Chapter 8: Rotational Equilibrium and Dynamics Force vs. Torque Forces cause accelerations Torques cause angular accelerations Definition of Torque Torque,, is the tendency of a force to rotate an object about some axis Torque = (Magnitude of Force) * (Lever arm) = F d Lever arm is the perpendicular distance from the axis of rotation to a line drawn along the direction of the force. Torque is a vector quantity. Direction of torque: right-hand rule = r F sin points from the point of reference (rotational axis) to the location of the force. Obviously, the magnitude of a torque depends on where we assume the axis of rotation to be. F sin is the tangential component of the force. Unit of torque: N m 1
Multiple torques, net torque, and an example When two or more torques are acting on an object, the torques are added as vectors. The sum is the net torque. Find the net torque (magnitude and direction) produced by the forces F 1 and F 2 about the rotational axis shown in the drawing. The forces are acting on a thin rigid rod, and the axis is perpendicular to the page. Center of Gravity The net torque produced by the force of gravity: The object is divided up into a large number of very small particles of weight (m i g) Each particle will have a set of coordinates indicating its location (x i, y i ) The torque produced by each particle is m i gx i The total (net) torque is (m i gx i ) Now locate a point (x cg, y cg ), so that the torque produced by the gravity of the whole mass on that point is equal to the total torque: ( m i g) x cg = (m i gx i ) 2
Finding the Center of Gravity The center of gravity of a homogenous, symmetric body must lie on the axis of symmetry. Often, the center of gravity of such an object is the geometric center of the object. If a single force can be used to support an object in equilibrium, it must exactly counteract the effect of gravitational force. The force must lie on a line through the center of gravity of that object. CG could be in a hollow! Two Conditions for Equilibrium A rigid body is in equilibrium if it has zero translational acceleration and zero angular acceleration. In equilibrium, the sum of the externally applied forces is zero, and the sum of the externally applied torques is zero: and Note that if the net force is zero, the net torque becomes independent of the choice of rotational axis. In this case, we can use any rotational axis to sum up the net torque. The nature of the problem will often suggest a convenient location for the axis. A zero net torque does not mean the absence of rotational motion. An object that rotates at uniform angular velocity can be under the influence of a zero net torque. This is analogous to the translational situation where a zero net force does not mean the object is not in motion. 3
Solving Equilibrium Problems Draw a diagram of the system Include coordinates and choose a rotation axis Isolate the object being analyzed and draw a free body diagram showing all the external forces acting on the object For systems containing more than one object, draw a separate free body diagram for each object Apply the Second Condition of Equilibrium This will yield a single equation, often with one unknown which can be solved immediately Apply the First Condition of Equilibrium This will give you two more equations Solve the resulting simultaneous equations for all of the unknowns Example of Equilibrium A woman of mass m=55.0 kg sits on the left end of the seesaw, a plank of length L=4.00 m and mass m=12.0 kg, pivoted in the middle. A man of mass M=75.0 kg sits on the right side of the pivot to balance the seesaw. (a) Where should the man sit? (b) Find the normal force exerted by the pivot. 4
More Example If the 70-kg fireman goes any higher than as shown, the ladder, which is 8.0-m long and has an evenly distributed mass of 20 kg, will slip. The wall is smooth. What is the coefficient of static friction between the ladder and the floor? Consequence of Net Torques: Angular Accel. For a point object rotating about an axis a distance r away from the object, F T = ma = mr rf T = = mr 2 Now define Moment of Inertia of a point object going around an external: then: I = mr 2 = I For a rigid body rotating about a fixed axis Net external torque = ( moment of intertia) * (angular acceleration) = I Note that a centripetal force leads to no angular acceleration. The relationship is analogous to F = ma 5
Moment of Inertia For a rigid system consisting of many particles, 1 = (m 1 r 12 ) 2 = (m 2 r 22 ) 3 = (m 3 r 32 )... The moment of inertia for a (rigid) system consisting of many particles is found to be Obviously, the moment of inertia of an object depends on the chosen axis of rotation. Note: distance to the axis of rotation (a line)! Distance parallel to the axis does not matter. Moments of Inertia Moment of inertia depends on the location of the rotational axis 6
Example A solid, frictionless cylindrical reel of mass M=3.00 kg and radius R=0.400 m is used to draw water from a well. A bucket of mass m=2.00 kg is attached to a cord that is wrapped around the cylinder. (a) Find the tension T in the cord and the acceleration a of the bucket. (b) If the bucket starts from rest at the top of the well and falls for 3.00 s before hitting the water, how far does it fall? Rotational Kinetic Energy The rotational kinetic energy KE R of a rigid object rotating with an angular speed about a fixed axis and having a moment of inertia I is 7
Total Mechanical Energy The total kinetic energy of an object that moves with both a linear velocity and a rotational velocity is the sum of the two contributions. Here v is the linear velocity of the center of mass of the object and is the angular velocity about its center of mass. of center of mass The total mechanical energy of the object is then the sum of its kinetic energy and potential energy. In the absence of work due to non-conservative forces, the total mechanical energy is conserved. of center of mass Energy concepts can be useful for simplifying the analysis of rotational motion. Conservation, and Work-Energy Theorem in a Rotating System Conservation of Mechanical Energy (KE t + KE r +PE) i = (KE t + KE r +PE) f Remember, this is for conservative forces, no dissipative forces such as friction can be present Potential energies of any other conservative forces could be added In the case where there are dissipative forces such as friction, use the generalized Work-Energy Theorem instead of Conservation of Energy: W nc = KE t + KE r + PE 8
Example Problems A thin-walled hollow cylinder (mass = m h, radius = r h ) and a solid cylinder (mass = m s, radius = r s ) start from rest at the top of an incline. Both cylinders start at the same vertical height h 0. All heights are measured relative to an arbitrarily chosen zero level that passes through the center of mass of a cylinder when it is at the bottom of the incline. Ignoring energy losses due to retarding forces, determine which cylinder has the greater translational speed upon reaching the bottom. Angular Momentum The angular momentum L of a rigid body rotating about a fixed axis is the product of the body s moment of inertia I and its angular velocity with respect to the axis: L = I, and For a point object, L=I =mr 2 =rmv. Changes in angular momentum are due to rotational impulse: t. Internal forces of a system cannot lead to a net torque on the system, because of the action and the reaction act in opposite directions, but ALONG THE SAME LINE. If the net external torque acting on the system is zero, the total angular momentum of a system remains constant, i.e. it is conserved. 9
Conservation of Angular Momentum, Examples With hands and feet drawn closer to the body, the skater s angular speed increases L is conserved, I decreases, increases Example Problems of Angular Momentum Movie The puck has a mass of 0.120 kg. Its original distance from the center of rotation is 40.0 cm, and the puck is moving with a speed of 80.0 cm/s. The string is pulled downward 15.0 cm through the hole in the frictionless table. Determine the work done on the puck. (Hint: Consider the change of kinetic energy of the puck.) 10
Rotational Analog of Translational Concepts Item Rotational Translational Displacement s Velocity v Acceleration a Cause of Accel. Torque Force F Inertia Moment of Inertia I Mass m Newton s 2 nd Law = I F = ma Work F s Kinetic Energy I 2 / 2 mv 2 / 2 Momentum Angular momentum I Linear Momentum mv Chapter 8 Summary Torque; Conditions for equilibrium; Center of gravity; Newton s second law for rotational motion; Moment of inertia; Rotational work; Rotational kinetic energy; Conservation of total mechanical energy; Angular momentum; Conservation of angular momentum. 11