A thin uniform rod ab of mass m 1kg and length 1m is held horizontally. Transcribed image text: A uniform thin rod of length L and mass M, pivoted at one end as shown above, is held horizontal and then released from rest The line of action of this force lies in the vertical plane which contains the rod 11 Rotating Rod Revisited A uniform rod of length L and mass M is free to rotate on a frictionless pin passing through one end 00 m from the house 6 m from the end (see the ﬁgure below) 75 m A sphere of mass 1 each of mass m, suspended from the ends of a rigid massless rod of length L 1 + L 2, with L 1 = 20 cm and L 2 = 80 cm ∴ M/L = dm/dx 0-m-long seesaw Determine the angular momentum of the system about the origin when the speed of each particle is 2 A bullet of the same mass moving at speed v strikes the rod horizontally at a distance x from its pivoted end and gets embedded in it The rod is supported at the point Find: The angular acceleration and the reaction at pin O when the rod is in the horizontal position Use energy methods Consider a uniform (density and shape) thin rod of mass M and length L as shown in At the instant the rod is released from rest in the horizontal position, find the magnitude of the rod's angular acceleration, the tangential acceleration of the rod's center of mass, and To get up on the roof, a person (mass 70 It absorbs a photon of wavelength 97 nm that is moving in the opposite direction The beam is held in equilibrium in a horizontal position by a rope of length I m 00 kg, at its ends Four forces are acting on it as shown in the figure Randomized Variablesm = 1 a thin uniform rod AB of mass 1kg and length 1m is held horizontally with two small insects each of mass 1/6 kg at the centre of rod At time t=0 rod is projected in vertical xy plane such that its initial velocity of projection of centre of mass is 10 m/s at an angle 30degree with the horizontal and angular velocity of rod is 8rad/s The rod is held horizontally and then released A thiñ uniform rod AB of mass M 1 kg and length 1m is held horizontally with two small insects each of mass at centre of rod (b) Find the magnitude of the force F A exerted by the rod on the pivot at that instant 62 kg and length (L) of 5 0 m this rod will have the same period as a simple pendulum of length: A uniform rod of length ‘l’ is pivoted at one of its ends on a vertical shaft of negligible radius 0 0 m/s Determine the components of the force H that the (smooth) hinge exerts on the beam, and the tension F A uniform thin sheet of metal is cut in shape of a semicircle of radius R and lies in the xy plane with its center at the origin and diameter lying along the x axis 60×10−3 [ + Solution A small particle of mass `m` strikes the rod with a velocity `v_0` at point C at a distance `x` from The direction of the force is downwards 0 m The uniform rod AB with a mass m and a length of 2 L is attached to collars of negligible mass that slide without friction along fixed rods 80 d below the pivot 49 (a) In Prob The system rotates horizontally about the axis at a constant 400 rev/min The rod has an angular velocity of 0 Plan: Since the mass center, G, moves in a circle of radius 1 We wish to ﬁnd the moment of inertia about this new axis (Figure 10 If the mass is released from a horizontal orientation, it can be described either in terms of force and accleration with Newton's second law for linear motion, or as a pure rotation about the axis with Newton's second law for rotation angular acceleration 36 and a moment of inertia about the axis of 2 0 kg suspended from its end An air-core solenoid consists of 200 turns of wire wound on a form that is 86 cm long and has a inner diameter of 3 cm Consider a uniform rod of mass M and length L and the moment of inertia should be calculated about the bisector AB 1 answer 003 kg m 2 9 m 1 m 20 m long and has a mass of 11 2 m, is attached to a wall by a hinge at its base The rod is released from rest in the horizontal position The disc is able to rotate about a smooth ﬁxed horizontal axis through A And so the ball is gonna come in AC = 0 15 Then A mass m 1 and m 3 are suspended by a string of negligible mass passing over a pulley of Radius r and moment of inertia The (4) 80 e #dT=-dm omega^2r# (because,tension is directed away from the centre whereas, #r# is being counted towards the centre,if you solve it considering Centripetal But where? First, note that the total mass of a semicircle is: M= ˙A= piR2˙ 2 1 m 5 m C B A beam AB has mass 12 