JL IV

PHY 1301, Physics I 1

Course Learning Outcomes for Unit IV Upon completion of this unit, students should be able to:

4. Apply the concept of momentum conservations to daily life. 4.1 Relate impulse-momentum theorem to Newton’s second law. 4.2 Show the relationship between linear momentum conservation and Newton’s third law. 4.3 Apply momentum conservation to rotational kinematics.

5. Identify the total mechanical energy conservation.

5.1 Interpret total kinetic energy conservation in elastic collision. Required Unit Resources Chapter 7: Impulse and Momentum Chapter 8: Rotational Kinematics Unit Lesson

Impulse-Momentum Relation and Newton’s Second Law

Momentum, or the linear momentum (p) of an object, is defined by its mass (m) times velocity (v): p = mv. Mass is an intrinsic property of an object, and it is the quantity of matter in the object. Mass is a scalar quantity and velocity is a vector. Thus, the direction of momentum is the same as the velocity points. Momentum can be related to force. The average force <F> acting on a body can be obtained by the momentum change ∆p per a time interval ∆t. That is, <F> = m <a> = m ∆v / ∆t = ∆p / ∆t. The product of <F> and ∆t is the impulse J: J = <F> ∆t = ∆p. The impulse is the same as the change in momentum, which is the final momentum minus initial momentum; this is the impulse-momentum theorem. The impulse-momentum theorem is another expression of Newton’s second law. Note that both are measured in kg m/s, which is equal to N s.

Suppose a stuntman is trying to jump from a tall building, and he has a choice to land on sand or water. If he lands on water, his stopping time is much greater than if he lands on the sand. The water exerts a smaller average force according to the impulse-momentum theorem because the sand and the water exert the same impulse on him to halt the motion. Therefore, he should land on water to diminish damage.

UNIT IV STUDY GUIDE Kinematics of Collisions

Sample Question 1: What impulse occurs when an average force of 20 N is exerted on a boxcar for 3 seconds? Solution: The impulse J = <F> ∆t. The average force <F> is given by 20 N, and the time interval ∆t is 3 s. So, the impulse J = 20 x 3 = 60 N s.

Impulse = Average Force x Time Interval = Change in Momentum

PHY 1301, Physics I 2

UNIT x STUDY GUIDE Title

Linear Momentum Conservation and Newton’s Third Law

The total linear momentum of a system without external forces is conserved, that is, constant. This is one of the most important and fundamental principles to support the action-reaction law of Newton. Suppose that there are two colliding balls, A and B, whose masses are mA and mB. There are two forces: the force exerted on A by B, FAB and the force exerted on B by A, FBA.

According to Newton’s second law, the total force of this system is F = FAB + FBA. Notice that F is 0 due to Newton’s third law; FAB = -FBA. That is, F = ma = m ∆v / ∆t = ∆p / ∆t = 0; momentum is constant (Cutnell et al., 2022). The total momentum before the collision is the same as the total momentum after collision; the total linear momentum is conserved. If it is not conserved, Newton’s law of action-reaction will no longer be valid.

Elastic and Inelastic Collision

In general, there are two kinds of collision: elastic and inelastic. If the total kinetic energy of the system is conserved, it is called an elastic collision. If it is not conserved, it is called inelastic collision (Cutnell et al., 2022). In any case, the total linear moment is conserved. When two atoms collide, the total kinetic energy remains constant before and after the collision in most cases. This is an example of an elastic collision. If the first atom gains kinetic energy, the second atom lost the energy. On the other hand, when two cars collide, the kinetic energy is lost mainly because of friction. The total kinetic energy is not constant in this inelastic collision.

Sample Question 3: Boxcar A whose mass mA = 5,000 kg and velocity vA = 1 m/s will link to boxcar B whose mass mB = 15,000 kg and velocity vB = 4 m/s. Find the final velocity V after they are connected. Assume that there is no friction in this system.

Solution: According to the momentum conservation, the total momentum before linking, mAvA + mBvB should be equal to the total momentum after linking, (mA+ mB)V. That is, V = (mAvA + mBvB) / (mA + mB) = (5000 x 1 + 15000 x 4) / 20000 = 13 / 4 = 3.25 m/s.

