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Exam Review and Friction

•  Exam Thursday at 7:30pm – Bring a #2 pencil – Will be long answer

questions in addition to multiple choice

– Room assignments will be on web page (Exam info)

– Calculators and 1 double sided sheet of notes allowed

Web page: http://www.colorado.edu/physics/phys1110/phys1110_sp12/

Announcements:

The Thinker by August Rodin – museum in Paris

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An Atwood's machine is a pulley with two masses connected by a string as shown. The mass of object A, mA, is twice the mass of object B, mB. What is the tension T in the string ?

Atwood’s machine

A B

A B

Since the weights are unequal, it should be clear that there will be some non-zero acceleration. Thus, since and are not zero, neither A nor B can be correct.

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A B

A B

We defined up as positive for both masses which means & give

Full solution to Atwood’s machine

Solving for acceleration: so the tension is

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How does the force exerted on the cart by the string (T) compare with the weight of body B?

(Assume all surfaces are frictionless)

A: T = mB g. B: T < mB g. C: T > mB g.

The system WILL accelerate, Ma will accelerate to the right, Mb will accelerate down (at the same rate, they're tied together!) So look at Mb: it accelerates down, so the net force on it is DOWN (by N-II). There are only 2 forces acting on it: mB*g down, and T up. If the NET force on Mb is down, then mB*g must "win". . If you think mB*g = T, then that means you think there is ZERO net force on Mb. Which means it would just sit there.... but there's no friction, so it really will be pulled down, and start to accelerate.

Clicker question 1 Set frequency to BA

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Set frequency to BA Clicker question 2 In the 17th century, Otto von Güricke, a physicist in Magdeburg, fitted two hollow bronze hemispheres together and removed the air from the resulting sphere with a pump. Two eight-horse teams could not pull the halves apart even though the hemispheres fell apart when air was readmitted.

Suppose von Güricke had tied both teams of horses to one side and bolted the other side to a heavy tree trunk. In this case, the tension on the hemispheres would be

A: twice B: exactly the same as C: half

what it was before

Draw a force diagram! Let's call the force of an 8-horse team "F". Before, you had "F" to the left, and "F" to the right. They balanced, but there was a tension in the system. Now, you put both teams on one side, which means you have "2F" on that side. The other side is connected to a tree. Assuming the tree doesn't budge, there must be "2F" acting the opposite direction. Once again, these forces balance (the sphere doesn't accelerate), but the TENSION is twice as much as before.

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Set frequency to BA Clicker question 3

Net force = ma, that's Newton's law. "Air track" implies no friction, so this glider is clearly going to be accelerating. Assuming it stays on the track (implied by "pulled along an air track"), it must be accelerating to the right. So by N-II, Net force is horizontal! Apparently, the vertical component of F, plus N (normal force), minus mg, must all cancel - the net result is OBSERVED, the glider is moving to the right, so the net force must be to the right!

A glider is pulled along an air track with a string at an angle (theta) from the horizontal, with a constant force.

The direction of the net force on the glider is

A: Horizontal. B: At an angle theta above the horizontal. C: At some angle above the horizontal, but not necessarily theta.

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Review of what is covered Chapter 1 •  Significant figures •  Units and unit conversion •  Vectors

– Getting components from trigonometry – Getting magnitude and direction from components – Addition and subtraction

The magnitude of is

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Review of what is covered Chapter 2 & 3 – Motion •  Displacement is a vector measuring change in

position •  Velocity measures displacement per time

•  Acceleration measures change of velocity per time

•  Often deal with constant acceleration such as free fall where in the downward direction

Average velocity: Instantaneous velocity:

Average acceleration: Instantaneous acceleration:

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Review of what is covered Chapter 2 & 3 – Motion

No displacement in this equation

No final velocity in this equation

No time in this equation

No acceleration in this equation

One dimensional equations of motion

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Review of what is covered Chapter 2 & 3 – Motion

Equation for projectile motion

Constant velocity in x direction:

Free fall in y direction:

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Review of what is covered Chapter 3 – Circular motion & relative velocity Circular motion, even at constant speed, is an example of acceleration because the velocity vector direction changes

Can divide acceleration into tangential and radial

components: and

Relative velocity: A reference frame which moves at constant velocity (can be 0) is an inertial reference frame.

Use vector addition of velocities to transfer from one reference frame to another

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Review of what is covered Chapter 4 & 5 – Forces & Newton’s Laws Newton’s First Law: A body maintains a constant velocity (which can be 0) unless acted upon by a net force

Newton’s Second Law: or &

Newton’s Third Law: If body A exerts a force on body B then there is a force which is equal in magnitude but opposite in direction by body B on body A.

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What chapters are included Basically, all of chapters 1,2,3, and 4 plus the first section of chapter 5 are covered

Chapter 1.10 on vector products is not included

Chapter 2.8 on velocity and position from integration is not included

However, some parts were not covered and will not be on the exam

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Exam tips •  Remember that displacement, velocity,

acceleration, and force are vectors and vectors have a direction and magnitude

•  Understand the CAPA problems, clicker questions, and tutorials

•  You have nearly two hours to answer multiple choice plus long answer questions. Take your time and check your work.

A rider in a "barrel of fun” finds herself stuck with her back to the wall. Which diagram correctly shows the forces acting on her?

Pink Blue Green Purple Yellow A) B) C) D) E)

Clicker question 4 Set frequency to BA

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What is going on is that the ride is pushing on the people to make them go in a circle. For every action there is an equal but opposite reaction, and so the people push back on the ride. The centripetal force the ride exerts on people becomes the normal force that causes the friction that keeps them from sliding down the walls, and that friction as such must be larger than their weight, or down the wall they go. Nothing is pushing them to the outside; acceleration is toward the middle. Remember you feel "thrown" the opposite direction of the acceleration. (i.e. if your car accelerated forward, you are thrown backward)

Barrel of Fun

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A bucket containing a brick is swung in a circle at constant speed in a vertical plane as shown. The bucket is swung fast enough that the brick does not fall out.

The net force on the brick as it is swung has maximum magnitude at position. A) Top. B) Bottom. C) Right D) The net force has the same magnitude at all positions.

N R

B

T

The net force has the same magnitude at all positions. Since the magnitude of the acceleration is constant (a = v2/R), the net force must have constant magnitude (Fnet = ma).

Clicker question 5 Set frequency to BA

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A bucket containing a brick is swung in a circle at constant speed in a vertical plane as shown.

A) Top. B) Bottom. C) Right D) The net force has the same magnitude at all positions.

N R

B

T

Clicker question 6 Set frequency to BA

Consider the normal force exerted on the brick by the bucket when the bucket is at the three positions shown: R, T, B. The magnitude of the normal force is a minimum at position...

At the top. Draw free-body diagrams and keep in mind the net force has the same magnitude at the top and the bottom.