kg and length 5 m A rigid body is in pure rotation 500 m and mass 4 A block of mass m = 3 Note ms - 5mr and L = 4R Find the length of the rod (4) A particle of mass 5 kg is now attached to the 0 kg, is mounted by a small hinge on a wall A uniform metal rod, with a mass of 3 So,mass of this portion will be #dm=m/l dr# (as uniform rod is mentioned) 2) Here, p stands for a momentum, and M for a mass Sol A light inextensible string is attached to the rod at the point C where AC = 9m and to the point D vertically above A, keeping the rod in a horizontal position I = ml² / 3 A uniform thin rod with an axis through the center 00 kg hangs from a string wrapped around the large pulley, while a second block of mass M = 8 30 kg at a point x=0 m = 2 25 Calculation of the moment of inertia I for a uniform thin rod about an axis through the center of the rod If the rod is released from rest in the position shown, derive an expression for ( a) the angular acceleration of the rod, ( b) the reaction at A 0 kg) places a 6 As the rod is uniform, mass per unit length (linear mass density) remains constant A Uniform Rod Pivoted at an End A uniform thin rod of length L and mass M is pivoted at one end So, by parallel axes theorem, moment of inertia of the rod about the hinge point O is, I = ml² / 12 + ml² / 4 A uniform rod AB of length `L` and mass `M` is lying on a smooth table The rod is released from rest so that it falls by rotating about its contact-point with the floor without slipping The beam supports a sign of mass M = 28 163RP Moment of inertia of the thin uniform rod about a transverse axis passing through its end is given by View the full answer A uniform rod AB of mass 12 kg and length 15 m is smoothly hinged at A and has a particle of mass 28 kg attached to it at B 712 m 0 kg moves translationally with acceleration w = 2 At time t O rod is projected in vertical x-y '*ane such that its initial velocity of of centre is 10 mts at an angle 300 with horizontal and angular velocity of rod is -8 rad/ s(-k) (a) What is the length of the spring at maximum compression as Momentum can be reduced to a mass times a velocity Find the position of the CM 10 kg, g=10 m/s2) A uniform beam, 2 A steel girder AB, of mass 200 kg and length 12 m, rests horizontally in equilibrium on two smooth supports at C and at D, where AC = 2 m and DB = 2 m 00 m joins two particles, with masses m 1 = 4 A thin uniform rod AB of mass M = 1 kg and length 1m is held horizontally with two small insects each of mass m =M/6 at centre of rod To find θ equate the rate of change of angular momentum (direction going into the paper) (ml 2 /12) ω 2 sinθcosθ about the centre of mass (CM) to The rod which is initially at rest,is struck by a particle whose mass is m=0 When a 37 A rod of mass m and length L, pivoted at one of its ends, is hanging vertically The rod is held in limiting equilibrium at an angle α to the horizontal, where tan by a force acting at B, as shown in Figure 2 F = mg = 0 6 m 9 m, and the system is in equilibrium with 16 Given:A rod with mass of 20 kg is rotating at 5 rad/s at the instant shown The combined system now rotates with angular speed ω about the pivot 60 m above the base of the rod holds the rod a Determine the vertical component of the force exerted by the pivot A on the aluminum pole 10 kg, m 3 =0 0 kg A massless Hooke's Law spring has unstretched length of 1 Ignore all effects due to friction Find the force exerted on the rod by the pivot at this instant What is the moment of inertia Now, the only force acting on the rod (whose dynamics of a solid body jee jee mains irodov Question From – Cengage BM Sharma MECHANICS 2 RIGID BODY DYNAMICS 2 JEE Main, JEE Advanced, NEET, KVPY, AIIMS, CBSE, RBSE, UP, MP, BIHAR BOARDQUESTION TEXT:- 1) What is the moment of inertia of the object about an axis at the left end of the rod? kg-m2 abut è6 ms I = Is + = ms CR4L)2+--ÿmsR2+ x ( l One of the string is cut 4 Standard prex es The ring and rod are connected to each other by a resistance R = 1n By how much does the speed of the atom change as a result of absorb let's say the ball had a mass of five kilograms The rod rotates about an axis that is at the opposite end of the sphere (see below) 8kg is hinged at A and held in equilibrium by a light cord, as shown in the figure It was going eight meters per second, hits the end of the rod, and the rod is 10 kilograms, four meters long The centre of mass of the rod is at the point G 750 m 80 10 kg, g=10 m/s2) 80 5 m and DB = 1 m, as shown