis A bucket of water with a mass of 20kg is suspended by a rope wrapped around a windlass in the form of a solid cylinder 0.4m in radius, also with a mass of 20kg. The cylinder is pivoted on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 20m to the water. a) What is the tension in the rope while the bucket is falling? b) With what speed does the bucket strike the water? c) What is the time of fall? −TR = − ½ MRa T = ½ Ma >eq. 4 a = (9.8m/s2) [20kg/(10kg + 20kg)] eq. 4 = ea. 1 a = 6.533 m/s2 ½ Ma = mg – ma a in ea. 4 ½ Ma + ma = mg T = ½ (20kg)(6.533 m/s2) a (½ M+m) = mg T = 65.33 N a = g [m/(0.5M + m)] Rotation about a FIXED Axis A bucket of water with a mass of 20kg is suspended by a rope wrapped around a windlass in the form of a solid cylinder 0.4m in diameter, also with a mass of 20kg. The cylinder is pivoted on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 20m to the water. a) What is the tension in the rope while the bucket is falling? b) With what speed does the bucket strike the water? c) What is the time of fall? From Kinematics VF2 = VO2 + 2ah a = 6.533 m/s2 h = 20 m VF2 = (0)2 + 2(6.533)(20) VF2 = 261.333 m2/s2 VF = 16.167 m/s VO = 0 Rotation about a FIXED Axis A bucket of water with a mass of 20kg is suspended by a rope wrapped around a windlass in the form of a solid cylinder 0.4m in radius, also with a mass of 20kg. The cylinder is pivoted on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 20m to the water. a) What is the tension in the rope while the bucket is falling? b) With what speed does the bucket strike the water? c) What is the time of fall? From Kinematics VF = VO + at a = 6.533 m/s2 t = (VF – VO)/ a t = (16.167 m/s – 0)/ 6.533 m/s2 t = 2.475 s VF = 16.167 m/s VO = 0 Work, Energy & Power in Rotational Motion F M R W = FS, but S = Rθ W = FRθ , but τ = FR W = τθ K = ½ MV2 ,but ω = V/R and V = ωR K = ½ M(ωR)2 = ½ MR2ω2 , but I = MR2 K=½ Iω2 Average Power (PAVE) PAVE = W/t P = (τθ)/t , but ωAVE = θ/t PAVE = τωAVE R θ S M Instantaneous Power P(t) = dW/dt P(t) = τ (dθ/dt) , but ω(t) = dθ/dt P(t) = τω(t) Rotation about a FIXED Axis A playground merry-go-round has a radius of 4.4 m and a moment of inertia of 245 kg-m2 and turns with negligible friction about a vertical axle through its center. A child applies a 25 N force tangentially to the edge of the merry-go-round for 20 seconds. a)If the merry-go-round is initially at rest what is the angular velocity after this 20 second interval? b) How much work did the child do on the merry-go-round? c) What is the average power supplied by the child? R = 4.4 m I = 245 kg-m2 F = 25 N t = 20 s Rotation about a FIXED Axis A playground merry-go-round has a radius of 4.4 m and a moment of inertia of 245 kg-m2 and turns with negligible friction about a vertical axle through its center. A child applies a 25 N force tangentially to the edge of the merry-go-round for 20 seconds. a)If the merry-go-round is initially at rest what is the angular velocity after this 20 second interval? Using NSLM b) How much work did the child do on the merry-go-round? c) What is the average power supplied by the child? τ net = Iα < eq -2 R = 4.4 m eq-1 = eq-2 2 I = 245 kg-mFrom Kinematics FR = Iα ω F = ω O + αt ω O = 0 t = 20 s α = ? Using Torque F = 25 N t = 20 s τ net = FnetR = (F – f)R = FR < eq -1 α = FR/I α = (25N)(4.4m)/(245 kg-m2) α = 0.45 rad/s2 ωF = 0 + (0.45 rad/s2)(20s) ωF = 9 rad/s Rotation about a FIXED Axis A playground merry-go-round has a radius of 4.4 m and a moment of inertia of 245 kg-m2 and turns with negligible friction about a vertical axle through its center. A child applies a 25 N force tangentially to the edge of the merry-go-round for 20 seconds. a)If the merry-go-round is initially at rest what is the angular velocity after this 20 second interval? b) How much work did the child do on the merry-go-round? c) What is the average power supplied by the child? R = 4.4 m Work From Kinematics I = 245 kg-m2 2 W=τθ Using Torque F = 25 N t = 20 s θ = ωOt + ½ αt θ = (0)(20s) + ½ (0.45rad/s2)(20s)2 θ = 90 rad τ net = FnetR = (F – f)R = FR τ = (25N)(4.4m) = 110 Nm W = (110 Nm)(90 rad) W = 9,900 J = 9.9 KJ Rotation about a FIXED Axis A playground merry-go-round has a radius of 4.4 m and a moment of inertia of 245 kg-m2 and turns with negligible friction about a vertical axle through its center. A child applies a 25 N force tangentially to the edge of the merry-go-round for 20 seconds. a)If the merry-go-round is initially at rest what is the angular velocity after this 20 second interval? b) How much work did the child do on the merry-go-round? c) What is the average power supplied by the child? R = 4.4 m From Average Power I = 245 kg-m2 F = 25 N t = 20 s PAVE = τ ω AVE Γ = 110 Nm θ = 90 rad ωAVE = θ/t ωAVE = 90 rad/20s = 4.5 rad/s PAVE = (110 Nm)(4.5 rad/s) PAVE = 495 Watts Rotation about a MOVING Axis A string wrapped several times around the rim of a small hoop with radius 0.08 m and a mass of 0.28 kg (I = MR2). If the free end of the string is held in place and the hoop is released from rest, calculate : a) The tension in the string while the hoop descends as the string unwinds; b) The time it takes the hoop to descend 0.5 m; c) The angular velocity of the rotating hoop after it has descended 0.5 m. Using Torque h = 0.5 m τ net = FnetR = (T – f)R = TR < eq -1 t=? Using NSLM T T ω=? τ net = Iα < eq -2 eq-1 = eq-2 TR = Iα T = Iα/R T = (MR2)(a/R)/R T = Ma < eq-3 Rotation about a MOVING Axis A string wrapped several times around the rim of a small hoop with radius 0.08 m and a mass of 0.28 kg (I = MR2). If the free end of the string is held in place and the hoop is released from rest, calculate : a) The tension in the string while the hoop descends as the string unwinds; b) The time it takes the hoop to descend 0.5 m; c) The angular velocity of the rotating hoop after it has descended 0.5 m. Using LCE KE1 + PE1 = KE2 + PE2 0 + MgH = ½ Iω 2 + Mg(H-h) 2 h = 0.5 m 0 + MgH = ½ Iω + MgH – Mgh 2 – Mgh 0 = ½ Iω t=? ½ Iω 2 = Mgh T T H ω=? ω 2 = 2Mgh/I ω 2 = 2Mgh/(MR2) ω 2 = 2gh/(R2) = 2(9.8)(0.5) / (0.08)2 = 1,531.25 ω = 39.13 rad/s Rotation about a MOVING Axis A string wrapped several times around the rim of a small hoop with radius 0.08 m and a mass of 0.28 kg (I = MR2). If the free end of the string is held in place and the hoop is released from rest, calculate : a) The tension in the string while the hoop descends as the string unwinds; b) The time it takes the hoop to descend 0.5 m; c) The angular velocity of the rotating hoop after it has descended 0.5 m. Using Kinematics ω F = ω O + αt T ω F = ω = 39.13 rad/s ω O = 0 h = 0.5 m T t=? ω=? ω F = ω O + αt 39.13 = 0 + (122.5)t α=? ω F2 = ω O2+ 2αθ (solve first for α) θ = s/r = h/r = 0.5 m / 0.08 m = 6.25 rad 1531.25 = 0 + 2α(6.25) α = 122.5 rad/s2 1531.25 = 12.5α t = 39.13 rad/s / (122.5 rad/s2) t = 0.319 s Rotation about a MOVING Axis A string wrapped several times around the rim of a small hoop with radius 0.08 m and a mass of 0.28 kg (I = MR2). If the free end of the string is held in place and the hoop is released from rest, calculate : a) The tension in the string while the hoop descends as the string unwinds; b) The time it takes the hoop to descend 0.5 m; c) The angular velocity of the rotating hoop after it has descended 0.5 m. Using Torque h = 0.5 m τ net = FnetR = (T – f)R = TR < eq -1 t=? Using NSLM T T ω=? τ net = Iα < eq -2 eq-1 = eq-2 TR = Iα T = Iα/R T = (MR2)(a/R)/R T = Ma < eq-3 a = αR a = (122.5 rad/s2)(0.08m) a = 9.8 m/s2 a = g = 9.8 m/s2 T = (0.28 kg)(9.8 m/s2) T = 2.744 N ROTATION ABOUT A MOVING AXIS : ROLLING OBJECT (Rolling without Slipping) MS = 1 kg RS = 0.2 m h = 0.6 m θ = 25° 1 2 3 s = 5.5 m 4 A sphere (as shown) starts from rest and rolls without slipping down an inclined surface that is 0.6 m above the ground. Determine : (a) Linear Acceleration of the sphere as it slides down the slope (a) & the frictional force (f) (b) V3 = Velocity at pt. 3 (c) The μ of the flat surface for it to stop at pt. 4 In order that there will be NO SLIPPING. Friction on the slope is NOT zero. f θ N wS= MSg τnet = ∑τ (CW+) = + τf τnet = + (f)(RS) :eq. 1 eq.1 = eq.2 fRS = +(2/5)aMSRS τnet = Iα (CW+) = +(2/5)MSRS2(a/RS) τnet = +(2/5)aMSRS :eq. 2 f = (0.4)aMS : eq.3 ROTATION ABOUT A MOVING AXIS : ROLLING OBJECT (Rolling without Slipping) MS = 1 kg RS = 0.2 m h = 0.6 m θ = 25° s = 5.5 m 1 2 3 4 A sphere (as shown) starts from rest and rolls without slipping down an inclined surface that is 0.6 m above the ground. Determine (a) Linear Acceleration of the sphere as it slides down the slope (a) & the frictional force (f) N ∑Fx = ma (+→) wSX – f = +Ms a wSX = (MSg)sinθ f = wSX – Msa : eq.4 a Consider the forces acting on the surface of slope f f eq.3 = eq.4 f wS= MSg N θ N wS= MSg θ θ wS= MSg wSX = (MSg)cosθ (0.4)aMS = wSX – MSa (1.4)aMS = wSX a = (MSgsinθ)/(1.4)MS ROTATION ABOUT A MOVING AXIS : ROLLING OBJECT (Rolling without Slipping) MS = 1 kg RS = 0.2 m h = 0.6 m θ = 25° s = 5.5 m 1 2 3 4 A sphere (as shown) starts from rest and rolls without slipping down an inclined surface that is 0.6 m above the ground. Use LCE and Rotational Dynamics to determine (a) Linear Acceleration of the sphere as it slides down the slope (a) & the frictional force (f) a a = (MSgsinθ)/(1.4)MS N f wSX = (MSg)sinθ a = (gsinθ)/(1.4) a = (9.8m/s2)(sin25°)/(1.4) wS= MSg θ wSX = (MSg)cosθ a = 2.958 m/s2 f = (0.4)aMS : using eq.3 f = 0.4(2.958 m/s2)(1kg) f = 1.1832 N ROTATION ABOUT A MOVING AXIS : ROLLING OBJECT (Rolling without Slipping) MS = 1 kg RS = 0.2 m h = 0.6 m θ = 25° s = 5.5 m 1 2 3 4 A sphere (as shown) starts from rest and rolls without slipping down an inclined surface that is 0.6 m above the ground. Determine (b) V3 = Velocity at pt.3 WO = (K3 – K1) + (U3 – U1) K1 = 0, since v1 = 0 (from rest) U1 = MSgh = (1kg)(9.8m/s2)(0.6m) = 5.88 J K3 = KLinear + KRotational = ½MSv32 + ½ Iω2 = 0.5MSv32 + 0.5[(2/5)MSRS2][v3/RS]2 U3 = 0, since h3 = 0 K3 = 0.5MSv32 + 0.2MSv32 = 0.7MSv32 = 0.7(1kg)v32 S1 = ? WO = WF + Wf = 0 + (– f)(s1) = – f (h/sinθ) WO = (– 1.1832N)(0.6m/sin25°) = – 1.678 J h = 0.6 m θ = 25° ROTATION ABOUT A MOVING AXIS : ROLLING OBJECT (Rolling without Slipping) MS = 1 kg RS = 0.2 m h = 0.6 m θ = 25° s = 5.5 m 1 2 3 4 A sphere (as shown) starts from rest and rolls without slipping down an inclined surface that is 0.6 m above the ground. Determine (b) V3 = Velocity at pt.3 WO = (K3 – K1) + (U3 – U1) – 1.678 J = [(0.7)(1kg)(v32) – 0] + (0 – 5.88J) v32 = 6 m2/s2 v3 = 2.45 m/s ROTATION ABOUT A MOVING AXIS : ROLLING OBJECT (Rolling without Slipping) MS = 1 kg RS = 0.2 m ΣFy = 0 (at flat surface) N –WS = 0 h = 0.6 m N = WS = MSg θ = 25° f2 = ? & μ = ? s = 5.5 m 1 2 3 4 A sphere (as shown) starts from rest and rolls without slipping down an inclined surface that is 0.6 m above the ground. Determine (c) The μ of the flat surface for it to stop at pt.4 WO = (K4 – K3) + (U4 – U3) K3 = 0.7 MSv32 = 0.7(1kg)(2.45m/s)2 = 4.2 J K4 = 0, since v4 = 0 (sphere stops) U3 = 0 , since h3 = 0 U4 = 0 , since h4 = 0 WO = WF + Wf = 0 + (– f2)(s) = (–μN)s WO = –μMSgs = –μ(1kg)(9.8m/s2)(5.5m) WO = –μ(53.9 J) WO = (K4 – K3) + (U4 – U3) – μ(53.9 J) = [ 0 – 4.2 J] + [0 – 0] μ = 4.2 J /53.9 J μ = 0.0779 Impulse & Momentum in Rotational Motion Angular Momentum (L) p = MV ,but ω = V/R and V = ωR p = MωR L = pR = MωR(R) = MωR2 but I = MR2 L = Iω M R Angular Impulse (Jθ) Jθ = τ Δt = ΔL Jθ = τ Δt = Iω2 – Iω1 Conservation of Angular Momentum “When two bodies interact with each other and nothing else (isolated system), their total angular momentum is conserved or constant” ΣLbefore = ΣLafter Iaωa + Ibωb = (Ia + Ib)ω V Impulse & Momentum in Rotational Motion A solid wood door 1 m wide and 2 m high is hinged along one side (2m side) and has a total mass of 35 kg. Initially open and at rest, the door is struck with a hammer at its center. During the blow, an average force of 2,000 N acts for 5 ms. Find the angular velocity of the door after the impact F = 2,000 N R ωO = 0 F = 2,000 N Δt = 5x10-3 s R ωA= ? Angular Impulse (Jθ) Jθ = τ Δt = Iω2 – Iω1 τ = FR = (2,000N)[½ (1 m)] τ = 1,000 Nm Jθ = (1,000 Nm)(5x10-3s) = 5 Nms Impulse & Momentum in Rotational Motion b b a a Inertia of a rectangular plane – axis at its geometric center I= (1 / 12 )M(a2 + b2) Inertia of a rectangular