Sample Question 2: A 0.2 kg model rocket is constructed with a motor that can provide a total impulse of 30 N s. What is the speed that this rocket achieves when launched from rest? Ignore the effects of gravity and air resistance. Solution: According to the impulse-momentum theorem, the rocket’s final momentum mV differs from its initial momentum mv by an amount equal to the impulse J of the net force exerted on it. That is J = mV - mv. Here, m is 0.2 kg, and v is 0 m/s because the rocket starts from rest. We are ignoring gravitational and frictional forces, so this impulse is due entirely to the force generated by the motor. The magnitude of the motor’s impulse is given as J = 30 N s. Therefore, the rocket’s final speed V is the magnitude of the motor impulse divided by the rocket’s mass: V = J / m = 30 / 0.2 = 150 m/s.

PHY 1301, Physics I 3

UNIT x STUDY GUIDE Title

Angular Velocity, Angular Acceleration, and Angular Displacement

When describing a rotational motion such as spinning top, rotating fan blades, and rolling wheels, we use angular velocity, angular acceleration, and angular displacement. Remember that we have studied the relationship between velocity, acceleration, and displacement in the first unit to describe the linear motion of an object. The angular displacement has an analogous idea of linear displacement. The angular displacement is defined as the angular velocity multiplied by the elapsed time of the motion.

Angular Displacement = Angular Velocity x Elapsed Time

Sample Question 4: Consider an elastic head-on collision between a green ball and an orange ball. The mass m of the green ball is 1 kg and the mass M of the orange ball is 3 kg. Before the collision, the orange ball was at rest, and the initial velocity of the green ball was 5 m/s. After the collision, the final speed V of the orange ball is identical to that of the green ball. However, they have a different velocity; the direction of the green ball is the opposite to that of the orange ball. What are the velocities after the collision?

Solution: The linear momentum is conserved during the collisional process. The linear momentum before the collision is the same as that after the collision: mv = MV + (-mV) So, the final speed V = mv / (M - m) = (1 x 5) / (3 - 1) = 5 / 2 = 2.5 m/s. The final velocity of the orange ball is +2.5 m/s and that of the green ball is -2.5 m/s. In an elastic collision, the total kinetic energy is conserved. That is, total kinetic energy before the collision is equal to that after the collision.

Total kinetic energy before the collision is ½ mv2 = 0.5 x 1 x 25 = 12.5 J. Total kinetic energy after the collision is ½ mV2 + ½ MV2 = (0.5 x 1 x 6.25) + (0.5 x 3 x 6.25) = 12.5 J. That is, total kinetic energy of the system is conserved.

PHY 1301, Physics I 4

UNIT x STUDY GUIDE Title

The SI unit of angular displacement is radian or rad. Please note, 1 radian is equivalent to 360 degrees divided by 2π radians. So, 1 radian is equal to 57.3 degrees. The SI unit of angular velocity is radian per second or rad/s. The angular velocity is related to the angular acceleration as follows. The change in angular velocity is the product of average angular acceleration and the elapsed time of the motion. The SI unit of angular acceleration is radian per second squared or rad/s2.

Small Angle Formula

What is parallax? In order to see what parallax is, hold your index finger in front of your nose, a few centimeters away, and look at it first with the left eye closed, and then with the right eye closed. If you compare its position with some background objects, you will notice that, apparently, the finger moves when you switch from one eye to the other! This explanation is obvious: your eyes are not in the same place, and hence they have a different line of sight to the finger and have to look at a different angle in order to see the finger. The angles at which your eyes have to look obviously depend on the distance of the finger to your nose. What is more important is that the difference in the angles depends on the distance; the angle gets smaller as the distance gets greater. Therefore, the difference in angles, the parallax, can be used to measure the distance. This method is known as trigonometric or heliocentric parallax and can be used to measure up to 100 parsecs (pc) or 0.01 arcseconds. The distance between your eyes is called the baseline; the bigger the baseline (here, D), the bigger the parallax, for the same distance, d, of the observed object to the baseline.