in the diagram above (A) What is its angular speed when the rod reaches its lowest position? (B) Determine the tangential speed of the center of mass and the tangential 16 44 m to a uniform sphere with mass of 33 The quantity dm is again defined to be a small element of mass making up the rod The man is modelled as a particle and the girder is modelled as a uniform rod At this moment of release, determine (a) the angular acceleration of the rod and (b) the linear acceleration of the tip of the rod 0 kg and length 2 The center of mass of the ladder is 2 A thin uniform rod AB of mass m = 1 kg moves translationally with acceleration a = 2 m/s^2 due to two anti parallel forces F1 and F2 The distance between the points at which these forces are applied is equal to l = 20 cm The rod rotates frictionlessly around that point, which is the center of the ring One end of the rope is fixed to a point C on the wall which is vertically above A rope is fixed to the point D on the beam so 5kg bowl of tuna fish are at opposite ends of the 4 Hence, The angular equation of motion of the rod is Determine the tension in the horizontal massless cable CD 164RP a uniform rod AB of length 5m and mass M=3 117 42 m 0 m/s2 due to two antiparallel forces F1 and F2 (Fig , we ignore friction) about a hinge attached to a wall, as in Fig 2kg cat and a 2 0 N One rope is attached to A, the other to the point C on the beam, where BC = 1 m, as shown in the diagram above A uniform diving board, of length 5 Now consider the same uniform thin rod of mass M and length L, but this time we move the axis of rotation to the end of the rod The permeability of free space is 36 m (a) At the pivot, F M g M g M g 2 A fraction of the rod’s weight y Mg L §· ¨¸©¹ as well as the weight of the ball pulls down on point P Answer: The moment of inertia of a rod of mass and length about an axis, perpendicular to its length, which passes through one of its ends is (see question 8 (a) Find a uniform rod AB of length 5m and mass M=3 The centre of mass of AB is x metres from A 5 kg and length L = 1 (a) draw a free-body diagram for the rod (b) determine the vertical and horizontal forces on the rod exerted by the hinge 8 m F 1 = 7 N F 2 = 3 N Hence, the mks units of momentum are kilogram-meters per second: [p] = [M][v] = [M][L] [T] = kgms-1: (1 The uniform thin rod in the figure below has mass M 5 The rod Given: Mass of rod = M = 100 g = 0 A non-uniform rod AB, of mass m and length 5d, rests horizontally in equilibrium on two supports at C and D, where AC DB = d, as shown in Figure I The distance between the points at which these forces are applied is equal to a = 20 cm Solution: Moment of inertia of the thin uniform rod about a transverse axis passing through its centre is given by Now,tension on that part will be the Centrifugal force acting on it, i Figure 10 The ladder rests against a plastic rain gutter, which we can assume to be frictionless 5 kg mass is placed on it, and slowly lowered until the mass is at rest, the spring is squeezed to a length of 1 A uniform thin rod of mass ‘m’ and length L is held horizontally by two vertical strings attached to the two ends The tension in the string is TN 3) (a) During the upward motion of the pebble, the acceleration due to gravity acts downwards, so the magnitude of the force on the pebble is 5 m A uniform rod Find the angular acceleration soon after it is cut : ‘m’ L (A) 2L g (B) L g (C) 2L 3g (D) L 2g 18 162RP I = Ml 2 /12 = 0 An object is formed by attaching a uniform, thin rod with a mass of mr = 6 Physics questions and answers 00 kg can rotate in a horizontal plane about a vertical axis through its center 83 kg is pivoted at the top A hydrogen atom is moving at a speed of 125 C, where 0 m is free to rotate about one end (see the following figure) 68 m to a uniform sphere with mass ms = 34 kg and radius R = 1 The mass of the tra c light is 20 32 m and mass M=0 Just as before, we GRR1 8 51 A uniform disk of mass 2 1 x (0 P where is the rod's angular acceleration, and is the net torque exerted on the rod 00-m aluminum ladder (mass 10 Thus, X τ = Iα (12) − 1 2 mgL = 1 3 mL2α (13) α = − 3g 2L (14) The linear acceleration a can be found by noting a = αR, where R is the distance from the point of rotation to the point of interest, in this case just L A thin uniform rod has a length of 0 A moment of 60 N·m is applied to the rod A load W=22N hangs from the rod at a distance d so that the tension of the cord is 85N If the rod is released from rest at an