plane – axis at one edge I = (1/3)Ma2 Impulse & Momentum in Rotational Motion A solid wood door 1 m wide and 2 m high is hinged along one side (2m side) and has a total mass of 35 kg. Initially open and at rest, the door is struck with a hammer at its center. During the blow, an average force of 2,000 N acts for 5 ms. Find the angular velocity of the door after the impact M = 35 kg F = 2,000 N R R ωA= ? ωO = 0 F = 2,000 N Δt = 5x10-3 s Jθ = 5 Nms = Iω2 – Iω1 I = (1/3)Ma2 I = (1/3)(35 kg)(1 m)2 I = 11.667 kg-m2 ω1 = ωO = 0 ω2 = ωA = ? Impulse & Momentum in Rotational Motion A solid wood door 1 m wide and 2 m high is hinged along one side (2m side) and has a total mass of 35 kg. Initially open and at rest, the door is struck with a hammer at its center. During the blow, an average force of 2,000 N acts for 5 ms. Find the angular velocity of the door after the impact M = 35 kg F = 2,000 N R R ωO = 0 F = 2,000 N Δt = 5x10-3 s 5 Nms = IωA – IωO 5 Nms = (11.667 kg-m2)ωA ωA = 5 Nms / 11.667 kg-m2 ωA = 0.429 rad/s ωA= ? Conservation of Angular Momentum A large turntable rotates about a fixed vertical axis, making 1 revolution in 6 seconds. The moment of inertia about this axis is 1,200 kg-m2. A man with a mass of 80 kg, then steps into the turntable and stops at a 2m distance from the axis. What is the new angular velocity of the turntable due to the presence of the man at that distance? (Treat the man as a point mass when solving for his inertia) ωO = 1 rev / 6 sec = 0.1667 rev/s ωN = ? r=2m M = 80 kg I T = 1,200 kg-m2 I T = 1,200 kg-m2 ωO = (0.1667 rev/s) (2π rad / 1 rev) = 1.047 rad/s I M = MR2 = (80kg)(2m)2 = 320 kg-m2 Conservation of Angular Momentum ωO = 1 rev / 6 sec = 0.1667 rev/s ωN = ? I M = 320 kg-m2 r=2m ωO = 1.047 rad/s M = 80 kg I T = 1,200 kg-m2 I T = 1,200 kg-m2 Lbefore = Lafter Iaωa + Ibωb = (Ia + Ib)ω ITωOT + IMωOM = (IT + IM)ωN (1,200)(1.047) + 0 = (1,200+320)ωN 1,256.4 kg-m2/s= (1,520 kg-m2)ωN ωN = (1,256.4 kg-m2/s) / (1,520 kg-m2) ωN = 0.827 rad/s A circular disc rotates on a thin air film with a period of 0.3s. Its moment of inertia about its axis of rotation is 0.06 kg m2. A small mass is dropped onto the disc and rotates with it. The moment of inertia of the mass about the axis of rotation is 0.04 kg m2. Determine the final period of the rotating disc and mass T1 = 0.3 s T2 = ? IM = 0.04 kg-m2 M ID = 0.06 kg-m2 ID = 0.06 kg-m2 I1 ω1 = I2 ω2 ID ω1 = (ID + IM)ω2 (0.06)(2π/0.3) = (0.06 + 0.04)( 2π/T2) (0.06/0.3) = (0.1/T2) 0.2 = 0.1/T2 T2 = 0.5 sec Given : BEFORE ω2 = 20 rad/s ω=? AFTER ω1 = 20 rad/s M2 = 5 kg M1 = 2 kg R2 = 0.6 m R1 = 0.2 m ω =−15.294 /s ΣLbefore = ΣLafter + I1ω1 + I2ω2 =(I1+I2)ω (−52kg·m2/s) = (3.4kg·m2)ω ω = (−52kg·m2/s)/(3.4kg·m2) I = ½ MR2 =−15.294 rad/s ω =15.2954 rad/s ,CCW ½M1R12ω1 + ½M2R22ω2 =(½M1R12+ ½M2R22)ω ½(M1R12ω1 +M2R22ω2) = ½(M1R12+ M2R22)ω M1R12ω1 +M2R22ω2 = (M1R12+ M2R22)ω (2kg)(0.2m)2(+20rad/s) +(5kg)(0.6m) 2(−20rad/s) = [(2kg)(0.2m)2+(5kg)(0.6m)2]ω (8kg·m2/s)+(−60kg·m2/s) = (3.4kg·m2)ω SECOND CONDITION OF EQUILIBRIUM Recall : FIRST CONDITION : “At Equilibrium the vector sum of a force, must be zero” ΣFx = 0 , ΣFy = 0 , Fnet = R = 0 SECOND CONDITION : “The sum of all Torques of all forces acting on the body with respect to ANY specified axis, must be zero” Στ = 0 (about any axis) , τnet = 0 Or Total CCW Torque equals Total CW Torque ΣτCCW = ΣτCW SECOND CONDITION OF EQUILIBRIUM CENTER OF GRAVITY (C.G.) This is the point on the body that the entire force of gravity is concentrated. For a body of definite shape, it is found at its geometric center. For a composite body the center of gravity can be calculated using torques. Center of Gravity of Common Shapes Circle Semi-Circle Y-axis Y-axis XCG = r YCG = r XCG = r (0.0) X-axis (0.0) YCG = 4r/3π = 0.424 r X-axis Note : All shapes are made from the same material and uniform thickness Center of Gravity of Common Shapes Rectangles Square Y-axis Height (H) Y-axis XCG = B/2 side (s) YCG = s/2 YCG = H/2 (0.0) X-axis Base (B) XCG = s/2 (0.0) X-axis side (s) Note : All shapes are made from the same material and uniform