NEWTONS LAWS PHY 1012 F ROTATION Gregor Leigh

NEWTON’S LAWS PHY 1012 F ROTATION Gregor Leigh gregor. leigh@uct. ac. za

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY Learning outcomes: At the end of this chapter you should be able to… Extend the ideas, skills and problem-solving techniques developed in kinematics and dynamics (particularly with regard to circular motion) to the rotation of rigid bodies. Calculate torques and moments of inertia. Apply appropriate mathematical representations (equations) in order to solve rotation problems. 2

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY ROTATIONAL KINEMATICS Rigid body model: A rigid body is an extended object whose shape and size do not change as it moves. Neither does it flex or bend. Types of motion: Translational motion: Rotational motion: Combination motion: 3

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY ANGULAR ACCELERATION Linear motion relationship Rotational motion s s = r vt = r at = r A body’s angular acceleration, , is the rate at which its angular velocity changes. Units: [rad/s 2] 4

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY ANGULAR VELOCITY and ANGULAR ACCELERATION Angular velocity, Angular acceleration, + – >0 <0 faster slower >0 >0 <0 slower faster <0 5

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY VELOCITY GRAPHS ACCELERATION GRAPHS Angular acceleration is equivalent to the slope of a -vs-t graph. (rad/s) 3 0 Eg: A wheel rotates about its axle… 3 6 9 12 t (s) – 3 For the first 6 s the slope acceleration is (rad/s 2) ½ 0 –½ 3 6 9 12 t (s) – 6

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY VELOCITY GRAPHS ACCELERATION GRAPHS Angular acceleration is equivalent to the slope of a -vs-t graph. (rad/s) 3 0 Eg: A wheel rotates about its axle… 3 6 9 12 t (s) – 3 For the last 6 s the slope acceleration is (rad/s 2) ½ 0 –½ 3 6 9 12 t (s) – 7

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY KINEMATIC EQUATIONS The following equations apply for constant acceleration: Linear motion Rotational motion vf = vi + a t f = i + t xf = xi + vi t + ½a ( t)2 f = i + i t + ½ ( t)2 vf 2 = vi 2 + 2 a x f 2 = i 2 + 2 8

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY CENTRE OF MASS While the various points on a flipped spanner describe different (complicated) trajectories, one special point, the centre of mass, follows the usual, simple parabolic path. The centre of mass (CM) of a system of particles… is the weighted mean position of the system’s mass; is the point which behaves as though all of the system’s mass were concentrated there, and all external forces were applied there; is the point around which an unconstrained system (i. e. one without an axle or pivot) will naturally rotate. 9

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY CENTRE OF MASS To find the centre of mass of two masses… 1. place the masses on an x-axis, with one of the masses at the origin; x m 2 m 1 x. CM x 2 x 1 = 0 2. apply the formula: In general, for many particles on any axis: And for a continuous distribution of mass in which mi = M: 10

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY TORQUE The rotational analogue of force is called torque, . Torque may be regarded as… the “amount of turning” required to rotate a body around a certain point called an axis or a pivot; the effectiveness of a force at causing turning. E. g. To push open a heavy door around its hinge (as seen from the top) requires a force applied at some point on the door. hinge Consider the effectiveness of each of the (equal) forces shown… 11

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY TORQUE Three factors determine the amount of torque achieved: magnitude of the applied force; distance between the point of application and the pivot; angle at which the force is applied. Only the tangential component of y the applied force produces any turning… Ft = Fsin Fr = r. Ft Hence: r r. Fsin Units: [N m] ( joule!) pivot radial line point of application x 12

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY TORQUE Torque can also be defined as the product of the force, F, and the perpendicular distance between the pivot and the line of action of the force, d, known as the torque arm, moment arm, or lever arm: | | = d. F Torque… is positive if it tries to rotate the object anticlockwise about the pivot; negative if rotation is clockwise; torque arm d = r sin r pivot line of action x must be measured relative to a specific pivot point. 13

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY NET TORQUE Several forces act on an extended object which is free to rotate around an axle. The axle prevents any translational movement, so… axle And the net torque is given by: net = 1 + 2 + 3 + … = i ( causes no torque since it is applied at the axle. ) 14

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY GRAVITATIONAL TORQUE For an object to be balanced, its centre of mass must lie either directly above or directly below the point of support. centre of mass If this is not so, the body’s own weight, acting through its centre of mass (as if all its mass were concentrated there), causes a net torque due to gravity: grav = –Mgx. CM where x. CM is the distance between the centre of mass and the pivot CM Mg 0 x. CM x 15

