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ROTATIONAL AND EQUILIBRIUM OF RIGID BODY Standard Competency Apply the concept and principle of classical mechanics of continuum system by problem solving Base Competency Formulate the relation of torsion concept, angular momentum and moment inertia base on Newton’s II Law and their application in rigid body problem Learning Objectives 1 Formulate the effect of torsion in an object related to its rotational motion 2 Reveal the analogy of Newton’s II law about translational and rotational motions 3 Using the concept of moment of inertia for several shapes of rigid body 4 Formulate the conservative law of angular momentum on rotational motion 5 Apply the concept of the center of mass of an object in daily life 6 Analyze the rotational dynamic of rigid body for several situations 7 Show the ability of psychomotoric and in experiment groups References: [1] John D Cutnell and Kenneth W. Johnson (2002). Physics 5th Ed with Compliments. John Wiley and Sons, Inc. pp. 212-248 [2] Sunardi dan Etsa Indra Irawan (2007). Fisika Bilingual SMA/MA untuk SMA/MA Kelas X. CV Yrama Widya pp. 255-298 ROTATIONAL VARIABLES Consider a rigid body A surrounding a center of xy- coordinate. One complete of revolution object A has elapsed distance (d) as d = 2πr The angle of rotation (θ) equal to 2π radian, s = arc length r = radius θ = 2π (rad) or Base on the picture, θ is equal to distance elapsed by the rigid body A. It means that θ is equal to distance d in Kinematics. If d = v.t or s = v.t, then θ = (v).t. The velocity v in rotational motion is known as angular velocity, simbolized by ω, hence θ = ω.t Refer to the scientific terms, - v denote for linier velocity, whilst - ω denote for angular velocity Relation between linier (v) and angular velocity (ω) s θ s=rθ ; v = ; ω = t t rθ v = → v =r ω t Angular acceleration (α) is derived from linier acceleration v a= t ω hence α = t Relation between linier (a) and angular acceleration (α) s θ s=rθ ; v = ; ω = t t rθ ω v = → v =r ω ; α = t t v r ω a= → a= t t Acceleration in rotational motion has both the radial and the tangential components, namely radial (ar) and tangential acceleration (at) The radial component is always present as long as [omega] is not equal to zero The tangential component is only present if the angular acceleration is not zero Tangential acceleration (at) v r ω ω at = → at = → at = r t t t at = r α Radial acceleration (ar) Every part of the body when a rigid body is rotating around a fixed axis has: - the same angular velocity [omega] and the same angular acceleration - points that are located at different distances from the rotation axis have different linear velocities and different linear accelerations THE EQUATIONS OF ROTATIONAL KINEMATICS Basically, rotational variables are similar to translational (linier) variables described in Kinematics, namely position, velocity and acceleration. This is apply as well since the equations of rotational kinematics have similar form with the equation of translational kinematics. Linier Motion (Translational) Circular Motion (a = constant) (Rotational) (α = constant) v = vo + a t ω = ωo + α t x = ½ (vo + v)t θ = ½ (ωo + ω)t x = vo t + ½ at2 θ = ωo t + ½ αt2 v2 = vo2 + 2 a x ω2 = ωo2 + 2 α θ MOMENT OF FORCE (TORQUE) The measure of a force's tendency to produce a rotation about an axis is called torque. 3.0 m If a force is used to begin 4.0 N to spin something, a torque is generated. A torque would also be generated if a force was used to stop something from spinning. Torque is simbolized by T or τ (Greek letter tau). Torque arm or lever arm is defined as distance from applied force (line of action) to fulcrum or axis of rotation perpendiculary Torque Direction of Rotation Positive (+) torque direction Negative (−) torque direction as it rotates counter as it rotates clockwise clockwise Torque Equation Torque angle φ is measured counter clockwise from the radial line, see the picture below. Pivot point = titik tumpu ccw = counter clockwise Radial line is drawn from pivot point to the point where force is applied Only the tangential component of force causes a torque The torque increases as the force increases and also as the distance increases The same force will be much more effective at rotating an object such as as nut or a door if our hand is not too close to the axis. This is why we have long handled wrench, and why doorknobs are not next to hinges. Torque is not a type of energy eventhough its unit is newton- meter, therefore it should not have a unit of joule. EQUILIBRIUM OF RIGID BODY Body