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Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.
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Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

Dec 14, 2015

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Page 1: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

Fig. 5.4

Centripetal Acceleration:

a = v2/rDirected inwards, towards the center.

Page 2: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.
Page 3: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

A lever is used to lift a rock. A small force F1 generates a larger force F2 to lift the rock, but

F1 acts through a larger distance d1 than does F2.

In other words, the work you do equals the work done on the rock. (Conservation of energy.)

Page 4: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

A crate is moved a distance d across a concrete floor under the influence of a constant horizontal force F.

WORK = Force times Distance

Can work be negative?

Page 5: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

The work done on an object by the net force acting on the object results in an increase in the object’s kinetic energy.

What about friction?

Page 6: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

Important concepts

1. Work = force x distance

2. Kinetic energy KE = (1/2)mv2

3. Total work done = change in KE

4. Potential energy from gravity PE = mgh

5. Energy is conserved:

E = KE + PE = constant! (Some restrictions apply!!)

Page 7: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

Conservation of mechanical energy:

If there are no forces doing work on a system, the total mechanical energy of the system remains constant.

If W=0, E = PE + KE = constant.

Page 8: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

Momentum is a vector:

Magnitude = mass times speed = mv

Direction = same as velocity

(To be more precise, this is the linear momentum.)

Page 9: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

If the net external force acting on a system of objects is zero, the total momentum of the system is conserved.

Page 10: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

We also want to know about changes in momentum.

Notation: initial momentum pi

final momentum pf

change in momentum p = pf - pi

Page 11: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

There is a close resemblance between quantities used to describe linear motion and those

used to describe rotational motion.

Page 12: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

Figure 8.15Expressions for the rotational inertia of

several objects, each with a uniform distribution of mass over its volume. The letter m is used to

represent the total mass of the object.

Page 13: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

Physics of Angular Motion

1. If the total torque on an object is zero, then it does not rotate, or it rotates at a constant angular velocity.

Page 14: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

Physics of Angular Motion

2. The net torque on an object is proportional to its angular acceleration. The constant of proportionality is the moment of inertia.

vectors

Page 15: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

Physics of Angular Motion

3. If the net torque on a system is zero, the total angular momentum of the system is conserved.

L=Ivectors

Page 16: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

CONCEPT Linear Motion Rotational

Inertia m I

Second Law F=ma

Momentum P=mv L=I

conservation P=constant,

if F=0.

L=constant,

if Kinetic energy KE=1/2mv2 KE=1/

Page 17: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

A woman’s high heels sink into the soft ground, but the larger shoes of the much bigger man do not.

Pressure = force/area

Page 18: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

Pascal’s Principle:

Any change in the pressure of a fluid is transmitted uniformly in all directions throughout the fluid.

Page 19: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

In Boyle’s experiment, adding mercury to the open side of the bent tube caused a decrease in the

volume of the trapped air in the closed side.

P1V1 = P2V2

Boyle’s Law

Page 20: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

Archimedes’ Principle:

The buoyant force acting on an object fully or partially submerged in a fluid is equal to the weight of the fluid displaced by the object.

Page 21: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

What is heat?

It is the energy that flows from a hot object to a cooler one.

Page 22: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

Will adding heat to a body always raise its temperature?

No, there can be a change of phase.

It takes 0.49 cal to raise 1 gram of ice from -1 C to 0 C.

But, it takes another 80 cal to go from ice at 0 C to water at 0 C.

Similarly, it takes 540 cal to go from 1 gram of water at 100 C to one gram of steam at 100 C.

Page 23: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

An object can get hotter by

• adding heat to it

• doing work on it!

Total energy is conserved. Heat is another form of energy.

Page 24: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

1. When an ice skater twirling on the point of a skate draws her arms in sheends up whirling faster. This is because --

A. rotational energy is conserved.B. angular momentum is conserved.C. linear momentum is conserved.D. a net torque acts on the skater.

Page 25: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

1. When an ice skater twirling on the point of a skate draws her arms in sheends up whirling faster. This is because --

B. angular momentum is conserved.

Page 26: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

2. Suppose water is boiled in a vessel open to the atmosphere. Compare the internal energy of 1 gram of steam at the boiling point to the internal energy of 1 gram of water at the same temperature.