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY COUPLES Rotation without translation is achieved by the application of a pair of equal but opposite forces at two different points on the object. d 1 d 2 pivot | net | = d 1 F + d 2 F = (d 1 + d 2)F | net | = l. F Notes: (sign by inspection) l The pivot is immaterial – a couple will exert the same net torque l. F about any point on the object. Unless the rotation is constrained to act around a specific pivot, it will occur around the body’s centre of mass. 16

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY ROTATIONAL DYNAMICS A rocket is constrained to move in a circle by a lightweight rod… y Ft Newton II (linear): Force causes linear acceleration. rod Linear acceleration is “limited” by inertial mass. Ft = mat Ft = mr pivot r. Ft = mr 2 Fr x = mr 2 Newton II (rotational): Torque causes angular acceleration. Angular acceleration is “limited” by the particle’s rotational inertia, mr 2, about the pivot. 17

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY ROTATIONAL INERTIA A rotating extended object can be modelled as a collection of particles, each a certain distance from the pivot. The body’s rotational inertia… m 1 r 1 pivot m 2 r 2 m 3 r 3 is also known as its moment of inertia, I; is the aggregate of the individual (mr 2)’s: I = m 1 r 12 + m 2 r 22 + m 3 r 32 + … = miri 2 gives an indication of how the mass of the body is distributed about its axis of rotation (pivot); is the rotational equivalent of mass. Thus: 18

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY ROTATIONAL DYNAMICS Summary of corresponding quantities and relationships: Linear dynamics force Rotational dynamics Fnet torque inertial mass m moment of inertia I acceleration a angular acceleration Newton II Fnet = ma Newton II net = I 19

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY MOMENTS OF INERTIA For an extended system with a continuous distribution of mass, the system is divided into equal-mass elements, m. Then, by allowing these to shrink in size, the moment of inertia summation is converted to an integration: y m For complex distributions of mass, r is usually replaced by x and y components… r pivot x y x …and, before integration, dm is replaced by an expression involving a coordinate differential such as dx or dy. 20

NEWTON’S LAWS MOMENTS OF INERTIA For simple, uniform distributions of mass, however, the integration can be trivial. E. g. in a wheel, or hoop, where all the mass lies at a distance R from the axis… becomes R , where the integral is simply the sum of all the mass elements, i. e. the total mass, M, of the wheel. So for open wheels (hoops) I = MR 2. In practice, the rotational inertias of certain common shapes (of uniform density) are looked up in tables…

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY PARALLEL AXIS THEOREM Moments of inertia are always calculated/stated with respect to a specific axis of rotation. However (assuming the moment of inertia for rotation around the centre of mass is known), the moment of inertia around any offcentre rotation axis lying parallel to the axis through the centre of mass can be found using the parallel-axis theorem: CM d M (mass) I = ICM + Md 2 22

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY ROTATION ABOUT A FIXED AXIS Problem-solving strategy: 1. 2. 3. Model the object as a simple shape. 4. Identify all the significant forces acting on the object and determine the distance of each force from the axis. Determine all torques, including their signs. Apply Newton II: net = I. (I-values from tables and the parallel-axis theorem. ) 5. 6. 7. Identify the axis around which the object rotates. Draw a picture of the situation, including coordinate axes, symbols and known information. Use rotational kinematics to find angular positions and velocities. 23

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY ROTATION ABOUT A FIXED AXIS A 70 g metre stick pivoted freely at one end is released from a horizontal position. At what speed does the far end swing through its lowest position? 0 x pivot M = 0. 07 kg 1 -3. Model the stick as a uniform rod rotating around one end, and draw a picture with pivot, x-axis and data. 24

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY ROTATION ABOUT A FIXED AXIS A 70 g metre stick pivoted freely at one end is released from a horizontal position. At what speed does the far end swing through its lowest position? 0 x. CM = 0. 5 m x pivot Mg M = 0. 07 kg 4. Identify all significant forces acting on the object and determine the distance of each force from the axis. 25

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY ROTATION ABOUT A FIXED AXIS A 70 g metre stick pivoted freely at one end is released from a horizontal position. At what speed does the far end swing through its lowest position? 0 x x. CM = 0. 5 m pivot Mg M = 0. 07 kg –ve grav = –Mgx. CM = – 0. 07 9. 8 0. 5 = – 0. 34 Nm 5. Determine all torques, including their signs. 26

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY ROTATION ABOUT A FIXED AXIS A 70 g metre stick pivoted freely at one end is released from a horizontal position. At what speed does the far end swing through its lowest position? pivot Mg x M = 0. 07 kg net = I grav = – 0. 34 Nm = – 15 rad/s 2 6. Apply Newton II: net = I. 27