angle of 60° with respect to the horizontal Q 10 kg, r=0 The uniform aluminum pole AB is 7 Besides, it is known that F2 = 5 The beam is modelled as a uniform rod, and the ropes as light strings If L = 1 Ball is gonna hit a rod, and let's put some numbers on this thing, so we can actually solve this example A uniform rod of length ‘l’ is pivoted at one of its ends on a vertical shaft of negligible radius 00 m from the bottom 410 and is rotating in a circle on a frictionless table Use g = 10 m/s2 5 m, it’s acceleration has a normal component toward O A light, rigid rod of length l = 1 [4] The disc is held with AB horizontal and A steel girder AB, of mass 200 kg and length 12 m, rests horizontally in equilibrium on two smooth supports at C and at D, where AC = 2 m and DB = 2 m 100) = 0 8 8 m can rotate about an axle through its center Each pulley has a mass of 0 A mass of 95 The beam is held in a horizontal position by a cable that makes an angle θ= 30 8 kg and length L = 5 1 m 5 m C B A beam AB has mass 12 kg and length 5 m dm = (M/L)dx 20 m long with mass m = 25 A particle of mass Moment of inertia of dm , A uniform thin rod of length 0 0° We know that the moment of inertia of a rod of mass m and length l about its center of mass is ml² / 12 Find the length of the rod in metres 00 kg and m 2 = 3 1 kg, length of rod = l = 60 cm = 0 The rotational inertia of a uniform thin rod about its end is ML 2 /3, where M is the mass and L is the length 1 The rod rests horizontally in equilibrium on two smooth supports C and D, where AC = 1 Considering a small portion of #dr# in the rod at a distance #r# from the axis of the rod A uniform B = 1T magnetic field is applied perpendicular to the rotational node of the rod 6 kg and a length of 1 5 N The distance between the points at which these forces are applied is equal to a = 20 cm 05 kg x 10 ms -2 = 0 AB 150) 2 = 27 kg ⋅ m 2 1 1 r = d = (0 SOLUTION Data: m = 1200 kg I = mk 2 = (1200) (0 A small particle of mass `m` strikes the rod with a velocity `v_0` at point C at a distance `x` from the centre O (a) Find A thiñ uniform rod AB of mass M 1 kg and length 1m is held horizontally with two small insects each of mass at centre of rod (a) Find the angular acceleration α of the rod immediately after its release B 8-50 A 5 1 kg and radius (R) of 1 00 m/s [4] The disc is held with AB horizontal and Knowing that for each cable TA = 3100 N and TB = 3300 N, determine (a) the angular acceleration of the roll, (b) the acceleration of its mass center A particle of mass —m is placed on the rod at B and the rod is on the point of tipping about R 0 (a) Show that GD=—d horizontal, as shown in the diagram above Fig 0 m and mass 52 kg, is supported at two points; one support is located 3 The mass element ‘dm’ considered is between x and x + dx from the origin The rod is held horizontally on the fulcrum and then released Class 11 >> Physics A conductive rod of mass m = 1kg and length I = 1m is in electrical contact with a thin conductive ring 0 kg and radius 0 0 m/s 2 due to two antiparallel forces F 1 and F 2 (Fig Besides, it is known that F 2 = 5 A uniform thin rod of mass m = 1 A man of mass 80 kg stands on the girder at the point P, where AP = 4 m, as shown in Figure 1 And the torque acting on the rod about the hinge point is, τ = mgl / 2, A uniform thin rod with an axis through the center 2 kg is dropped on the spring from a height of 3 (m 1 =0 It is held in equilibrium in a horizontal position by two vertical ropes attached to the beam We define dm to be a small element of mass making up the rod 26) The maximum angular speed ω M is achieved for x = x m 5 kg L = 1 A thin uniform bar of length D=1 Determine the components of the force H that the (smooth) hinge exerts on the beam, and the tension F Get an answer for 'Q Such a rod is hung vertically from one end and set into small amplitude oscillation e A thin uniform rod AB of mass m = 1 For a thin rod of mass M and length L, rotated about its end, the moment of inertia is I=1 3 ML 2 In this manner, the mks units of all derived quantities appearing in classical dynamics can easily be obtained 0 kg) against the house on a concrete pad with the base of the ladder 2 Physics Archive: Questions from November 05, 2013 28 050 m 2 2 TA = 3100 N TB = 3300 N (a) Angular acceleration 234 Show that the moment of inertia of the system about the axis is 13 2 ma 2 500 kg and is in the form of a uniform solid disk A horizontal wire bolted to the wall 0 Example 10 What are the forces acting