A. The internal energy of the water and steam are the same.B. The internal energy of the water will be higher.C. The internal energy of the steam will be higher.

Page 27: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

2. Suppose water is boiled in a vessel open to the atmosphere. Compare the internal energy of 1 gram of steam at the boiling point to the internal energy of 1 gram of water at the same temperature.

C. The internal energy of the steam will be higher.

Page 28: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

3. A bullet is fired into a wall and comes to rest. Considering the bullet as the system we can say that --

A. the total mechanical energy is conserved.

B. the momentum is conserved.C. both total mechanical energy and

momentum are conserved.D. neither total mechanical energy nor

momentum are conserved.

Page 29: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

3. A bullet is fired into a wall and comes to rest. Considering the bullet as the system we can say that --

D. neither total mechanical energy nor momentum are conserved.

Page 30: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

4. Anthea exerts a force of 500 N against a 100-kg desk which does not move. Campaspe exerts a force of 400 N against a 60-kg desk which moves 2 m in the direction of the push. Fiorinda exerts a force of 200 N against a 50-kg desk which moves 4 m in the direction of the push. The most work is done by

A. Anthea.B. Campaspe.C. Fiorinda.D. Campaspe and Fiorinda, who do equal

amounts of work.

Page 31: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

4. Anthea exerts a force of 500 N against a 100-kg desk which does not move. Campaspe exerts a force of 400 N against a 60-kg desk which moves 2 m in the direction of the push. Fiorinda exerts a force of 200 N against a 50-kg desk which moves 4 m in the direction of the push. The most work is done by

D. Campaspe and Fiorinda, who do equal amounts of work.

Page 32: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

5. A truck of mass 4000 kg is at rest but free to roll with no resistance. If you push with a force of 500 N, the momentum at the end of 5 s of pushing will be

A. zero.B. 500 N.C. 800 kg/s.D. 2500 kg m/s.

Page 33: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

5. A truck of mass 4000 kg is at rest but free to roll with no resistance. If you push with a force of 500 N, the momentum at the end of 5 s of pushing will be

D. 2500 kg m/s.

Page 34: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

6. A merry-go-round is set in motion by students applying a a force of 500 N tangential to the rim of the wheel having a radius of 2.0 m. The merry-go round reaches a rotational velocity of 2.0 rad/s after 10 seconds, starting from rest. The rotational inertia of the merry-go-round is

A. 1.0 x 105 kg m2B. 5.0 x 103 kg m2C. 5.0 x 104 kg m2D. 5.0 x 105 kg m2E. 1.0 x 104 kg m2

Page 35: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

6. A merry-go-round is set in motion by students applying a a force of 500 N tangential to the rim of the wheel having a radius of 2.0 m. The merry-go round reaches a rotational velocity of 2.0 rad/s after 10 seconds, starting from rest. The rotational inertia of the merry-go-round is

B. 5.0 x 103 kg m2

Page 36: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

7. A hard rubber ball is thrown against a wall and rebounds with nearly the same speed that it had initially. A tomato of the same mass is thrown against the wall with the same speed, splatters and hardly rebounds. The impulse exerted by the wall on each object--

A. is larger for the tomatoB. is the same for both and not zeroC. is zero for bothD. is larger for the ball.

Page 37: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

7. A hard rubber ball is thrown against a wall and rebounds with nearly the same speed that it had initially. A tomato of the same mass is thrown against the wall with the same speed, splatters and hardly rebounds. The impulse exerted by the wall on each object--

D. is larger for the ball.

Page 38: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

8. In using a long steel rod as a lever to move a large rock, where should the fulcrum be placed?

A. Halfway between the rock and the applied force.

B. Close to the rock.C. Close the point where the force is applied.D. Anywhere in between -- it makes no

difference.

Page 39: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

8. In using a long steel rod as a lever to move a large rock, where should the fulcrum be placed?

B. Close to the rock.

Page 40: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

9. A ball at the end of a string is started swinging as a simple pendulum. Assuming no loss in energy due to friction we can say for the ball that

A. the potential energy is maximum at the center of motion.

B. the potential energy is maximum where the kinetic energy is a

minimum.C. the potential energy is maximum

where the kinetic energy is maximum.D. the kinetic energy is maximum at

each end of the motion.

