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Chapter 8 – Rotational Equilibrium and Rotational Dynamics This chapter deals with the forces that produce rotational motion. It builds onto the material developed in Ch 6 on rotational kinematics. Torque F Torque causes an object to rotate. It depends on the force and the point at which the force is applied relative to the axis of rotation. Specifically, torque is the product of the force and the nearest distance between the line of action of the force and the axis or rotation, Fd pivot r d (units are N-m) F F = F sin If r is the distance from the point of application of the force to the pivot, then d = r sin (called the ‘lever arm’). We can also write pivot r F = F cos F r where F = F sin is the component of F that is perpendicular to r. The torque due to the force shown in the diagram would produce a counter-clockwise rotation about the pivot and is assumed to be positive. A torque that would produce a clockwise rotation is negative. Example: A disk pivoted about an axis through its center is subjected to two forces, as shown to the right. Given: F1 = 10 N, F2 = 5 N, d1 = 8 cm, d2 = 5 cm. What is the net torque? = F1d1 – F2d2 = (10 N)(0.08 m) – (5 N)(0.05 m) = 0.8 N-m – 0.25 N-m = 0.55 N-m 1 d1 d2 F1 F2 Center of Gravity Gravity acts on all the particles that make up an extended object. The net torque due to gravity and the resulting rotation of the object about a pivot depend on how the mass is distributed. Effectively, we can think of all these individual weights as acting at a single point located so as to give the same net torque. In a uniform gravitational field this ‘center of gravity’ is the same as the ‘center of mass’. In 2-D the location of the center of mass is given by y (m) m2 mi xi mi yi 3 xcm , ycm m3 mi mi 2 m1 Example: 1 What is the location of the center of mass of the three particles in the figure to the right? Given: 1 2 3 4 x (m) m1 = 10 g, m2 = 20 g, m3 = 30 g. m x m2 x 2 m3 x3 xcm 1 1 m1 m2 m3 (0.01kg)(1m) (0.02kg)( 2m) (0.03kg)( 4m) 2.83 m 0.01kg 0.02kg 0.03kg m y m2 y 2 m3 y3 (0.01kg)(1m) (0.02kg)(3m) (0.03kg)( 2m) y cm 1 1 m1 m2 m3 0.01kg 0.02kg 0.03kg 2.17 m If a regularly shaped object has a uniform mass density, then its center of mass is the same as its geometrical center. Equilibrium An object is in equilibrium if both the net force and net torque on it are zero. Thus, for equilibrium F 0 1st condition for equilibrium 0 2nd condition for equilibrium (Note: The location of the axis of rotation is arbitrary when the object is in equilibrium.) 2 Example: Masses m1 = 1 kg and m2 = 2 kg hang from opposite ends of a 1-m rod. The mass of the rod is 3 kg. Where would you lift the rod to keep it in balance? x Pick the axis of rotation to be the left end (any position will work). The resulting forces are shown below. F x L/2 m1 g mrod g m2 g From the 1st condition for equilibrium, F 0 F m1 g m2 g mrod g F (m1 m2 mrod ) g (6kg)(9.8m / s 2 ) 58.8 N From the 2nd condition, using the left end as the pivot, 0 Fx m1 g (0) m2 gL mrod L / 2 m2 gL mrod gL / 2 (2kg)(9.8m / s 2 )(1m) (3kg)(9.8m / s 2 )(0.5m) x 0.58 m F 58.8 N 3 Example: A sign hangs from a horizontal beam as shown to the right. The left end of the beam is attached to a building and a cable supports the right end of the beam at an angle of 30o with respect to the beam. The beam is 2 m long and its mass is 5 kg. The sign is attached 1.5 m from the building and its mass if 10 kg. What is the tension in the cable? The forces acting on the beam are shown below. T R mbg msg Picking the left end as the pivot, 0 T sin (2m) mb g (1m) ms g (1.5m) mb g (1m) ms g (1.5m) (5kg)(9.8m / s 2 )(1m) (10kg)(9.8m / s 2 )(1.5m) T sin (2m) (sin 30 o )( 2m) 196 N Note: By picking the left end as the pivot, we don’t need to worry about solving for R, the force exerted on the left end by the building. Torque and Angular Acceleration For a single point mass going in a circle subject to a tangential acceleration, Ft mat mr (it also has a centripetal component of force and acceleration) The torque is then Ft r mr 2 , or mr 2 4 For a rigid body consisting of many masses, each mass goes in a circle and has the same angular