RUN PHY101 Part C PDF - Angular Momentum & Physics Questions

1 Angular Momentum 1.1 The Vector Product and Torque An important consideration in defining angular momentum is the process of multiplying two vectors by means of the operation called the vector product. We will introduce the vector product by considering the vector nature of torque....

1 Angular Momentum 1.1 The Vector Product and Torque An important consideration in defining angular momentum is the process of multiplying two vectors by means of the operation called the vector product. We will introduce the vector product by considering the vector nature of torque. → → → The torque vector τ is related to the two vectors r and F. We can establish a mathematical → → → relationship between τ , r , and F using a mathematical operation called the vector product: → → → τ = r ×F (1.1) → → We now give a formal definition of the vector product. Given any two vectors A and B, the vector → → → product A × B is defined as a third vector C, which has a magnitude of AB sin θ, where θ is the → → → angle between A and B. That is, if C is given by → → → C=A×B (1.2) its magnitude is C = AB sin θ (1.3) Some properties of the vector product that follow from its definition are as follows: ˆ Unlike the scalar product, the vector product is not commutative. Instead,the order in which the two vectors are multiplied in a vector product is important: → → → → A × B = −B × A (1.4) Therefore, if you change the order of the vectors in a vector product, you must change the sign. You can easily verify this relationship with the right hand rule → → → → → → ˆ If A is parallel to B (θ = 0 or 180◦ ), then A × B = 0; therefore, it follows that A × A = 0. → → → → ˆ If A is perpendicular to B, then |A × B| = AB ˆ The vector product obeys the distributive law: → → → → → → → A × (B + C) = A × B + A × C (1.5) ˆ The derivative of the vector product with respect to some variable such as t is → → d → → dA → → dB (A × B) = ×B+A× (1.6) dt dt dt From the definition of unit vectors that the cross products of the unit vectors î, ĵ,and k̂ obey the following rules: î × î = ĵ × ĵ = k̂ × k̂ = 0 î × ĵ = −ĵ × î = k̂ (1.7) ĵ × k̂ = −k̂ × ĵ = î k̂ × î = −î × k̂ = ĵ 1  → → The cross product of any two vectors A and B can be expressed in the following determinant form: → → î ĵ k̂ A Az A Ax A Ay A × B = Ax Ay Az = y î + z ĵ + x k̂ By Bz Bz Bx Bx By Bx By Bz Expanding these determinants gives the result → → A × B = (Ay Bz − Az By )î + (Az Bx − Ax Bz )ĵ + (Ax By − Ay Bx )k̂ (1.8) → → Example 1.1. Two vectors lying in the xy plane are given by the equations A = 2î + 3ĵ and B = → → → → → → −î + 2ĵ. Find A × B and verify that A × B = −B × A Solution 1.1. → → A × B = (2î + 3ĵ) × (−î + 2ĵ) → → A × B = 2î × (−î) + 2î × 2ĵ + 3ĵ × (−î) + 3ĵ × 2ĵ → → A × B = 0 + 4k̂ + 3k̂ + 0 = 7k̂ To verify → → B × A = (−î + 2ĵ) × (2î + 3ĵ) → → B × A = (−î) × 2î + (−î) × 3ĵ + 2ĵ × 2î + 2ĵ × 3ĵ → → B × A = 0 − 3k̂ − 4k̂ + 0 = −7k̂ → → → → Therefore, A × B = −B × A → Example 1.2. A force of A = (2.00î + 3.00ĵ) N is applied to an object is pivoted about a fixed axis → aligned along the z coordinate axis. The force is applied at a point located at r = (4.00î + 5.00ĵ) m. → Find the torque τ applied to the object. Solution 1.2. Set up the torque vector → → → τ = r × F = [(4.00î + 5.00ĵ) m] × [(2.00î + 3.00ĵ) N ] → τ = [(4.00)(2.00)î × î + (4.00)(3.00)î × ĵ + (5.00)(2.00)ĵ × î + (5.00)(3.00)ĵ × ĵ]N · m → τ = [0 + 12.0k̂ − 10.0k̂ + 0]N · m = 2.0k̂ N · m Angular momentum is a measure of the rotational motion of an object around a point or an axis. It plays a similar role in rotational motion to what linear momentum plays in linear motion and is a conserved quantity in closed systems, meaning it remains constant if no external torque acts on the → → system. For a particle with mass m moving with velocity v at a position r relative to an origin, the → angular momentum L with respect to that origin is defined as: → → → → → L = r × p = r × mv (1.9) 2  The SI unit of angular momentum is kg · m2 /s. Notice also that both the magnitude and the di- → rection of L depend on the choice of axis. In → → → Figure 1, r and p are in the xy plane, so L → → points in the z direction. Because p = m v, the → magnitude of L is L = mvr sin ϕ (1.10) → → where ϕ is the angle between r and p. It follows → → that L is zero when r is parallel to p (ϕ = 0 or 180◦ ). In other words, when the translational velocity of the particle is along a line that passes through the axis, the particle has zero angular momen- tum with respect to