# Version PREVIEW Practice 8 carroll (11108) 1

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1 Version PREVIEW Practice 8 carroll This print-out should have 12 questions. Multiple-choice questions may continue on the net column or page find all choices before answering. Inertia of Solids points A circular disk, a ring, and a square have the same mass M and width 2 r points A small metallic bob is suspended from the ceiling by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread describes a cone. The acceleration of gravity is 9.8 m/s 2. ring disk square 9.8 m/s m 2 r 2 r 2 r For the moment of inertia about their center of mass about an ais perpendicular to the plane of the paper, which statement concerning their moments of inertia is true? 1. I square > I ring > I disk correct v 5 kg Calculate the magnitude of the angular momentum of the bob about the supporting point. Correct answer: kg m 2 /s. 2. I disk > I square > I ring 3. I disk > I ring > I square Basic Concepts: Angular Momentum 4. I square > I disk > I ring 5. I ring > I disk > I square L = m r v. 6. I ring > I square > I disk In the ring, the same mass of the disk is concentrated at the maimum distance from the ais, so I ring > I disk. Note: In this problem r and v are perpendicular, where r = l sin. In the square, the same mass of the ring lies at distances which are between r and r 2 at least the radius of the ring, so I square > I ring and I square > I ring > I disk. Let : l = 2.4 m, = 34, g = 9.8 m/s 2, m = 5 kg. and Conical Pendulum 04 Use the free body diagram below.

2 T Version PREVIEW Practice 8 carroll bar is fastened by a pivot at one end to a wall which is at an angle with respect to the horizontal. The bar is held horizontal by a vertical cord that is fastened to the bar at a distance a distance from the wall. A mass m 2 is suspended from the free end of the bar. T mg Solution: The second Newton s law in the vertical and horizontal projections, respectively, in our case reads T cos mg = 0 T sin mω 2 l sin = 0, where T is the force with which the wire acts on the bob and the radius of the orbit is R = l sin. From this system of equations we find g ω = l cos = 9.8 m/s m cos34 = rad/s. The angular momentum L then will amount to L = mω l sin 2 g l = m sin 2 3 cos = 5 kg sin m/s m 3 cos34 = kg m 2 /s. L Find the tension T in the cord. 1. T = 0 2. T = 2 m 1 + m 2 m 1 m 2 g correct 3. T = m 1 + m 2 g cos 4. T = 2 m 1 + m 2 g cos cord L g 5. T = m 1 + m T = m L 2 m 2 g cord 7. T = m L 2 m 2 g cos cord 8. T = m L 2 m 2 g sin cord 9. T = 2 m 1 + m 2 g sin 10. T = m 1 + m 2 g sin Basic Concepts: Static Equilibrium: F = 0. Beam on a Slanted Wall part 1 of points A solid bar of length L has a mass m 1. The τ = 0. Torque: τ = r F = r F

3 Version PREVIEW Practice 8 carroll We can begin by either summing the torques or the forces and setting them equal to zero. In this case, since we know little about the reaction force between the bar and the wall, it is easiest to begin by eamining the torques around the connection between the bar and the wall, because the reaction forces will produce no torques around that point. τ = L m2 g + T L 2 m 1 g = 0. Solving this for T gives T = 2 m 1 + m 2 g. 004 part 2 of points Find the the horizontal component of the force eerted on the bar by the wall. Take the positive direction to be right. 1. F = m 1 + m 2 g cos 2. F = m 1 + m 2 g sin 3. F = 2 m 1 + m 2 g cos cord L g 4. F = m 1 + m F = m L 2 m 2 g sin cord 6. F = m L 2 m 2 g cord 7. F = 2 m 1 + m 2 g sin 8. F = 0 correct 9. F = 2 m 1 + m F = m L 2 m 2 g g cos In general, there will be a horizontal reaction force R at the connection between an object and a wall. However, none of the other forces in this situation act in the horizontal direction. Therefore: F = R = 0. Rotation of a Solid Disk points A uniform solid disk of radius 6.85 m and mass 34.8 kg is free to rotate on a frictionless pivot through a point on its rim. Pivot 6.85 m If the disk is released from rest in the position shown by the solid circle, what is the speed of its center of mass when the disk reaches the position indicated by the dashed circle? The acceleration of gravity is 9.8 m/s 2. Correct answer: m/s. Let : r = 6.85 m and m = 34.8 kg. From the parallel ais theorem I = 1 2 M R2 + M R 2 = 3 2 M R2. so the final kinetic energy is K f = M R2 = M v 2 ω 2 = 3 4 M v2 = 3 4 M R2 ω 2. From conservation of energy K + U i = K + U f 0 + M g R = 1 2 I ω2 M g R = 3 4 M R2 ω 2 4 g ω = 3 R

