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Rotational Mechanics Part III – Dynamics, Energy, Momentum

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Rotational Mechanics Part III – Dynamics, Energy, Momentum Rotational Mechanics Part IV – Angular Momentum Pre AP Physics Angular Momentum and Its Conservation In an analogy with linear momentum, we can define angular momentum L: L = I Since = I and since fi , We can then t write the total torque as being the rate of change of angular momentum. fi I fi L I t tt If the net torque on an object is zero, the total angular momentum is constant. Angular Momentum and Its Conservation Another way of expressing the angular momentum L of a particle relative to a point O is the cross product of the particle's position r relative to O with the linear momentum p of the particle. L r p r mv Angular momentum of a particle The value of the angular momentum depends on the choice of the origin O, since it involves the position vector relative to the origin The units of angular momentum: kgm2/s NOTE: If = v/r and I = mr2, then L = mr2 OR L = mvr Change in angular momentum τ (t) = Iωf – Iωi τ = torque t = time I = moment of inertia ωf = final angular velocity ωi = initial angular velocity The angular momentum of a system remains unchanged unless an external torque acts on it Spinning ice skater Arms extended, inertia is larger, velocity is smaller Pull arms in tight, inertia decreases, velocity increases Therefore, systems that can change their rotational inertia through internal forces will also change their rate of rotation: Example #1 If the ice cap at the South Pole melted and the water were uniformly distributed over the earth's oceans, what would happen to the earth’s angular velocity? Would it increase, decrease or stay the same? REASONING AND SOLUTION Consider the earth to be an isolated system. Note that the earth rotates about an axis that passes through the North and South poles and is perpendicular to the plane of the equator. If the ice cap at the South Pole melted and the water were uniformly distributed over the earth's oceans, the mass at the South Pole would be uniformly distributed and, on average, be farther from the earth's rotational axis. The moment of inertia of the earth would, therefore, increase. Example #1 con’t Since the earth is an isolated system, any torques involved in the redistribution of the water would be internal torques; therefore, the angular momentum of the earth must remain the same. If the moment of inertia increases, and the angular momentum is to remain constant, the angular velocity of the earth must decrease. Rotational Inertia, I, INCREASES BUT angular momentum, L, stays the same Since L = I Angular velocity, , DECREASES Example #2 – Pulling Through a Hole A particle of mass m moves with speed vi in a circle of radius ri on a frictionless tabletop. The particle is attached to a massless string that passes through a hole in the table as shown. The string is pulled slowly downward until the particle is a distance rf from the hole and continues to rotate in a circle of that radius. (A) Find the final velocity vf. (B) Find the tension T in the string when the particle moves in a circle of radius r in terms of the angular momentum L. vrii v f (A) Lf = Li mv r = mv r f f i i rf T = centripetal Force 2 mv2 m L L2 (B) T r r mr L = mvr v = L/mr mr3 Example #3 A skater has a moment of inertia of 3.0 kg m2 when her arms are outstretched and 1.0 kg m2 when her arms are brought in close to her sides. She starts to spin at the rate if 1 rev/s when her arms are outstretched, and then pulls her arms to her sides. (A) What is her final angular speed? (B) How much work did she do? Ii (A) fi Lf = Li Iff = Iii I f 2 3 kg m 3 rev/sec f 2 1rev / sec 1 kg m (B) Work (rotational) = KE (rotational) Work-Energy Theorem 1122 Work I 2 2 22f f i i W = (.5)(1.0)(3 x 2π) – (.5)(3)(1 x 2π) 118 J Angular Momentum Angular momentum depends on linear momentum and the distance from a particular point. It is a vector quantity with symbol L. If r and v are then the magnitude of angular momentum with respect to point Q is given by L = p∙r = m v r. In the case shown below L points out of the page. If the mass were moving in the opposite direction, L would point into the page. The SI unit for angular momentum is the kg m2 / s. (It