Equilibrium It’s a condition of a body (an object) which still remain as in initial condition after external forces applied vanished. According to Newton’s I law, an object to said as equlibry if resultant force applies on object is equal to zero. Rigid Body A characteristic of an object when one or more forces apply, it will remain as it was (no volume or shape change) Equilibrium An object is in equilibrium if the linear momentum of its center of mass is constant and if its angular momentum about its center of mass is constant: p = constant L = constant An object is in static equilibrium if its linear momentum and angular momentum is equal to zero: p = 0 kg m/s L = 0 kg m2/s Requirements for Equilibrium If a body is in translational equilibrium then dp/dt = 0, or If a body is in rotational equilibrium then dL/dt = 0, or In summary, the following equations must be satisfied for an object in static equilibrium If we restrict ourselves to two dimensions (the x-y plane) the following equations must be satisfied: Equilibrium and the Force of Gravity Figure left. Weight of an object balanced by a single force. It shows a body of arbitrary shape balanced by a single force. The origin of the coordinate system is defined such that it coincides with the center of gravity of the object, which is the point upon which the balancing force acts. An object that is supported at its center of gravity will be in static equilibrium, independent of the orientation of the object. If the body is in equilibrium, the net force acting on it must be zero. It shows that Since the body is in equilibrium and therefore In obtaining this result we have assumed that the gravitational acceleration is the same for every point of the body. The net torque acting on the body is given by Since the body is in static equilibrium and therefore This shows that rcm = 0 or rcm is parallel to g. We conclude that for a body to be in equilibrium, its center of mass must coincide with its center of gravity. Example A uniform beam of length L whose mass is m, rest with its ends on two digital scales. A block whose mass is M rests on the beam, its center one-fourth away from the beam's left end. What do the scales read ? Answer For the system to be in equilibrium, the net force and net torque must be zero. Here we have replaced the force acting on the beam with a single force acting on its center of gravity. The net torque of the system, with respect to the left scale, is This shows immediately that From the equation of the net force we obtain Example A ladder with length L and mass m rests against a wall. Its upper end is a distance h above the ground. The center of gravity of the ladder is one-third of the way up the ladder. A firefighter with mass M climbs halfway up the ladder. Assume that the wall, but not the ground, is frictionless. What is the force exerted on the ladder by the wall and by the ground ? Answer The wall exerts a horizontal force FW on the ladder (the normal force); it exerts no vertical force. The ground exerts a force Fg on the ladder with a horizontal component Fgx and a vertical component Fgy. If these two components were not present, the system would not be in equilibrium. The net force in the x and y directions is given by and The net torque, with respect to O (which is the contact point between the ladder and the ground), is given by This immediately shows that We can now calculate the force Fg: and We observe that Fgx depends on the position of the firefighter. Suppose that the firefighter is a distance f L up the ladder. In this case Fgx is given by If the coefficient of static friction between the ladder and the ground is us, than the maximum distance the firefighter can climb is reached when or This shows that Stacking Blocks Two bricks of length L and mass m are stacked. Using conditions of static equilibrium we can determine the maximum overhang of the top brick Two bricks of length L and mass m are stacked. Using conditions of static equilibrium we can determine the maximum overhang of the top brick. The two forces acting on the top brick are the gravitational force Fg and the normal force N, exerted by the bottom brick on the top brick. Both forces are directed along the y-axis. Since the system is in equilibrium, the net force acting along the y-axis must be zero. We conclude that The torque of the normal force and the gravitational force with respect to O is given by The net torque acting on the top brick is given by If the system is in equilibrium, then the net torque acting on the top brick with respect to O must be zero. This implies that or