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY ROTATION ABOUT A FIXED AXIS A 70 g metre stick pivoted freely at one end is released from a horizontal position. At what speed does the far end swing through its lowest position? pivot cs i t 2 2 a i = 0 rad f = i + 2 n em i k l na i = 0 rad/s t? o i o t n a t 2 y h ef ro= 0 + !2(– 15)(– 0. 5 ) W s u m T le O b = – 15 rad/s 2 N o n r a p c his t You e v l to so vt = r f = – 0. 5 rad vt = – 6. 8 1 = 6. 8 m/s f = ? 7. Use rotational kinematics to find angular positions and velocities. (Not this time!) 28

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY CONSTRAINTS DUE TO ROPES and PULLEYS Provided it does not slip, a rope passing over a pulley moves in the same way as the pulley’s rim, and thus also objects attached to the rope. rim acceleration = | |R rim speed = | |R non-slipping rope R As before, the constraints are given as magnitudes. Actual signs are chosen by inspection. vobj = | |R aobj = | |R 29

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY T 2 Two blocks are connected by a light string which passes over two identical pulleys, each with a moment of inertia I, as shown. Find the acceleration of each mass and the tensions T 1 , T 2 and T 3. T 1 m 1 g –ve T 1 w +ve T 3 y m 1 m 2 x T 2 n 1 T 1 n 2 +ve w T 3 –ve T 3 m 2 g F 1 y = T 1 – m 1 g = m 1 a (1) F 2 y = T 3 – m 2 g = –m 2 a (2) net = T 1 R – T 2 R = –I (3) net = T 2 R – T 3 R = –I (4) 30

PHY 1012 F NEWTON’S LAWS F 1 y = T 1 – m 1 g = m 1 a (1) F 2 y = T 3 – m 2 g = –m 2 a (2) net = T 1 R – T 2 R = –I (3) net = T 2 R – T 3 R = –I (4) ROTATION OF A RIGID BODY T 2 T 1 m 1 (3) + (4): T 1 R – T 3 R = – 2 I (1) – (2): T 1 – T 3 + m 2 g – m 1 g = m 1 a + m 2 a T 3 y x m 2 …etc 31

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY EQUILIBRIUM Problem-solving strategy: 1. Model the object as a simple shape. 2. Draw a picture of the situation, including coordinate axes, symbols and known information. 3. Identify all the significant forces acting on the object. 4. Choose a convenient pivot point and determine the moment arm of each force from it. 5. Determine the sign of each torque around the pivot. 6. Write equations for Fx = 0; Fy = 0; net = 0; and solve. 32

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY EQUILIBRIUM A 3 -metre ladder leans against a frictionless wall, making an angle of 60° with the ground. What minimum coefficient of friction is needed to prevent the foot of the ladder from slipping? y L CM x 1 -2. Model the ladder as a rigid rod, and draw a picture with axes, symbols and known information. 33

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY EQUILIBRIUM A 3 -metre ladder leans against a frictionless wall, making an angle of 60° with the ground. What minimum coefficient of friction is needed to prevent the foot of the ladder from slipping? y L CM x 3. Identify all the significant forces acting on the object. 34

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY EQUILIBRIUM A 3 -metre ladder leans against a frictionless wall, making an angle of 60° with the ground. What minimum coefficient of friction is needed to prevent the foot of the ladder from slipping? y L d 2 CM d 1 pivot x 4. Choose a convenient pivot point and determine the moment arm of each force from it. 35

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY EQUILIBRIUM A 3 -metre ladder leans against a frictionless wall, making an angle of 60° with the ground. What minimum coefficient of friction is needed to prevent the foot of the ladder from slipping? –ve y L CM d 2 +ve d 1 pivot x 5. Determine the sign of each torque around the pivot. 36

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY EQUILIBRIUM Fx = nw – fs = 0 (1) Fy = nf – Mg = 0 (2) net = d 1 w – d 2 nw = 0 (3) net = ½(Lcos 60°)Mg – (Lsin 60°)nw= 0 –ve y L CM d 2 +ve d 1 pivot x 6. Write equations for Fx = 0; Fy = 0; net = 0; and solve. 37

PHY 1012 F NEWTON’S LAWS ROTATION OF A RIGID BODY Learning outcomes: At the end of this chapter you should be able to… Extend the ideas, skills and problem-solving techniques developed in kinematics and dynamics (particularly with regard to circular motion) to the rotation of rigid bodies. Calculate torques and moments of inertia. Apply appropriate mathematical representations (equations) in order to solve rotation problems. 38