thickness Center of Gravity of Common Shapes Right Triangle Triangles Isosceles and Equilateral Triangle Y-axis Height (H) Y-axis XCG = B/3 Height (H) YCG = H/3 (0.0) YCG = H/3 X-axis Base (B) XCG = B/2 (0.0) X-axis Base (B) Note : All shapes are made from the same material and uniform thickness Center of Gravity of Common Shapes Generally all CG of triangles are determined by ½c ½A C A C.G ½B B Note : All shapes are made from the same material and uniform thickness Center of Gravity of Common Shapes The Center of Gravity is NOT always INSIDE the object Ring Y-axis XCG = r YCG = r (0.0) X-axis Note : All shapes are made from the same material and uniform thickness Center of Gravity of Composite Shapes H= 9.5 c m Given : Find the CG of this figure of uniform density & thickness. B = 10 c m Center of Gravity of Composite Shapes 1st Step : Place the figure in an x-y plane, and identify the basic shape it contains. Use the formulas given earlier to determine the INDIVIDUAL C.G.’s Y-axis X1 = r Y2 = H/3 H= 9.5 c m X2 = B/2 Y1 = 0.424r B = 10 c m (0.0) X-axis Center of Gravity of Composite Shapes 2nd Step : Make a tabulation for your measurements and calculations. The number of rows depends on the number of basic shapes Y-axis Shape Semicircle Triangle Total Area xCG from origin yCG from origin Ax Ay Center of Gravity of Composite Shapes 3rd Step : Compute for the individual areas and also the total area Y-axis Shape Area Semicircle A1 Triangle A2 Total xCG from origin yCG from origin Ax Atotal A1 = ½ πr2 : Area of a Semicircle A2 = ½BH : Area of a Triangle A1 = ½ π(5cm)2 A2= ½ (10cm)(9.5cm) A1 = 39.27 cm2 A2 = 47.5 cm2 AT = 39.27 cm2 + 47.5 cm2 AT = 86.77 cm2 Ay Center of Gravity of Composite Shapes 4th Step : Measure the x-coordinate of the individual shape FROM the origin Y-axis Shape Area xCG from origin Semicircle 39.27 X1 Triangle 47.5 X2 Total 86.77 x1 = ½ B = ½(10 cm) = 5 cm x2 = ½ B = 5 cm yCG from origin Ax Ay Center of Gravity of Composite Shapes 5th Step : Measure the y-coordinate of the individual shape FROM the origin Y-axis Shape Area xCG from origin yCG from origin Semicircle 39.27 5 Y1 Triangle 47.5 5 Y2 Total 86.77 y1 = (4r/3π) + H = (0.424r)+H y1 = 0.424(5 cm) + 9.5 cm = 11.62 cm y2 = (2/3) H = (2/3)(9.5cm) = 6.33 cm Ax Ay Center of Gravity of Composite Shapes 6th Step : Compute the area “torque” Y-axis Shape Area xCG from origin yCG from origin Ax Semicircle 39.27 5 11.62 A1x1 A1y1 Triangle 47.5 5 6.33 A2x2 A2y2 Total 86.77 A1x1 = (39.27cm2)(5cm) = 196.35 cm3 A1y1 = (39.27cm2)(11.62cm) = 456.32 cm3 A2x2 = (47.5cm2)(5cm) = 237.5 cm3 A2y2 = (47.5cm2)(6.33cm) = 300.675 cm3 Ay Center of Gravity of Composite Shapes 7th Step : Compute the total “torque” per axis Y-axis Shape Area xCG from origin yCG from origin Ax Semicircle 39.27 5 11.62 196.35 456.32 Triangle 47.5 5 6.33 237.5 300.675 Total 86.77 ΣAx ΣAy ΣAx = 196.35 cm3 + 237.5 cm3 = 433.85 cm3 ΣAy = 456.32 cm3 + 300.675 cm3 = 757 cm3 Ay Center of Gravity of Composite Shapes 8th Step : Compute the composite’s center of gravity coordinates: Y-axis Shape Area xCG from origin yCG from origin Ax Ay Semicircle 39.27 5 11.62 196.35 456.32 Triangle 47.5 5 6.33 237.5 300.675 Total 86.77 433.85 757 XCG = ΣAx/Atotal = ( 433.85 cm3 ) / (86.77cm2) = 5 cm YCG = ΣAy/Atotal = ( 757 cm3 ) / (86.77cm2) = 8.724 cm Center of Gravity of Composite Shapes last Step : Locate the C.G. on your composite figure XCG = 5 cm YCG = 8.724 cm (0.0) Center of Gravity of Composite Shapes r = 1 cm H= 6 c m 3cm Given : Find the CG of this figure of uniform density & thickness. B= 4cm Center of Gravity of Composite Shapes 1st Step : Place the figure in an x-y plane, and identify the basic shape it contains. Use the formulas given earlier to determine the INDIVIDUAL C.G.’s Let : Object 1 : Rectangle X 2= B/2 X 1= B/2 B= 4cm (0.0) r = 1 cm r = 1 cm H= 6 c m 3cm Object 2 : Circle (Hole) Y 2= r Y 1= H/2 X-axis Center of Gravity of Composite Shapes 2nd Step : Make a tabulation for your measurements and calculations. The number of rows depends on the number of basic shapes Y-axis Shape Rectangle