the axis. On the other → → hand, if r is perpendicular to p (ϕ = 90◦ ), then L = mvr. At that instant, the particle moves exactly as if it were on the rim of a wheel rotat- → ing about the axis in a plane defined by r and → → p. Figure 1: The angular momentum L of a particle → → → is a vector given by L = r × p 1.2 Conservation of Angular Mo- mentum The conservation of angular momentum states that if no external torque acts on a system, the total angular momentum of that system remains constant. This principle is analogous to the conser- vation of linear momentum in linear motion and is particularly important in rotational dynamics. Mathematically, it can be expressed as: → dL =0 (1.11) dt → where L is the total angular momentum of the system. Examples of conservation of angular momentum includes planetary orbits, spinning tops and gyro- scopes, neutron stars, figure skater pulling in their arms etc. 1.3 Circular Motion Circular motion is the motion of an object along a circular path. This type of motion occurs when a force continuously acts on an object perpendicular to its velocity, causing it to change direction but not speed. Circular motion can be uniform or non-uniform. ˆ Uniform Circular Motion: The object moves with a constant speed along a circular path. Although the speed is constant, the direction changes continuously, causing a continuous change in velocity. 3  ˆ Non-uniform Circular Motion: The speed of the object changes as it moves along the circular path. This results in both tangential and centripetal acceleration. Concepts in Circular Motion ˆ Angular Displacement (θ): The angle, in radians, through which the object has rotated around the center of the circle. For a complete revolution, θ = 2π radians. ˆ Angular Velocity (ω): This is the rate of change of angular displacement. For uniform circular motion, it is constant and given by ω = θt. Its unit is radians per second (rad/s). ˆ Tangential Velocity (v): This is the linear speed of the object along the circular path. It is related to the angular velocity by v = rω, where r is the radius of the circular path. ˆ Centripetal Acceleration (ac ): Is the acceleration directed towards the center of the circle, causing the change in direction of the object’s velocity. For an object moving at v along a path 2 of radius r, centripetal acceleration is ac = vr = rω 2. This acceleration does not change the speed of the object, only its direction. ˆ Centripetal Force (Fc ): The net force required to keep an object moving in a circular path, 2 directed towards the center of the circle. It is given by Fc = mac = mvr = mrω 2. Common sources of centripetal force include gravitational force (e.g planets orbiting the sun), tension in a string (e.g an object being swung in a circle) and friction (e.g car turning around a curve). Example 1.3. A model car moves round a circular track of radius 0.3 m at 2 revolutions per second. What is a. the angular speed ω, b. the period T, c. the speed v of the car? d. Find also the angular speed of the car if it moves with a uniform speed of 2 m/s in a circle of radius 0.4 m. Solution 1.3. a. For 1 revolution, angle turned θ = 2π rad (360◦ ). So ω = 2 × 2π = 4π rad/s b. Period T 2π 2π = time for 1 rev = = = 0.5 s ω 4π c. Speed v = rω = 0.3 × 4π = 1.2π = 3.8 m/s d. From v = rω v 2 ω= = = 5 rad/s r 0.4 4 2 Newton’s Gravitation Law Newton’s Law of Universal Gravitation states that every mass in the universe attracts every other mass with a force that is directly proportional to the product of their masses and inversely propor- tional to the square of the distance between them. Mathematically, it can be expressed as: Gm1 m2 F = (2.1) r2 where: F is the gravitational force between the two masses, m1 and m2 are the masses of the two objects, G is the gravitational constant (approximately 6.674 × 10−11 N · m2 /kg 2 ), and r is the distance between the centre of the two masses. Example 2.1. Determine the force of attraction between the sun (ms = 1.99 × 1030 kg) and the earth (me = 5.98 × 1024 kg). Assume the sun is 1.50 × 108 km from the earth. Solution 2.1. Gms me F = r2 r = 1.5 × 108 km = 1.5 × 108 × 103 m 6.67 × 10−11 × 1.99 × 1030 × 5.98 × 1024 F = (1.5 × 108 × 103 )2 F = 3.53 × 1022 N Example 2.2. Two 5.0 kg spherical balls are placed so that their centers are 50.0 cm apart. What is the magnitude of the gravitational force between the two balls? Solution 2.2. Gm1 m2 F = r2 r = 50 cm = 0.5 m 6.67 × 10−11 × 5.0 × 5.0 F = (0.5)2 F = 6.67 × 10−9 N 2.1 Kepler’s Laws of Planetary Motion Kepler’s