acceleration, . Thus, the net torque is mr 2 The moment of inertia (or rotational inertia) is defined as I mr 2 Thus, I This is in essence Newton’s 2nd law of motion for rotation of a rigid body. The torque drives the angular acceleration and the rotational inertia resists the angular acceleration. The rotational inertia depends not only on the mass but on where the mass is located with respect to the axis or rotation. y (m) (0, 3m) Example: m1 Three masses, each 2 kg, are located at the corners of a triangle consisting of light rigid rods, as shown to the right. The location (x, y) of each mass is shown. What is the rotational inertia about the x-axis? x (m) m2 (-3m, -2m) m3 (3m, -2m) What matters here is the nearest distance of each mass from the axis of rotation. So, I x mr 2 m1 y12 m2 y 2 2 m3 y3 2 (2)(3) 2 (2)( 2) 2 (2)( 2) 2 34 kg m 2 Suggested exercise: Find the rotational inertia about the y-axis and about the z-axis (which is perpendicular to the page). Example: What is the rotational inertia of a thin cylindrical shell or radius R about its symmetry axis? Since all the mass is the same distance from the axis, then I mr 2 ( m) R 2 mR 2 5 Example: m What is the rotational inertia of a uniform rod about an axis perpendicular to the rod at an end of the rod? x L I mx 2 The mass of the rod is distributed uniformly from x = 0 to x = L, so one must use the average value of x2. It can be shown that <x2> = L2/3. Thus, I 1 M L2 3 Rotational Inertia of Various Rolling Objects: The rotational inertia of various uniform ‘round’ objects can be obtained using calculus. As they roll, they rotate about a symmetry axis through the center of the object. Some results for rolling objects: Object cylindrical shell Rotational Inertia solid cylinder I 1 M R2 I M R2 spherical shell I solid sphere I 2 2M 3 2M 5 R2 R2 Example: A flywheel of radius R = 0.2 m and mass M = 25 kg is mounted to rotate about its symmetry axis. A rope is wrapped around the wheel and pulled with a force of 10 N. What is the angular acceleration of the wheel? (Assume the wheel is a solid cylinder.) I 1 MR 2 1 (25kg)(0.2m) 2 0.5 kg m 2 2 I 2 FR (10 N )(0.2m) 0.4 rad / s 2 2 I 0.5kg m 6 Rotational Kinetic Energy For a rigid object rotating about an axis, all the pieces that make up the object rotate in a circle with the same angular velocity, . Thus, the kinetic energy is KErot 1 mv 2 1 m 2 r 2 1 mr 2 2 2 2 2 or, KE rot 1 I 2 2 A rolling object will have kinetic energy related to the translational speed of its center of mass, ½ mv2, and kinetic energy related to its rotation about an axis through its center of mass, ½ I 2. It can also have gravitational potential energy. Thus, the total energy is E KEt KEr PE g 1 Mv 2 1 I 2 Mgh 2 2 Example: A ball rolls from rest down an incline of height 3 m without slipping. What is its speed when it reaches the base of the incline? Use conservation of energy – E f Ei 1 Mv 2 1 I 2 Mgh 2 2 v I 2 MR 2 , 5 R 1 Mv 2 1 2 MR 2 v 2 2 5 R 2 1 Mv 2 1 Mv 2 7 Mv 2 Mgh 2 5 10 v 10 gh 10 (9.8)(3) 6.5 m / s 7 7 Note that the final speed does not depend on either the mass of the ball or its radius. This will be true for all rolling objects. Question: What fraction of the final KE is due to rotation? 1 Mv 2 KErot 2 5 7 KEtotal Mv 2 7 10 7 Angular Momentum From Newton’s 2nd law of motion applied to rotation, I I 0 I I0 I t t t The angular momentum is defined as L I Thus, we can write L t This leads to the principle of conservation of angular momentum: If net = 0, then L = constant. Example: A figure skater goes into a spin with her arms outstretched. She then pulls her arms in against her body. If she is initially rotating at 2 rev/s, what is her final rotational speed? Assume that by pulling in her arms she reduces her rotational inertia from 2 kg-m2 to 1.5 kg-m2. L f Li I f f I i i f I i i (2kg m 2 )( 2rev / s) 2.67 rev / s If 1.5kg m 2 How much kinetic energy was gained or lost when she pulled in her arms? KE f 1 I f f 2 1 1.5kg m 2 (2 )( 2.67) rad / s 2 210.5 J 2 2 KEi 1 I f f 2 1 2kg m 2 (2 )( 2) rad / s 2 157.9 J 2 2 KE KE f KEi 210.5 J 157.9 J 52.6 J Thus, there was a gain in KE. Note that we had to convert angular velocity to rad/s when calculating KE. 8