We define the trigonometric parallax of the star as the angle θ subtended as seen from the star by the orbital radius of the Earth, 1 AU: in other words D = 1 AU. One parsec (pc), parallax second, is the distance d when the star has the parallax angle θ = 1 arcsecond = 1 / 206,264.81 radian. That is, d = D / θ = 206, 265 AU = 3.086 × 1016 m = 3.26 ly = 1 pc

Sample Question 5: The nearest star, Proxima Centauri, has a parallax of 0.76". Calculate the

distance of the star from us.

Solution: θ = 0.76” = 0.76/206264.81 = 3.684 × 10-6 radian

d = 206264.81 / 0.76 = 271,401 AU = 1.32 pc = 4.3 ly

PHY 1301, Physics I 5

UNIT x STUDY GUIDE Title

Centripetal Acceleration and Tangential Acceleration

When an object moves in a curved path, its total acceleration is the vector sum of the centripetal acceleration and the tangential acceleration. The magnitude of the total acceleration squared is equal to the centripetal acceleration squared plus the tangential acceleration squared. As we have learned about centripetal acceleration in Unit III, it maintains the circular motion. The direction of the centripetal acceleration points to the center of the circle. The direction of the tangential acceleration is perpendicular to that of the centripetal acceleration. The tangential acceleration exists because the tangential velocity of the object is increasing as it moves. The tangential acceleration is the product of the orbital radius and the angular acceleration in rad/s2. The tangential velocity is the product of the orbital radius and the angular velocity in rad/s.

Motions of Celestial Bodies

We can determine some obvious results from short observations of the sky. All the celestial objects move from the direction where the sun rises to the direction where the sun sets every day; this is known as diurnal motion. Also, the entire sky turns around any fixed points once a day. When one observes the night sky at the northern hemisphere, one would notice that the stars move counterclockwise around a fixed star in the north: the North Star. The North Star does not move because its direction aligns with the rotational axis of the Earth. The North Star is now the Polaris, but in the future, it will not be so, due to the Earth’s precession. The diurnal motion of stars is one of the consequences of the Earth’s rotation. The direction of the rotation of the Earth is the opposite of the diurnal motion’s direction. The common center of the circle of stars is the North Star, and stars are moving 15 degrees an hour in a counterclockwise direction. Diurnal motions of stars differ in different places. If one lives in the northern hemisphere at the mid-latitude area, then one would observe rising stars in the eastern sky, setting stars in the western sky, moving stars from east to west in the southern sky , and counterclockwise circling around Polaris in the northern sky at night. At the equator, the stars move from the east to west, perpendicular to the horizon. At the poles, the stars move parallel to the horizon. The sidereal day is defined as the time for the vernal equinox to return to your celestial meridian. It is the time, in fact, that it takes the Earth to complete one rotation with respect to the stars. It is the true rotation period of the Earth. The solar day is the time for the sun to return to your celestial meridian. Locally, it is the time from high noon to the next one. The Earth is also revolving around the sun, so the solar day is about 4 minutes longer than the sidereal day. The orbital period of the Earth around the sun is about 365 days, and the angle needed to complete one circle is 360 degrees. Therefore, the angle per day for the Earth to see the sun overhead position is 360 / 365 = 0.986 degrees. That is, the sun appears to move eastward by about 1 degree per day with respect to the background stars. There are 24 hours in a full circle (= 360o). Also, 1 hour is equal to 60 minutes. That is, 24 x 60 / 360 = 4 minutes. One degree is 4 minutes. After the Earth rotates from east to west for about 4 minutes (= 1 degree), the sun will rise over your head, your celestial meridian. However, if the Earth’s rotation direction suddenly changed from west to east, you will see the high sun 4 minutes before the complete 1 sidereal day. So, 4 + 4 = 8. The solar day will be about 8 minutes shorter than the original value. In the case of the moon, the sidereal period is 27.3 days. The moon must move eastward among the background stars to go completely around the sky (360 degrees) in the sidereal period. The moon moves about 13 degrees (= 360 / 27.3) a day. Since the celestial sphere appears to turn 1 degree about every 4 minutes, the moon crosses our celestial meridian about 13.2 x 4 = 52.8 minutes later each day. That is, the sun rises about one degree (about 4 minutes) and the moon rises about 13 degrees (about 53 minutes) later each day, and, in the long run, they are moving from the west to east along the ecliptic. The rotational axis of the Earth has an angle of 23.5 degrees from the celestial north. It is not fixed like a spinning top and undergoes a very slow turn, or precession, over time. Its precession period is about 26,000 years. The gravitational attraction of the sun and the moon on the Earth’s equatorial bulge is responsible for this effect. As a consequence, the vernal equinox, the intersection of the equator and ecliptic, regresses by about 50” per year along the ecliptic. In the present, the North Star is the Polaris, but it will change to Vega in 13,000 years. In another 13,000 years, Polaris will be the North Star again. The position of the equinoxes changes, as well, with respect to the background stars, so the coordinates of celestial objects change as well, over a period of 26,000 years. This might be a negligible amount, but it is significant for an exact calculation. Because the moon and the sun move above and below the Earth’s equatorial plane, periodic variations occur

