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**Announcements EXAM: October 13 (through Chapter 9)**

calculator (no stored functions, etc) #2 pencil

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Reading and Review

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Elastic Collisions Carefully place a small rubber ball (mass m) on top of a much bigger basketball (mass M) and drop these from the same height h so they arrive at the ground with the speed v. What is the velocity of the smaller ball after the basketball hits the ground, reverses direction, and then collides with the small rubber ball? a) zero b) v c) 2v d) 3v e) 4v m v v M

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Elastic Collisions Carefully place a small rubber ball (mass m) on top of a much bigger basketball (mass M) and drop these from the same height h so they arrive at the ground with the speed v. What is the velocity of the smaller ball after the basketball hits the ground, reverses direction, and then collides with the small rubber ball? a) zero b) v c) 2v d) 3v e) 4v Remember that relative velocity has to be equal before and after collision! Before the collision, the basketball bounces up with v and the rubber ball is coming down with v, so their relative velocity is –2v. After the collision, it therefore has to be +2v!! v 3v (a) (b) (c) m M

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Recoil Speed A cannon sits on a stationary railroad flatcar with a total mass of kg. When a 10-kg cannonball is fired to the left at a speed of 50 m/s, what is the recoil speed of the flatcar? a) 0 m/s b) 0.5 m/s to the right c) 1 m/s to the right d) 20 m/s to the right e) 50 m/s to the right Answer: b

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Recoil Speed A cannon sits on a stationary railroad flatcar with a total mass of kg. When a 10-kg cannonball is fired to the left at a speed of 50 m/s, what is the recoil speed of the flatcar? a) 0 m/s b) 0.5 m/s to the right c) 1 m/s to the right d) 20 m/s to the right e) 50 m/s to the right Because the initial momentum of the system was zero, the final total momentum must also be zero. Thus, the final momenta of the cannonball and the flatcar must be equal and opposite. pcannonball = (10 kg)(50 m/s) = 500 kg-m/s pflatcar = 500 kg-m/s = (1000 kg)(0.5 m/s)

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**Elastic Collisions in 2-D**

Two-dimensional collisions can only be solved if some of the final information is known, such as the final velocity of one object

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**CEBAF at JLAB Precision nuclear microscopy**

CEBAF at Jefferson Lab recirculation through continuous-wave superconducting RF linacs simultaneous beam delivery to 3 experimental halls with large complementary spectrometers cold RF = stable, clean, quiet up to 200 microAmps per hall, E ~ GeV, >80% polarization Let me briefly introduce CEBAF This photograph shows the characteristic “racetrack” shape of two cold RF linacs connected by recirculating arcs. There are three experimental halls with complementary standard spectrometer equipment. Beam can be simultaneously delivered to all 3. The machine operates in continuous wave mode, which maximizes the useful luminosity. The cold RF is key in producing a stable, quiet and clean beam. With up to 200 microAmps to a hall, an energy range of ~ GeV, and beam polarization routinely over 80%, there has never been an instrument like this for precision hadron microscopy. This facility has over 1200 outside users, performing a wide range of studies. An ideal machine for precision nuclear microscopy! 3 experimental Halls Project started in 1979 First experiments begun in 1994 User community 1200 members strong

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**v1 v0 = 3.5x105 m/s 37o 53o initial v2 final**

A proton collides elastically with another proton that is initially at rest. The incoming proton has an initial speed of 3.5x105 m/s and makes a glancing collision with the second proton. After the collision one proton moves at an angle of 37o to the original direction of motion, the other recoils at 53o to that same axis. Find the final speeds of the two protons. v0 = 3.5x105 m/s initial 37o 53o v2 v1 final

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**if we’d been given only 1 angle, would have needed **

A proton collides with another proton that is initially at rest. The incoming proton has an initial speed of 3.5x105 m/s and makes a glancing collision with the second proton. After the collision one proton moves at an angle of 37o to the original direction of motion, the other recoils at 53o to that same axis. Find the final speeds of the two protons. Momentum conservation: v0 = 3.5x105 m/s initial 37o 53o v2 v1 final if we’d been given only 1 angle, would have needed the other angle or conservation of energy also!

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**Is energy conserved in this collision?**

In an elastic collision between a moving mass incident on a stationary identical mass, the angle between the final momenta is always 90 degrees.

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**Center of Mass Treat extended mass as a bunch of small masses:**

In a uniform gravitational field you can treat gravitational force as if it acts at the “Center of Mass”

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**The center of mass is closer to the more massive object.**

The center of mass of a system is the point where the system can be balanced in a uniform gravitational field. For two objects: The center of mass is closer to the more massive object.