4 Version PREVIEW Practice 8 carroll Hence the velocity of the center of mass is v cm = ω R R g = m 9.8 m/s = = m/s. Alternate Solution The kinetic energy is K total f keywords: = K rotational f cm linear + Kf K f = M R2 ω M v2 = M v M v2 = 3 4 M v2 = 3 4 M R2 ω 2. Solid Sphere on an Incline part 1 of points A solid sphere of radius 34 cm is positioned at the top of an incline that makes 26 angle with the horizontal. This initial position of the sphere is a vertical distance 1.6 m above its position when at the bottom of the incline. The sphere is released and moves down the incline. 34 cm l µ M m Calculate the speed of the sphere when it reaches the bottom of the incline if it rolls without slipping. The acceleration of gravity is 9.8 m/s 2. The moment of inertia of a sphere with respect to an ais through its center is 2 5 M R2. Correct answer: m/s. From conservation of energy we have U i = K trans,f + K rot,f M g h = 1 2 M v I ω2 = 1 2 M v v M R2 R 2 = 7 10 M v2.v 1 = 10 = m/s2 1.6 m = m/s g h 007 part 2 of points Calculate the speed of the sphere if it reaches the bottom of the incline by slipping frictionlessly without rolling. Correct answer: 5.6 m/s. From conservation of energy we have U i = K trans,f M g h = 1 2 M v2 v 2 = 2 g h = m/s m = 5.6 m/s. keywords: /* If you use any of these, fi the comment symbols. g=9.8 ; * u=m/s 2 h = e 34; u = J s Someof yourvariablesresembleconstants. Airplane Momentum 008 part 1 of points An airplane of mass kg flies level to the ground at an altitude of 16 km with a constant speed of 172 m/s relative to the Earth. What is the magnitude of the airplane s angular momentum relative to a ground observer directly below the airplane in kg m 2 /s? Correct answer: kg m 2 /s.

5 Version PREVIEW Practice 8 carroll Since the observer is directly below the airplane, L = h mv 009 part 2 of points Does this value change as the airplane continues its motion along a straight line? 1. Yes. L changes with certain period as the airplane moves. 2. Yes. L decreases as the airplane moves. 3. Yes. L increases as the airplane moves. 4. No. L = constant. correct 5. Yes. L changes in a random pattern as the airplane moves. L = constant since the perpendicular distance from the line of flight to Earth s surface doesn t change. Asteroid Collision 010 part 1 of points Consider an Earth-like planet hit by an asteroid. The planet has mass M p = kg and radius R p = m, and you may approimate it as a solid ball of uniform density. It rotates on its ais once every T = 16 hr. The asteroid has mass M a = kg and speed v a = m/s relative to the planet s center; its velocity vector points = 70 below the Eastward horizontal. The impact happens at an equatorial location. The picture below shows the view from above the planet s North pole: ω R v m First, calculate the planet s angular momentum relative to its spin ais before the impact. Correct answer: kg m 2 /s. Basic Concept: A rigid body rotating around a fied ais has angular momentum L = I ω where I is the body s moment of inertia about the ais of rotation. Approimating the planet as a solid ball of uniform density, its moment of inertia is I p = 2 5 M R2 = kg m 2. Its angular velocity before the impact is ω = 2 π T = rad/s, so L = I ω = I ω = kg m rad/s = kg m 2 /s. 011 part 2 of points Calculate the asteroid s angular momentum relative to the planetary ais. Correct answer: kg m 2 /s. Approimating the asteroid as a pointlike particle, its angular momentum is L a = R M a v a where R is the asteroid s radius vector and M a v a is its linear momentum vector. At the moment of impact, both the radius vector R and the momentum vector M a v a of the asteroid lie in the planet s equatorial plane. Consequently, their vector product is perpendicular to the equatorial plane and

6 Version PREVIEW Practice 8 carroll parallel to the planet ais. Because the horizontal component of the asteroid s velocity is directed Eastward the same as the planet s rotation the asteroid s angular momentum L a has the same direction as the planet s angular momentum L p. In magnitude, L a = R p M a horizontal component of v a = R p M a v a cos = m kg m/s cos70 = kg m 2 /s. 012 part 3 of points The impact is totally inelastic the asteroid is stuck in the planet s crust. But thanks to the asteroid s angular momentum, the planet rotates faster after the impact than it did before. By how many seconds has the collision shortened the planetary day? For simplicity, ignore the effect of the asteroid s mass on the planet s moment of inertia and assume I after = I before planet. Warning: At intermediate stages of this calculation, do not round off intermediate results and keep at least 7 significant digits. Correct answer: s. The net angular momentum is conserved in this collision, so after the impact the planet with the asteroid stuck in its crust has and since we assume unchanged moment of inertia I p = I p, it follows that after the impact, the planet rotates at new angular velocity ω = L p I p In other words, = ω + R p M a v a cos I p. ω = ω ω = R p M a v a cos I p = 5 M a v a cos 2 M p R p = rad/s. In terms of the planetary day, this change means T = T T = 2 π ω 2 π ω = 2 πω ω ω ω = 2 π ω 1 ω + ω ω = 2 π rad/s rad/s rad/s = s, where ω + ω = rad/s rad/s = rad/s. L p = L p + L a. Focusing on the Northward component of L the other two components vanish, we have L p = L p + L a At the same time, = I p ω + R p M a v a cos. L p = I P ω,

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