laws of planetary motion state that: Law 1: the planets describe ellipses about the sun as one focus. Law 2: the line joining the sun and the planet sweeps out equal areas in equal times. Law 3: the squares of the periods of revolution of the planets are proportional to the cubes of their mean distances from the sun. The force acting on the planet of mass m is mrω 2 , where r is the radius of the circle and ω is the angular speed of the motion. Since ω = 2π/T , where T is the period of the motion,  2 2π 4π 2 mr force on planet = mr = T T2 5 This is equal to the force of attraction of the sun on the planet. Assuming an inverse-square law for the distance r, then, if k is a constant, km force on planet = r2 km 4π 2 mr ∴ = r2 T2 4π 2 3 ∴ T2 = r k since k, π are constants T 2 ∝ r3 (2.2) 2.2 Relation between the Gravitational Constant ‘G’ and the accelera- tion of gravity at the earth surface ‘g’ The relationship between the acceleration due to gravity (g) and the gravitational constant (G) is derived from Newton’s law of gravitation and relates these quantities to the mass and radius of a celestial body. Gme m F = = mg (2.3) (re )2 F The force per unit mass, m , is given by: F Gme = =g m (re )2 Hence, Gme g= (2.4) (re )2 This means that the acceleration due to gravity ‘g’ can be considered as the force per unit mass on the earth’s surface. If ‘g’ and ‘G’ are accurately known, the above equation can be use to calculate the mass of the earth, me. gr2 me = e (2.5) G Example 2.3. Determine the mass of the earth if the radius of the earth is approximately 6.38 × 106 m, G = 6.67 × 10−11 N m2 kg −2 and g = 9.80 ms−2 Solution 2.3. gre2 me = G 9.80 × (6.38 × 106 )2 me = 6.67 × 10−11 me = 5.98 × 1024 kg Example 2.4. The radius of the moon is one-fourth, and its mass is one eighty-first that of the earth. If the acceleration due to gravity on the surface of the earth is 9.8 ms−2 , what is the value on the moon’s surface. 6 Solution 2.4. The relation between g and G is given by: Gme g= re2 For the earth, Gme ge = re2 For the moon, Gmm gm = 2 rm Since G is the universal constant of gravitation it is constant for both equations. re rm = 4 me mm = 81 From Gme ge re2 ge = , G = re2 me Substituting this into Gmm gm = 2 rm we have: ge re2 mm gm =. 2 me rm re mm = 9.8.( )2. rm me 1 1 gm = 9.8 × (4)2 × = 9.8 × 16 × 81 81 −2 = 1.9 ms 2.3 Gravitation Potential Energy Gravitational potential pnergy at a point is defined as the workdone in taking unit mass from infinity to that point. Unit is Jkg −1. The gravitational potential V is given by: Gm V =− (2.6) r where m is the mass producing the gravitational field and r is the distance of the point to the mass. The gravitational potential decreases as r increases and becomes zero where r is infinitely large. The negative sign indicates that the potential at infinity (zero) is higher than the potential close to the mass. Example 2.5. Find the gravitational potential at a point on the earth’s surface. Take mass of earth as 5.98 × 1024 kg, its radius as 6.38 × 106 m and G = 6.67 × 10−11 N m2 kg −2. Solution 2.5. Gm V = r −11 6.67 × 10 × 5.98 × 1024 V = 6.38 × 106 V = 6.25 × 107 Jkg −1 7 2.4 Escape Velocity Escape velocity (ve ) can be define as the minimum velocity required for an object to just escape or leave the gravitational influence or field of an astronomical body permanently. Escape velocity (ve ) is given by: p ve = 2gR (2.7) where R is the earth’s radius. Example 2.6. Calculate the escape velocity of a satellite from the earth’s gravitational field. g = 9.8 ms−2 , radius of earth = 6.4 × 106 m Solution 2.6. p ve = 2gR √ ve = 2 × 9.8 × 6.4 × 106 ve = 11 × 103 ve = 11 kms−1 Thus, to escape the gravitational influence of the earth, the satellite must be projected with a velocity greater than 11 km−1. 2.5 Satellite Motion and Orbits Satellite motion refers to the movement of a satellite under the influence of gravitational forces. A satellite is any object that orbits around a larger body, such as a planet or star. Understanding satellite motion requires concepts from Newtonian mechanics and gravitation. A satellite stays in orbit due to the balance between: ˆ Gravitational Force: The force pulling the satellite toward the center of the planet (or larger body). ˆ Centripetal Force: The force required to keep the satellite moving in a circular or elliptical orbit. The radius of the orbit is r, measured from the center of the earth: the acceleration of the satellite has magnitude arad = v 2 /r and is always directed toward the centre of the circle. By the law of gravitation, the net force (gravitational force) on the satellite of mass m has a magnitude Fg = Gmr2E m P→ → and is in the same direction as the acceleration. Newton’s second law ( F = m a) then tells us that GmE m mv 2 = (2.8) r2 r solving for v gives r GmE v= (2.9) r This relationship shows that we can’t choose the orbit radius r and the speed v independently: for a given radius r, the speed v for a circular orbit is determined. 