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Center of Mass In general: Symmetry often very useful in determining the Center of Mass

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**The center of mass need not be within the object**

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**Motion about the Center of Mass**

The center of mass of a complex or composite object follows a trajectory as if it were a single particle - with mass equal to the complex object, and experiencing a force equal to the sum of all external forces on that complex object

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**Motion of the center of mass**

Action/Reaction pairs inside the system cancel out

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The total mass multiplied by the acceleration of the center of mass is equal to the net external force The center of mass accelerates just as though it were a point particle of mass M acted on by

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**Momentum of a composite object**

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Recoil Speed II A cannon sits on a stationary railroad flatcar with a total mass of kg. When a 10-kg cannonball is fired to the left at a speed of 50 m/s, what is the speed of the center of mass (of the flatcar + cannonball)? a) 0 m/s b) 0.5 m/s to the right c) 1 m/s to the right d) 20 m/s to the right e) 50 m/s to the right Answer: b

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Recoil Speed II A cannon sits on a stationary railroad flatcar with a total mass of kg. When a 10-kg cannonball is fired to the left at a speed of 50 m/s, what is the speed of the center of mass (of the flatcar + cannonball)? a) 0 m/s b) 0.5 m/s to the right c) 1 m/s to the right d) 20 m/s to the right e) 50 m/s to the right Because the initial momentum of the system was zero, the final total momentum must also be zero, regardless of the release of internal energy, internal forces, etc. If no external forces act, the motion of the center of mass does not change

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**Center of Mass a) higher b) lower c) at the same place**

The disk shown below in (1) clearly has its center of mass at the center. Suppose the disk is cut in half and the pieces arranged as shown in (2). Where is the center of mass of (2) as compared to (1) ? a) higher b) lower c) at the same place d) there is no definable CM in this case Answer: a (1) X CM (2)

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**Center of Mass CM a) higher b) lower c) at the same place**

The disk shown below in (1) clearly has its center of mass at the center. Suppose the disk is cut in half and the pieces arranged as shown in (2). Where is the center of mass of (2) as compared to (1) ? a) higher b) lower c) at the same place d) there is no definable CM in this case The CM of each half is closer to the top of the semicircle than the bottom. The CM of the whole system is located at the midpoint of the two semicircle CMs, which is higher than the yellow line. (1) X CM (2) CM

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A charging bull elephant with a mass of 5240 kg comes directly toward you with a speed of 4.55 m/s. You toss a kg rubber ball at the elephant with a speed of 7.81 m/s. When the ball bounces back toward you, what is its speed?

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A charging bull elephant with a mass of 5240 kg comes directly toward you with a speed of 4.55 m/s. You toss a kg rubber ball at the elephant with a speed of 7.81 m/s. When the ball bounces back toward you, what is its speed? Our simplest formulas for speed after an elastic collision relied on one body being initially at rest. So lets try a frame where one body (the ball) is at rest! What is the speed of the elephant relative to the ball? In the frame where this is the elephant’s speed, what is the final speed of the ball? Back in the frame of the ground:

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A charging bull elephant with a mass of 5240 kg comes directly toward you with a speed of 4.55 m/s. You toss a kg rubber ball at the elephant with a speed of 7.81 m/s. When the ball bounces back toward you, what is its speed? Our simplest formulas for speed after an elastic collision relied on one body being initially at rest. So lets try a frame where one body (the ball) is at rest! What is the speed of the elephant relative to the ball? NOTE: Formulas for 1-D elastic scattering with non-zero initial velocities are given in end-of-chapter problem 88. In the frame where this is the elephant’s speed, what is the final speed of the ball? Back in the frame of the ground:

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**Rotational Kinematics**

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Angular Position

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θ > 0 θ < 0 Angular Position Degrees and revolutions:

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**Arc Length s = rθ Arc length s, from angle measured in radians:**

- What is the relationship between the circumference of a circle and its diameter? C / D = π C = 2 π r - Arc length for a full rotation (360o) of a radius=1m circle? s = 2 π (1 m) = 2 π meters 1 complete revolution = 2 π radians 1 rad = 360o / (2π) = 57.3o

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Angular Velocity

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**Instantaneous Angular Velocity**

Period = How long it takes to go 1 full revolution Period T: SI unit: second, s

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**Linear and Angular Velocity**

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**Greater translation for same rotation**

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Bonnie and Klyde Bonnie sits on the outer rim of a merry-go-round, and Klyde sits midway between the center and the rim. The merry-go-round makes one revolution every 2 seconds. Who has the larger linear (tangential) velocity? a) Klyde b) Bonnie c) both the same d) linear velocity is zero for both of them Answer: b ω Bonnie Klyde

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Bonnie and Klyde Bonnie sits on the outer rim of a merry- go-round, and Klyde sits midway between the center and the rim. The merry-go-round makes one revolution every 2 seconds. Who has the larger linear (tangential) velocity? a) Klyde b) Bonnie c) both the same d) linear velocity is zero for both of them Answer: b Their linear speeds v will be different because v = r and Bonnie is located farther out (larger radius r) than Klyde. ω Klyde Bonnie

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Angular Acceleration

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**Instantaneous Angular Acceleration**

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**Rotational Kinematics, Constant Acceleration**

If the angular acceleration is constant: If the acceleration is constant: v = v0 + at

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**Analogies between linear and rotational kinematics:**

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**½ Angular Displacement I ¾ ¼ ½ a) b) c) d) 2 e) 4 **

An object at rest begins to rotate with a constant angular acceleration. If this object rotates through an angle in the time t, through what angle did it rotate in the time t? Answer: b

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**½ Angular Displacement I ¾ ¼ ½ a) b) c) d) 2 e) 4 **

An object at rest begins to rotate with a constant angular acceleration. If this object rotates through an angle in the time t, through what angle did it rotate in the time t ? The angular displacement is = t 2 (starting from rest), and there is a quadratic dependence on time. Therefore, in half the time, the object has rotated through one-quarter the angle.