Circle Total Area xCG from origin yCG from origin Ax Ay Center of Gravity of Composite Shapes 3rd Step : Compute for the individual areas and also the total area Y-axis Shape Area Rectangle A1 Circle A2 Total xCG from origin yCG from origin Ax Atotal A1 = BH : Area of a rectangle A2 = πr2 : Area of a Circle A1 = (4cm)(6cm) A2= π(1cm)2 A1 = 24 cm2 A2 = 3.1416 cm2 AT = 24 cm2 + ( – 3.1416 cm2) AT = 20.858 cm2 Ay Center of Gravity of Composite Shapes 4th Step : Measure the x-coordinate of the individual shape FROM the origin Y-axis Shape Area xCG from origin Rectangle 24 X1 Cicle -3.1416 X2 Total 20.858 x1 = ½ B = ½(4 cm) = 2 cm x2 = ½ B = 2 cm yCG from origin Ax Ay Center of Gravity of Composite Shapes 5th Step : Measure the y-coordinate of the individual shape FROM the origin Y-axis Shape Area xCG from origin yCG from origin Rectangle 24 2 Y1 Cicle -3.1416 2 Y2 Total 20.858 y1 = H/2 = (6cm/2) = 3 cm y2 = (H/2) + r = 3 cm + 1 cm = 4 cm Ax Ay Center of Gravity of Composite Shapes 6th Step : Compute the area “torque” Y-axis Shape Area xCG from origin yCG from origin Ax Rectangle 24 2 3 A1x1 A1y1 Cicle -3.1416 2 4 A2x2 A2y2 Total 20.858 A1x1 = (24cm2)(2cm) = 48 cm3 A1y1 = (24cm2)(3cm) = 72 cm3 A2x2 = (-3.1416cm2)(2cm) = – 6.2832 cm3 A2y2 = (-3.1416cm2)(4cm) = 300.675 – 12.5664 cm3 Ay Center of Gravity of Composite Shapes 7th Step : Compute the total “torque” per axis Y-axis Shape Area xCG from origin yCG from origin Ax Rectangle 24 2 3 48 72 Cicle -3.1416 2 4 – 6.2832 – 12.5664 Total 20.858 ΣAx ΣAy ΣAx = 48 cm3 – 6.2832 cm3 = 41.7168 cm3 ΣAy = 72 cm3 – 12.5664 cm3 = 59.4336 cm3 Ay Center of Gravity of Composite Shapes 8th Step : Compute the composite’s center of gravity coordinates: Y-axis Shape Area xCG from origin yCG from origin Ax Ay Rectangle 24 2 3 48 72 Cicle -3.1416 2 4 – 6.2832 – 12.5664 Total 20.858 41.7168 59.4336 XCG = ΣAx/Atotal = ( 41.7168cm3 ) / (20.858cm2) = 2 cm YCG = ΣAy/Atotal = ( 59.4336 cm3 ) / (20.858cm2) = 2.85 cm Center of Gravity of Composite Shapes last Step : Locate the C.G. on your composite figure r = 1 cm H= 6 c m 3cm B = 3 cm XCG YCG (0.0) B= 4cm XCG = 2 cm YCG = 2.85 cm SECOND CONDITION OF EQUILIBRIUM Calculation of Center of Gravity using Weights ΣτX = w1x1 + w2x2 + . . . . ΣτX = Σwx ΣτY = w1y1 + w2y2 + . . . . ΣτY = Σwy XCG = Σwx / ΣW YCG = Σwy / ΣW Where w1 , w2 - weight of individual element x1, x2 – lever arm of individual element with respect to x-axis y1, y2 – lever arm of individual element with respect to y-axis ΣW – total weight of composite body X,Y - coordinate of the center of gravity For simplicity : One of the axes can be graphically set to zero, IF the composite body is PURELY horizontal or vertical in structure, such that we only need to solve for one coordinate SECOND CONDITION OF EQUILIBRIUM Calculation of Center of Gravity using Weights Given : A 14 kg beam is hinged at one end. A 6 kg triangular object and a 7.5 kg object are positioned as shown. Dots indicate the individual centers of gravity of the beam and the two objects. What is the distance from the axis of rotation to the center of gravity for this system? M2 = 7.5 kg M1 = 6 kg MB = 14 kg Solution : Since the body is purely horizontal. We can set the Y coordinate for the C.G. to zero and solve only for X coordinate. Place the composite body at the center of the Y= 0 and one end of it at X=0 : Hence YCG = 0 SECOND CONDITION OF EQUILIBRIUM Calculation of Center of Gravity using Weights M2 = 7.5 kg M1 = 6 kg MB = 14 kg Using the origin as axis x1 = 0.36m x2 =0.36 m + 0.53m = 0.89m x3 = 0.89m + 1.92m = 2.81m Weights : w1 = m1g = (6kg)(9.8m/s2) = 58.8 N wB = mBg = (14kg)(9.8m/s2) = 137.2 N w2 = m2g = (7.5kg)(9.8m/s2) = 73.5 N WT = w1 + wB + w2 WT = 58.8N + 137.2N + 73.5N = 269.5 N Στx = Σwx = w1x1 + wBx2 + w2x3 Σwx = (58.8N)(0.36m)+(137.2N)(0.89m)+(73.5N)(2.81m) Σwx = 349.811 Nm XCG = Σwx / W = (349.811Nm) / (269.5N) XCG = 1.298 m SECOND CONDITION OF EQUILIBRIUM Calculation of Center of Gravity using Weights Given : A 14 kg beam is hinged at one end. A 6 kg triangular object and a 7.5 kg object are positioned