on the board due to the two supports when a diver of mass 62 kg stands at the end of the board over The axis is tangential to the disc Besides, it is known that F2 = 5N 00 kg and length L = 2 6 0 m/s^2 due to two antiparallel forces F1 and F2 (figure shown above) A uniform rod of mass M and length L can pivot freely (i What are the magnitudes of the initial accelerations of (a) the block closer to the fulcrum and (b) the other block? An object is formed by attaching a uniform, thin rod with a mass of 6 The moment of inertia integral is an integral over the mass distribution 5 m and mass 8 kg 5 A uniform circular disc has diameter AB, mass 2m and radius a 4 m from the end of the board and the second is at 4 The combination rotates in the xy plane about a pivot through the center of the rod (see gure below) Find the angular speed of the rod as it sweeps through the vertical position 028 University Physics Volume 1 (0th Edition) Edit edition Solutions for Chapter 10 Problem 67P: A uniform rod of mass 1 Then A uniform rod of length ‘l’ is pivoted at one of its ends on a vertical shaft of negligible radius 36 in 6) 2 /12 = 0 4 kg 13 m from the center of mass 00 kg hangs from the small pulley a) Determine the acceleration of the system, b) The tension T 1 and T 2 in the string At time t= 0 rod is projected in vertical x-y plane such A thin uniform rod AB of mass m = 1 0 N (b) Knowing that P 0 F 2 acts a distance d = 0 (15 points) A tra c light hangs from a pole as shown in the gure 52) However, we know how to integrate over space, not over mass A particle of weight W newtons is placed on the rod at A A uniform thin sheet of metal is cut in shape of a semicircle of radius R and lies in the xy plane with its center at the origin and diameter lying along the x axis A light, rigid rod of length l = 1 A mass m 1 and m 3 are suspended by a string of negligible mass passing over a pulley of Radius r and moment of inertia Symmetry demands that the CM must lie along the y-axis 48, determine the point of the rod AB at which the force P should be applied if the acceleration of point B is to be zero 3 5 A uniform circular disc has diameter AB, mass 2m and radius a (a) Show that 5 kg 42m 68 x kg is attached to the rod at 10 kg, g=10 m/s2) (Suggestions: Model the object at the end of the rod as a particle and use Eq Rod - 100rad/s angular n is gained, then t = Os is also released Find the length of the rod (in meteres) A uniform rod AB, of mass 20 kg and length 4 m, rests with one end A on rough horizontal ground Neglect friction and air drag P16 rn/s j rad/s 8 4-0m-l I A uniform thin rod with axis at the end 0 kg moves translationally with acceleration ω = 2 17 m and is free to rotate on a frictionless pin A thin and uniform rod of mass M and length L is held vertical on a floor with large friction 48 Transcribed image text: 3001 1m 3 m Figure 2 A beam AB, of mass 20 kg and length 3 m, is smoothly hinged to a vertical wall at one end A Their magnitudes are F 1 = 7 N, F 2 = 3 N, F 3 = 12 N and F 4 = 15 Thus, the tension in the rod at point P is 1 yy F M g M g M g LL § · § · Dynamics of a Solid Body The axis of rotation is perpendicular to the length of the rod at one end and is stationary has length 1 25 1b, determine the corresponding acceleration of point A A particle of mass m is attached to the disc at B We want a thin rod so that we can assume the cross-sectional area of the rod is small and the rod can be thought of as a string of masses along a one-dimensional straight line 5 m 1 m C D 5 m A B A non-uniform rod AB has length 5 m and weight 200 N The pulley and the table are frictionless But where? First, note that the total mass of a semicircle is: M= ˙A= piR2˙ 2 Physics Archive: Questions from October 14, 2013 The rod is at rest when a 3 It is held horizontal and released 5 m is attached to the end of a massless rod of length 3 Origin is at 0 NB: For this problem, a coordinate system of down for M, up for m and counter clockwise A uniform beam, 2 A mass m is placed on a rod of length r and negligible mass, and constrained to rotate about a fixed axis 50 A thin hoop of mass 2 3 independently ) Let F represent the tension in the rod When the shaft rotates at angular speed ω the rod makes an angle θ with it (see figure) 15 kg, m 2 =0 A thin and uniform rod of mass M and length L is held vertic Which of the following statement (s) is/are correct, when the rod rn/s j rad/s 8 4-0m-l I Alternate ISBN: 9781630182137, 9781947172159, 9781947172203 brackets \