8 We can derive a relationship between the radius r of a circular orbit and the period T , the time for one revolution. The speed v is the distance 2πr traveled in one revolution, divided by the period: 2πr v= (2.10) T Solving equation (2.10) for T and substitute v from equation (2.9): r s s 2πr r r3 r3 T = = 2πr = 2π = 2π (2.11) v GmE GmE grE Equations (2.9) and (2.11) show that larger orbits correspond to slower speeds and longer periods. If the period of the satellite in the orbit is exactly equal to the period of the earth as it turns about its axis, which is 24 hours, the satellite will stay over the same place on the earth while the earth rotates. This is sometimes called a ‘parking orbits’. Relay satellites can be placed in parking orbits, so that television programmes can be transmitted continously from one part of the world to another. Since T = 24 hours, the radius R can be found from 2.11. Its value is r 2 2 3 T grE R= and g = 9.8 m/s2 , rE = 6.4 × 106 m 4π 2 r (24 × 3600)2 × 9.8 × (6.4 × 106 )2 3 ∴R= = 42400 km 4π 2 The height above the earth’s surface of the parking orbit = R − rE = 42400 − 6400 = 36000 km In the orbit, assuming it is circular, the speed of the satellite 2πR 2π × 42400 km = = = 3.1 km/s T 24 × 3600 s Exercises → → → → 1. Given that M = 2î − 3ĵ + k̂ and N = 4î + 5ĵ − 2k̂, calculate the vector product M × N. → → 2. Two vectors are given by A = î + 2ĵ and B = −2î + 3ĵ. Find: → → i. A × B → → ii. the angle between A and B → 3. A particle is located at a vector position r = (4.00î + 6.00ĵ) m and a force exerted on it is → given by F = (3.00î + 2.00ĵ) N. i. What is the torque acting on the particle about the origin? ii. Can there be another point about which the torque caused by this force on this particle will be in the opposite direction and half as large in magnitude? iii. Can there be more than one such point? 9  iv. Can such a point lie on the y axis? v. Can more than one such point lie on the y axis? vi. Determine the position vector of one such point. → 4. A particle of mass 1.50 kg moves in the xy plane with a velocity of v = (4.20î − 3.60ĵ) m/s. Determine the angular momentum of the particle about the origin when its position vector is → r = (1.50î + 2.20ĵ) m. → 5. A particle of mass m moves in the xy plane with a velocity of v = (vx î + vy ĵ). Determine the → angular momentum of the particle about the origin when its position vector is r = (xî + y ĵ). 6. An object of mass 4 kg moves round a circle of radius 6 m with a constant speed pf 12 m/s. Calculate i. the angular speed ii. the force towards the centre 7. An object of mass 8.0 kg is whirled round rapidly in a vertical circle of radius 2 m with a constant speed of 6 m/s. Calculate the maximum and minimum tensions in the string. [224 N, 64 N] 8. Two objects attract each other with a gravitational force of magnitude 1.00 × 10−8 N when separated by 20.0 cm. If the total mass of the two objects is 5.00 kg, what is the mass of each? 9. The free-fall acceleration on the surface of the Moon is about one-sixth that on the surface of the Earth. The radius of the Moon is about 0.250RE (RE = Earth’s radius = 6.37 × 106 m). Find the ratio of their average densities, ρM oon /ρEarth. 10. A satellite of mass 200 kg is placed into Earth orbit at a height of 200 km above the surface. i. Assuming a circular orbit, how long does the satellite take to complete one orbit? ii. What is the satellite’s speed? iii. Starting from the satellite on the Earth’s surface, what is the minimum energy input necessary to place this satellite in orbit? Ignore air resistance but include the effect of the planet’s daily rotation 11. A satellite of mass m, originally on the surface of the Earth, is placed into Earth orbit at an altitude h. i. Assuming a circular orbit, how long does the satellite take to complete one orbit? ii. What is the satellite’s speed? iii. What is the minimum energy input necessary to place this satellite in orbit? Ignore air resistance but include the effect of the planet’s daily rotation. Represent the mass and radius of the Earth as ME and RE , respectively. 10