has v no special name.) Angular momentum is a conserved quantity. A torque is r m Q needed to change L, just a force is needed to change p. Anything spinning has angular has angular momentum. The more it has, the harder it is to stop it from spinning. If r and v are not then the angle between these two vectors must be taken into account. The general definition of angular momentum is given by a vector cross product: This formula worksL regardlessr p ofr them angle.v The magnitude of the angular momentum of m relative to point Q is: L = r p sin = m v r. v r m Q Now if a net force acts on an object, it must accelerate, which means its momentum must change. Similarly, if a net torque acts on a body, it undergoes angular acceleration, which means its angular momentum changes. Recall, angular momentum’s magnitude is given by if v and r are L = mvr mutually perpendicular So, if a net torque is applied, angular velocity must change, which changes angular momentum. So net torque is the rate of change of angular momentum, just as net force is the rate of change of linear momentum. Here is yet another pair of similar equations, one linear, one rotational. From the formula v = r , we get L = mvr = m r (r ) = m r2 = I This is very much like p = mv, and this is one reason I is defined the way it is. In terms of magnitudes, linear momentum is inertia times speed, and angular momentum is rotational inertia times angular speed. NOTE: If = v/r and I = mr2, then L = mr2 OR L = mvr Suppose Mr. Stickman is sitting on a stool that swivels holding a pair of dumbbells. His axis of rotation is vertical. With the weights far from that axis, his moment of inertia is large. When he pulls his arms in as he’s spinning, the weights are closer to the axis, so his moment of inertia gets much smaller. Since L = I and L is conserved, the product of I and is a constant. So, when he pulls his arms in, I goes down, goes up, and he starts spinning much faster. I = L = I Since an external net force is required to change the linear momentum of an object. An external net torque is required to change the angular momentum of an object. It is easier to balance on a moving bicycle than on one at rest. • The spinning wheels have angular momentum. • When our center of gravity is not above a point of support, a slight torque is produced. • When the wheels are at rest, we fall over. • When the bicycle is moving, the wheels have angular momentum, and a greater torque is required to change the direction of the angular momentum. Angular momentum is conserved when no external torque acts on an object. Angular momentum is conserved for systems in rotation. The law of conservation of angular momentum states that if no unbalanced external torque acts on a rotating system, the angular momentum of that system is constant. With no external torque, the product of rotational inertia and rotational velocity at one time will be the same as at any other time. Rotational speed is controlled by variations in the body’s rotational inertia as angular momentum is conserved during a forward somersault. This is done by moving some part of the body toward or away from the axis of rotation. Example #4 What is the angular momentum of a 0.210-kg ball rotating on the end of a thin string in a circle of radius 1.10 m at an tangential speed of 5.25 m/s? m = 0.210 kg; r = 1.10 m; v = 5.25 m/s L = mvr L = (0.210 kg)(5.25 m/s)(1.10 m) L = 1.21 kg∙m2/s Example #5 What is the angular momentum of a 0.210-kg ball rotating on the end of a thin string in a circle of radius 1.10 m at an angular speed of 10.4 rads/sec? m = 0.210 kg; r = 1.10 m; = 10.4 rads/sec L = I = mr2 L = (0.210 kg)(1.10 m)2(10.4 rads/s) L = 2.64 kg-m2/s Example #6 What is the angular momentum of a 2.8-kg uniform cylindrical grinding wheel of radius 18 cm when rotating at 1500 rpm? m = 2.8 kg; r = 0.18 m; = 1500 rpm L = I 1 2 Icylinder 2 mr 2 1500rpm= 50 rads/sec 60 1 2 L 2 2.8 kg 0.18 m 50 rads / sec L = 7.1 kg-m2/s Example #7 An artificial satellite is placed into an elliptical orbit about the earth. Telemetry data indicate that its point of closest 6 approach (called the perigee) is rP = 8.37 × 10 m from the center of the earth, and its point of greatest distance (called 6 the apogee) is rA = 25.1 × 10 m from the center of the earth.
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