PHY 1301, Physics I 6

UNIT x STUDY GUIDE Title

in the torques acting on the Earth’s equatorial bulge. These variations lead to a nutation or wobbling, of the Earth’s rotation axis. The nutation is a periodic oscillation of the rotational axis of the Earth around its “mean” position. Small contributions to the nutation have monthly and yearly periods, but the principal contribution (discovered by James Bradley) of amplitude 9” and period 18.6 years is from the regression of the nodes of the moon’s orbit.

Reference

Cutnell, J. D., Johnson, K. W., Young, D., & Stadler, S. (2022). Physics (12th ed.). Wiley. Learning Activities (Nongraded) Nongraded Learning Activities are provided to aid students in their course of study. You do not have to submit them. If you have questions, contact your instructor for further guidance and information.

1. Solve Questions 53–60 under Physics in Biology, Medicine, and Sports in Chapter 7 of your eTextbook.

2. Solve Questions 62–69 under Physics in Biology, Medicine, and Sports in Chapter 8 of your eTextbook.

PHY 1301, Physics I 1

Course Learning Outcomes for Unit IV Upon completion of this unit, students should be able to:

4. Apply the concept of momentum conservations to daily life. 4.1 Relate impulse-momentum theorem to Newton’s second law. 4.2 Show the relationship between linear momentum conservation and Newton’s third law. 4.3 Apply momentum conservation to rotational kinematics.

5. Identify the total mechanical energy conservation.

5.1 Interpret total kinetic energy conservation in elastic collision. Required Unit Resources Chapter 7: Impulse and Momentum Chapter 8: Rotational Kinematics Unit Lesson

Impulse-Momentum Relation and Newton’s Second Law

Momentum, or the linear momentum (p) of an object, is defined by its mass (m) times velocity (v): p = mv. Mass is an intrinsic property of an object, and it is the quantity of matter in the object. Mass is a scalar quantity and velocity is a vector. Thus, the direction of momentum is the same as the velocity points. Momentum can be related to force. The average force <F> acting on a body can be obtained by the momentum change ∆p per a time interval ∆t. That is, <F> = m <a> = m ∆v / ∆t = ∆p / ∆t. The product of <F> and ∆t is the impulse J: J = <F> ∆t = ∆p. The impulse is the same as the change in momentum, which is the final momentum minus initial momentum; this is the impulse-momentum theorem. The impulse-momentum theorem is another expression of Newton’s second law. Note that both are measured in kg m/s, which is equal to N s.

Suppose a stuntman is trying to jump from a tall building, and he has a choice to land on sand or water. If he lands on water, his stopping time is much greater than if he lands on the sand. The water exerts a smaller average force according to the impulse-momentum theorem because the sand and the water exert the same impulse on him to halt the motion. Therefore, he should land on water to diminish damage.

UNIT IV STUDY GUIDE Kinematics of Collisions

Sample Question 1: What impulse occurs when an average force of 20 N is exerted on a boxcar for 3 seconds? Solution: The impulse J = <F> ∆t. The average force <F> is given by 20 N, and the time interval ∆t is 3 s. So, the impulse J = 20 x 3 = 60 N s.