as shown. Dots indicate the individual centers of gravity of the beam and the two objects. What is the distance from the axis of rotation to the center of gravity for this system? M2 = 7.5 kg M1 = 6 kg MB = 14 kg XCG = 1.298m SECOND CONDITION OF EQUILIBRIUM Given : An 80 kg man balances (system is at equilibrium) the boy on a teeter-totter (similar to a seesaw) as shown. Note : Ignore the weight of the board they are standing on. a)What is the approximate mass of the boy b)What, approximately, is the magnitude of the downward force exerted on the fulcrum? m=? M = 80 kg 4m fulcrum 1m SECOND CONDITION OF EQUILIBRIUM Given : An 80 kg man balances (system is at equilibrium) the boy on a teeter-totter (similar to a seesaw) as shown. Note : Ignore the weight of the board they are standing on. a)What is the approximate mass of the boy b)What, approximately, is the magnitude of the downward force exerted on the fulcrum? m=? M = 80 kg L1 =4 m fulcrum L2 =1 m a) Using the fulcrum as axis ΣτCCW = ΣτCW mgL1=MgL2 mg(4m) =(80kg)g(1m) m = 20 kg SECOND CONDITION OF EQUILIBRIUM Given : An 80 kg man balances (system is at equilibrium) the boy on a teeter-totter (similar to a seesaw) as shown. Note : Ignore the weight of the board they are standing on. a)What is the approximate mass of the boy b)What, approximately, is the magnitude of the downward force exerted on the fulcrum? F=? m=? M = 80 kg L1 =4 m fulcrum L2 =1 m WB WM b) Using the first condition of equilibrium ΣFY = 0 +↑ – F – WBm – WM = 0 F = – (WB + WM) = – (mg + Mg) = –[(20 kg)(9.8m/s2)+(80kg)(9.8m/s2)] = – 980N Considering only the magnitude |F | = 980N SECOND CONDITION OF EQUILIBRIUM SCE on the ladder using red dot ( pt.A) as axis A ladder 10m long and has a weight of 400N, with the center of gravity at its center. It leans in equilibrium against a vertical frictionless wall. Dimensions are provided in the figure. Find the magnitudes forces C, N & f. ΣτA CCW = ΣτA CW ΣτA CCW = (C)(8m) = C(8m) ΣτA CW = (400N)(3m) = 1,200 Nm ΣτA CCW = ΣτA CW C(8m) = 1,200 Nm C 3m C = (1,200 Nm)/(8m) C = 150 N 8m W = 400N f 6m N SECOND CONDITION OF EQUILIBRIUM SCE on the wall corner using blue dot ( pt.B) as axis ΣτB CCW = ΣτB CW A ladder 10m long and has a weight of 400N, with the center of gravity at its center. It leans in ΣτB CCW = (C)(8m) + (400N)(3m) equilibrium against a vertical frictionless wall. ΣτB CCW = (150N)(8m) + 1200Nm = 2400 Nm Dimensions are provided in the figure. Find the magnitudes forces C, N & f. ΣτB CW = (N)(6m) = N(6m) ΣτB CCW = ΣτB CW 2400 Nm = N(6m) 3m N= 400 N C 3m SCE on the wall and ladder contact point using green dot ( pt.D) as axis ΣτD CCW = ΣτD CW 8m ΣτD CCW = (400N)(3m) + (f)(8m) ΣτD CCW = 1200Nm + f(8m) ΣτD CW = (N)(6m) = (400N)(6m) = 2400Nm W = 400N f 6m N ΣτD CCW = ΣτD CW 1200 Nm + f(8m) = 2400 Nm f = 150 N SECOND CONDITION OF EQUILIBRIUMAlternative Solution : for f & N SCE on the ladder using red dot ( pt.A) as axis A ladder 10m long and has a weight of 400N, with the center of gravity at its center. It leans in equilibrium against a vertical frictionless wall. Dimensions are provided in the figure. Find the magnitudes forces C, N & f. ΣτA CCW = ΣτA CW ΣτA CCW = (C)(8m) = C(8m) ΣτA CW = (400N)(3m) ΣτA CW = 1,200 Nm ΣτA CCW = ΣτA CW C(8m) = 1,200 Nm C 3m C = (1,200 Nm)/(8m) C = 150 N 8m W = 400N f 6m N Using FCE : ΣFx = 0 & ΣFy = 0 ΣFx = 0 (+ ) +f – C = 0 f=C ΣFx = 0 (+ ↑) +N – W = 0 N=W f = 150 N N = 400 N SECOND CONDITION OF EQUILIBRIUM ΣτA CCW = ΣτA CW Two men are carrying a uniform ladder that is 5m long & weighs 100N. If one man applies an upward (lifting) force equal to 42N at one end. For the system to be at equilibrium: a) at what point should the other man lift? b) what should be the magnitude of F? (100N)(x) = (42N)(2.5m+x) 5m b) SCE on the ladder using blue dot ( pt.B) as axis 2.5 m 42N 100N x (100N)X = 105Nm + (42N)X (58N)X = 105 Nm X = 1.81 m (from the center) ΣτB CCW = ΣτB CW F ΣτB CCW = (F)(2.5m+x) ΣτB CCW = (F)(2.5m+1.81m) = F(4.31m) A B a) SCE on the ladder using red dot ( pt.A) as axis ΣτA CCW = ΣτA CW ΣτA CCW = (100N)(x) ΣτA CW = (42N)(2.5m+x) ΣτB CW = (100N)(2.5m) = 250 Nm ΣτB CCW = ΣτB CW F(4.31m) = 250 Nm F = 58 N SECOND CONDITION OF EQUILIBRIUM a) SCE on the beam using red dot ( pt.A) as axis ΣτA CCW = ΣτA CW The horizontal beam weighs 200N and its C.G. is at its center. Find: a. The tension of the cable b. The horizontal (C) & vertical (f) components of the forces exerted on the beam by the wall. ΣτA CCW = (T)(4m)sin θ ΣτA CCW = (T)(4m)sin 36.87° ΣτA CCW = T(2.4m) ΣτA CW = (200N)(2m) + (300 N)(4m) FBD ΣτA CW = 1,600 Nm T T 3m 3m ΣτA CCW = ΣτA CW 5m 5m T(2.4m) = 1,600 Nm T = 666.667 N θ C 2m 4m 2m f 300N 200N 300N 5m 3m θ = tan -1 ( 3m/4m) = 36.87° θ 4m SECOND CONDITION OF EQUILIBRIUM b) SCE on the beam using blue dot ( pt.B) as axis The horizontal beam weighs 200N and its C.G. is at ΣτB CCW = ΣτB CW its center. Find: ΣτB CCW = (200N)(2m) = 400 Nm a. The tension of the cable ΣτB CW = (f)(4m) = f(4m) b. The horizontal (C) & vertical (f) components of the forces exerted on the beam by the wall. Στ = Στ B CCW 400Nm = f(4m) FBD f = 100 N T 5m 5m We have no more valid axis. Hence we use FCE on the entire beam to get the value of C T 3m 3m Tx C 2m 4m Using FCE : ΣFx = 0 & ΣFy = 0 θ 2m f 300N 200N 300N ΣFx = 0 (+ ) +C – Tx = 0 C = Tx = Tcosθ C = (666.667N)( cos 36.87°) 5m 3m θ = tan -1 ( 3m/4m) = 36.87° θ 4m B CW C = 533.333 N SECOND CONDITION OF EQUILIBRIUM Alternative Solution for friction (f) using FCE The horizontal beam weighs 200N and its C.G. is at its center. Find: a. The tension of the cable b. The horizontal (C) & vertical (f) components of the forces exerted on the beam by the wall. Using FCE : ΣFx = 0 & ΣFy = 0 ΣFy = 0 (+ ) +f – 200N – 300N + Ty = 0 f = 500N – Ty = 500N – Tsinθ FBD T 5m 5m f = 100 N T 3m 3m f = 500N – (666.667N)( sin 36.87°) Ty θ C 2m 4m 2m f 300N 200N 300N 5m 3m θ = tan -1 ( 3m/4m) = 36.87° θ 4m COUPLES Two forces with equal magnitude and opposite direction with a line of action that are parallel but does not coincide. F 0 X1 L F X2 τCouple = FL Στo + Στo = −Fx1+ Fx2 , But x2 = x1 + L Στo = −Fx1 + F(x1 + L) Στo = −Fx1 + Fx1 + FL Στo = + FL Στo = FL (CW) Couple principle : At any point at the plane the couple produces the SAME torque. COUPLES 1. Two equal parallel forces F1 = F2 = 5N are applied to a rod as shown : a. Calculate the net torque about pt. O due to these two forces by calculating the torque of each separate force. b. Calculate the net torque about pt. P due to these two forces by calculating the torque of each separate force. c. Compare your results using the couple equation. F1 F2 3m 37° 37° O P 1.5m 1.5m a) Getting net torque on the beam using green dot( pt. O) as axis ΣτO = τO net + ΣτO = -(F1)(3m)sin(180°-37°) + (F2)(3m+1.5m) sin(180°-37°) ΣτO = -(5N)(1.8054m) + (5N)(2.7082m) τO net = + 4.514 Nm = 4.514 Nm , CW COUPLES 1. Two equal parallel forces F1 = F2 = 5N are applied to a rod as shown : a. Calculate the net torque about pt. O due to these two forces by calculating the torque of each separate force. b. Calculate the net torque about pt. P due to these two forces by calculating the torque of each separate force. c. Compare your results using the couple equation. F1 F2 3m 37° 37° O P 1.5m 1.5m b) Getting net torque on the beam using red dot( pt. P) as axis ΣτP = τP net + ΣτP = +(F1)(3m)sin(37°) – (F2)(1.5m) sin(37°) ΣτP = +(5N)(1.8054m) – (5N)(0.9027m) τP net = + 4.514 Nm = 4.514 Nm , CW COUPLES 1. Two equal parallel forces F1 = F2 = 5N are applied to a rod as shown : a. Calculate the net torque about pt. O due to these two forces by calculating the torque of each separate force. b. Calculate the net torque about pt. P due to these two forces by calculating the torque of each separate force. c. Compare your results using the couple equation. F1 F2 3m 37° 37° 37° O P 1.5m sin 37° = L/1.5 m L = 1.5m(sin 37°) 1.5 m L = 0.9027 m 1.5m c) Using couple equation τCouple = FL τCouple = (5N)(0.9027m) τCouple = 4.514Nm, CW All the values are the same. Thus proving the couple principle. That at any point at the plane the couple produces the SAME torque.
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