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**Which child experiences a greater acceleration?**

(assume constant angular speed)

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**Larger r: - larger v for same ω - larger ac for same ω**

ac is required for circular motion. An object may have at as well, which implies angular acceleration

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**Angular acceleration and total linear acceleration**

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**Angular and linear acceleration**

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Rolling Motion If a round object rolls without slipping, there is a fixed relationship between the translational and rotational speeds:

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Rolling Motion We may also consider rolling motion to be a combination of pure rotational and pure translational motion: + =

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Truck Speedometer Suppose that the speedometer of a truck is set to read the linear speed of the truck but uses a device that actually measures the angular speed of the tires. If larger diameter tires are mounted on the truck instead, how will that affect the speedometer reading as compared to the true linear speed of the truck? a) speedometer reads a higher speed than the true linear speed b) speedometer reads a lower speed than the true linear speed c) speedometer still reads the true linear speed

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Truck Speedometer Suppose that the speedometer of a truck is set to read the linear speed of the truck but uses a device that actually measures the angular speed of the tires. If larger diameter tires are mounted on the truck instead, how will that affect the speedometer reading as compared to the true linear speed of the truck? a) speedometer reads a higher speed than the true linear speed b) speedometer reads a lower speed than the true linear speed c) speedometer still reads the true linear speed The linear speed is v = R. So when the speedometer measures the same angular speed as before, the linear speed v is actually higher, because the tire radius is larger than before.

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**Jeff of the Jungle swings on a vine that is 7. 20 m long**

Jeff of the Jungle swings on a vine that is 7.20 m long. At the bottom of the swing, just before hitting the tree, Jeff’s linear speed is 8.50 m/s. (a) Find Jeff’s angular speed at this time. (b) What centripetal acceleration does Jeff experience at the bottom of his swing? (c) What exerts the force that is responsible for Jeff’s centripetal acceleration?

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**(a) Find Jeff’s angular speed at this time. **

Jeff of the Jungle swings on a vine that is 7.20 m long. At the bottom of the swing, just before hitting the tree, Jeff’s linear speed is 8.50 m/s. (a) Find Jeff’s angular speed at this time. (b) What centripetal acceleration does Jeff experience at the bottom of his swing? (c) What exerts the force that is responsible for Jeff’s centripetal acceleration? a) b) c) This is the force that is responsible for keeping Jeff in circular motion: the vine.

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**Rotational Kinetic Energy**

For this mass m, and a massless rod

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**Provided in lecture notes on: 10/6**

PHYS2010 midterm 1, Fall 2008 13) If the net work done on an object is zero, then the object's kinetic energy A) is zero. B) decreases. C) increases. D) remains the same. E) cannot be determined without knowing the object mass. suggested time: 1 minute Please do not ask questions about this problem at discussion sessions before 10/7

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**Provided in lecture notes on: 10/6**

PHYS2010 midterm 1, Fall 2008 14) A horizontal force of 40 N pushes an object of mass 5.0 kg up an inclined plane through a distance of 1.6 m measured along the plane. The plane is inclined at an angle of 30o to the horizontal. Neglect friction and use g=10 m/s2, What is the work done by the normal force on the object? A) 53 J B) 15 J C) 0 J D) -25 J E) 37 J suggested time: 1 minute Please do not ask questions about this problem at discussion sessions before 10/7

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**Provided in lecture notes on: 10/6**

PHYS2010 midterm 2, Fall 2008 5) In a particle accelerator, protons are fired at a stationary target and collide elastically with the nuclei of the atoms in the target. In one such experiment, a beam of protons is fired with a speed of 2.10 x 107 m/s. The protons that undergo a collision rebound with a speed of 1.68 x 107 m/s. What is the mass of the nuclei that the protons are colliding against? A) 9.00 proton masses B) 10.0 proton masses C) 7.00 proton masses D) 8.00 proton masses E) 6.00 proton masses suggested time: 3-4 minutes Please do not ask questions about this problem at discussion sessions before 10/7

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**Provided in lecture notes on: 10/6**

PHYS2010 midterm 2, Fall 2008 6) A Ferris wheel with a radius of 8.00 m rotates at a constant rate, completing one revolution in 30.0 s. What is the apparent weight of a 60.0-kg passenger when she is at the top of the wheel? A) 615 N B) 589 N C) 568 N D) 325 N E) 432 N suggested time: 4-5 minutes Please do not ask questions about this problem at discussion sessions before 10/19

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Ch. 9 Rotational Kinematics

Ch. 9 Rotational Kinematics

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