Impulse = Average Force x Time Interval = Change in Momentum

PHY 1301, Physics I 2

UNIT x STUDY GUIDE Title

Linear Momentum Conservation and Newton’s Third Law

The total linear momentum of a system without external forces is conserved, that is, constant. This is one of the most important and fundamental principles to support the action-reaction law of Newton. Suppose that there are two colliding balls, A and B, whose masses are mA and mB. There are two forces: the force exerted on A by B, FAB and the force exerted on B by A, FBA.

According to Newton’s second law, the total force of this system is F = FAB + FBA. Notice that F is 0 due to Newton’s third law; FAB = -FBA. That is, F = ma = m ∆v / ∆t = ∆p / ∆t = 0; momentum is constant (Cutnell et al., 2022). The total momentum before the collision is the same as the total momentum after collision; the total linear momentum is conserved. If it is not conserved, Newton’s law of action-reaction will no longer be valid.

Elastic and Inelastic Collision

In general, there are two kinds of collision: elastic and inelastic. If the total kinetic energy of the system is conserved, it is called an elastic collision. If it is not conserved, it is called inelastic collision (Cutnell et al., 2022). In any case, the total linear moment is conserved. When two atoms collide, the total kinetic energy remains constant before and after the collision in most cases. This is an example of an elastic collision. If the first atom gains kinetic energy, the second atom lost the energy. On the other hand, when two cars collide, the kinetic energy is lost mainly because of friction. The total kinetic energy is not constant in this inelastic collision.

Sample Question 3: Boxcar A whose mass mA = 5,000 kg and velocity vA = 1 m/s will link to boxcar B whose mass mB = 15,000 kg and velocity vB = 4 m/s. Find the final velocity V after they are connected. Assume that there is no friction in this system.

Solution: According to the momentum conservation, the total momentum before linking, mAvA + mBvB should be equal to the total momentum after linking, (mA+ mB)V. That is, V = (mAvA + mBvB) / (mA + mB) = (5000 x 1 + 15000 x 4) / 20000 = 13 / 4 = 3.25 m/s.

Sample Question 2: A 0.2 kg model rocket is constructed with a motor that can provide a total impulse of 30 N s. What is the speed that this rocket achieves when launched from rest? Ignore the effects of gravity and air resistance. Solution: According to the impulse-momentum theorem, the rocket’s final momentum mV differs from its initial momentum mv by an amount equal to the impulse J of the net force exerted on it. That is J = mV - mv. Here, m is 0.2 kg, and v is 0 m/s because the rocket starts from rest. We are ignoring gravitational and frictional forces, so this impulse is due entirely to the force generated by the motor. The magnitude of the motor’s impulse is given as J = 30 N s. Therefore, the rocket’s final speed V is the magnitude of the motor impulse divided by the rocket’s mass: V = J / m = 30 / 0.2 = 150 m/s.

PHY 1301, Physics I 3

UNIT x STUDY GUIDE Title

Angular Velocity, Angular Acceleration, and Angular Displacement

When describing a rotational motion such as spinning top, rotating fan blades, and rolling wheels, we use angular velocity, angular acceleration, and angular displacement. Remember that we have studied the relationship between velocity, acceleration, and displacement in the first unit to describe the linear motion of an object. The angular displacement has an analogous idea of linear displacement. The angular displacement is defined as the angular velocity multiplied by the elapsed time of the motion.

Angular Displacement = Angular Velocity x Elapsed Time

Sample Question 4: Consider an elastic head-on collision between a green ball and an orange ball. The mass m of the green ball is 1 kg and the mass M of the orange ball is 3 kg. Before the collision, the orange ball was at rest, and the initial velocity of the green ball was 5 m/s. After the collision, the final speed V of the orange ball is identical to that of the green ball. However, they have a different velocity; the direction of the green ball is the opposite to that of the orange ball. What are the velocities after the collision?