Page 41: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

9. A ball at the end of a string is started swinging as a simple pendulum. Assuming no loss in energy due to friction we can say for the ball that

B. the potential energy is maximum where the kinetic energy is a

minimum.

Page 42: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

10. A penny, a quarter and a silver dollar lie at the bottom of a wishing well filled with water. The coin experiencing the largest downward force due to the water pressure is:

A. pennyB. quarterC. silver dollarD. all the same

Page 43: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

10. A penny, a quarter and a silver dollar lie at the bottom of a wishing well filled with water. The coin experiencing the largest downward force due to the water pressure is:

C. silver dollar

Page 44: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

11. An ice cube of mass 100 g and at 0 C is dropped into a styrofoam cup containing 200 g of water at 25 C. The heat of fusion of ice is 80 cal/g and the specific heat capacity of water is 1.0 cal/g C. Assuming the cup doesn't exchange any heat, the final temperature of the system will be which of the following?

A. -10 CB. 0 CC. +10 CD. +2.5 CE. +5.0 C

Page 45: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

11. An ice cube of mass 100 g and at 0 C is dropped into a styrofoam cup containing 200 g of water at 25 C. The heat of fusion of ice is 80 cal/g and the specific heat capacity of water is 1.0 cal/g C. Assuming the cup doesn't exchange any heat, the final temperature of the system will be which of the following?

B. 0 C

Page 46: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

12. A piece of aluminum has a mass of 0.25 kg and a density of 2700 kg/m3. The buoyant force exerted on this body when completely submerged in water of density 1000 kg/m3 is approximately:

A. 2.4 NB. 1.2 NC. 0.91 ND. 0.093 NE. 2.45 x 10-3 N

Page 47: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

12. A piece of aluminum has a mass of 0.25 kg and a density of 2700 kg/m3. The buoyant force exerted on this body when completely submerged in water of density 1000 kg/m3 is approximately:

C. 0.91 N

Page 48: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

13. A painter of mass 80 kg climbs 3.0 m up a ladder. The painter has gained potential energy in amount of --

A. 240 JB. 784 JC. 2352 JD. 261 JE. 1284 J

Page 49: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

13. A painter of mass 80 kg climbs 3.0 m up a ladder. The painter has gained potential energy in amount of --

C. 2352 J

Page 50: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

14. A body of rotational inertia 1.0 kg m2 is acted upon by a torque of 2.0 Nm. The angular acceleration of the body will be:

A. 0.50 rad/s2B. 0.5 rev/s2C. 1.0 rad/s2D. 2.0 rad/s2E. 2.0 rev/s2

Page 51: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

14. A body of rotational inertia 1.0 kg m2 is acted upon by a torque of 2.0 Nm. The angular acceleration of the body will be:

D. 2.0 rad/s2

Page 52: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

15. A box is moved 10 m across by a floor a force of 20 N acting along the direction of motion. The work done is

A. 2.0 JB. 20 JC. 30 JD. 100 JE. 200 J

Page 53: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

15. A box is moved 10 m across by a floor a force of 20 N acting along the direction of motion. The work done is

E. 200 J

Page 54: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

16. An object of mass 1.0 kg with a velocity of 4.0 m/s strikes head-on a second object of mass 2.0 kg at rest. After the collision the first body moves backward at 1.0 m/s and the struck body moves forward with a velocity of 2.5 m/s. In this collision --A. momentum is conserved but kinetic energy

is notB. kinetic energy is conserved but momentum

is notC. momentum and kinetic energy are conservedD. neither momentum nor kinetic energy are

conserved.

Page 55: Fig. 5.4 Centripetal Acceleration: a = v 2 /r Directed inwards, towards the center.

16. An object of mass 1.0 kg with a velocity of 4.0 m/s strikes head-on a second object of mass 2.0 kg at rest. After the collision the first body moves backward at 1.0 m/s and the struck body moves forward with a velocity of 2.5 m/s. In this collision --A. momentum is conserved but kinetic energy

is not.