Solution: The linear momentum is conserved during the collisional process. The linear momentum before the collision is the same as that after the collision: mv = MV + (-mV) So, the final speed V = mv / (M - m) = (1 x 5) / (3 - 1) = 5 / 2 = 2.5 m/s. The final velocity of the orange ball is +2.5 m/s and that of the green ball is -2.5 m/s. In an elastic collision, the total kinetic energy is conserved. That is, total kinetic energy before the collision is equal to that after the collision.

Total kinetic energy before the collision is ½ mv2 = 0.5 x 1 x 25 = 12.5 J. Total kinetic energy after the collision is ½ mV2 + ½ MV2 = (0.5 x 1 x 6.25) + (0.5 x 3 x 6.25) = 12.5 J. That is, total kinetic energy of the system is conserved.

PHY 1301, Physics I 4

UNIT x STUDY GUIDE Title

The SI unit of angular displacement is radian or rad. Please note, 1 radian is equivalent to 360 degrees divided by 2π radians. So, 1 radian is equal to 57.3 degrees. The SI unit of angular velocity is radian per second or rad/s. The angular velocity is related to the angular acceleration as follows. The change in angular velocity is the product of average angular acceleration and the elapsed time of the motion. The SI unit of angular acceleration is radian per second squared or rad/s2.

Small Angle Formula

What is parallax? In order to see what parallax is, hold your index finger in front of your nose, a few centimeters away, and look at it first with the left eye closed, and then with the right eye closed. If you compare its position with some background objects, you will notice that, apparently, the finger moves when you switch from one eye to the other! This explanation is obvious: your eyes are not in the same place, and hence they have a different line of sight to the finger and have to look at a different angle in order to see the finger. The angles at which your eyes have to look obviously depend on the distance of the finger to your nose. What is more important is that the difference in the angles depends on the distance; the angle gets smaller as the distance gets greater. Therefore, the difference in angles, the parallax, can be used to measure the distance. This method is known as trigonometric or heliocentric parallax and can be used to measure up to 100 parsecs (pc) or 0.01 arcseconds. The distance between your eyes is called the baseline; the bigger the baseline (here, D), the bigger the parallax, for the same distance, d, of the observed object to the baseline.

We define the trigonometric parallax of the star as the angle θ subtended as seen from the star by the orbital radius of the Earth, 1 AU: in other words D = 1 AU. One parsec (pc), parallax second, is the distance d when the star has the parallax angle θ = 1 arcsecond = 1 / 206,264.81 radian. That is, d = D / θ = 206, 265 AU = 3.086 × 1016 m = 3.26 ly = 1 pc

Sample Question 5: The nearest star, Proxima Centauri, has a parallax of 0.76". Calculate the

distance of the star from us.

Solution: θ = 0.76” = 0.76/206264.81 = 3.684 × 10-6 radian

d = 206264.81 / 0.76 = 271,401 AU = 1.32 pc = 4.3 ly

PHY 1301, Physics I 5

UNIT x STUDY GUIDE Title

Centripetal Acceleration and Tangential Acceleration

When an object moves in a curved path, its total acceleration is the vector sum of the centripetal acceleration and the tangential acceleration. The magnitude of the total acceleration squared is equal to the centripetal acceleration squared plus the tangential acceleration squared. As we have learned about centripetal acceleration in Unit III, it maintains the circular motion. The direction of the centripetal acceleration points to the center of the circle. The direction of the tangential acceleration is perpendicular to that of the centripetal acceleration. The tangential acceleration exists because the tangential velocity of the object is increasing as it moves. The tangential acceleration is the product of the orbital radius and the angular acceleration in rad/s2. The tangential velocity is the product of the orbital radius and the angular velocity in rad/s.

Motions of Celestial Bodies

We can determine some obvious results from short observations of the sky. All the celestial objects move from the direction where the sun rises to the direction where the sun sets every day; this is known as diurnal motion. Also, the entire sky turns around any fixed points once a day. When one observes the night sky at the northern hemisphere, one would notice that the stars move counterclockwise around a fixed star in the north: the North Star. The North Star does not move because its direction aligns with the rotational axis of the Earth. The North Star is now the Polaris, but in the future, it will not be so, due to the Earth’s precession. The diurnal motion of stars is one of the consequences of the Earth’s rotation. The direction of the rotation of the Earth is the opposite of the diurnal motion’s direction. The common center of the circle of stars is the North Star, and stars are moving 15 degrees an hour in a counterclockwise direction. Diurnal motions of stars differ in different places. If one lives in the northern hemisphere at the mid-latitude area, then one would observe rising stars in the eastern sky, setting stars in the western sky, moving stars from east to west in the southern sky , and counterclockwise circling around Polaris in the northern sky at night. At the equator, the stars move from the east to west, perpendicular to the horizon. At the poles, the stars move parallel to the horizon. The sidereal day is defined as the time for the vernal equinox to return to your celestial meridian. It is the time, in fact, that it takes the Earth to complete one rotation with respect to the stars. It is the true rotation period of the Earth. The solar day is the time for the sun to return to your celestial meridian. Locally, it is the time from high noon to the next one. The Earth is also revolving around the sun, so the solar day is about 4 minutes longer than the sidereal day. The orbital period of the Earth around the sun is about 365 days, and the angle needed to complete one circle is 360 degrees. Therefore, the angle per day for the Earth to see the sun overhead position is 360 / 365 = 0.986 degrees. That is, the sun appears to move eastward by about 1 degree per day with respect to the background stars. There are 24 hours in a full circle (= 360o). Also, 1 hour is equal to 60 minutes. That is, 24 x 60 / 360 = 4 minutes. One degree is 4 minutes. After the Earth rotates from east to west for about 4 minutes (= 1 degree), the sun will rise over your head, your celestial meridian. However, if the Earth’s rotation direction suddenly changed from west to east, you will see the high sun 4 minutes before the complete 1 sidereal day. So, 4 + 4 = 8. The solar day will be about 8 minutes shorter than the original value. In the case of the moon, the sidereal period is 27.3 days. The moon must move eastward among the background stars to go completely around the sky (360 degrees) in the sidereal period. The moon moves about 13 degrees (= 360 / 27.3) a day. Since the celestial sphere appears to turn 1 degree about every 4 minutes, the moon crosses our celestial meridian about 13.2 x 4 = 52.8 minutes later each day. That is, the sun rises about one degree (about 4 minutes) and the moon rises about 13 degrees (about 53 minutes) later each day, and, in the long run, they are moving from the west to east along the ecliptic. The rotational axis of the Earth has an angle of 23.5 degrees from the celestial north. It is not fixed like a spinning top and undergoes a very slow turn, or precession, over time. Its precession period is about 26,000 years. The gravitational attraction of the sun and the moon on the Earth’s equatorial bulge is responsible for this effect. As a consequence, the vernal equinox, the intersection of the equator and ecliptic, regresses by about 50” per year along the ecliptic. In the present, the North Star is the Polaris, but it will change to Vega in 13,000 years. In another 13,000 years, Polaris will be the North Star again. The position of the equinoxes changes, as well, with respect to the background stars, so the coordinates of celestial objects change as well, over a period of 26,000 years. This might be a negligible amount, but it is significant for an exact calculation. Because the moon and the sun move above and below the Earth’s equatorial plane, periodic variations occur

PHY 1301, Physics I 6

UNIT x STUDY GUIDE Title

in the torques acting on the Earth’s equatorial bulge. These variations lead to a nutation or wobbling, of the Earth’s rotation axis. The nutation is a periodic oscillation of the rotational axis of the Earth around its “mean” position. Small contributions to the nutation have monthly and yearly periods, but the principal contribution (discovered by James Bradley) of amplitude 9” and period 18.6 years is from the regression of the nodes of the moon’s orbit.

Reference

Cutnell, J. D., Johnson, K. W., Young, D., & Stadler, S. (2022). Physics (12th ed.). Wiley. Learning Activities (Nongraded) Nongraded Learning Activities are provided to aid students in their course of study. You do not have to submit them. If you have questions, contact your instructor for further guidance and information.

1. Solve Questions 53–60 under Physics in Biology, Medicine, and Sports in Chapter 7 of your eTextbook.

2. Solve Questions 62–69 under Physics in Biology, Medicine, and Sports in Chapter 8 of your eTextbook.