blueberry muffin nov. 29/30, 2016 period: names...blueberry muffin nov 29-30, 2016 mr. han's physics...

Blueberry Muffin Nov. 29/30, 2016 Period: Names:

Congratulations!

1.  To solve the problems, use your eTextbook, physical textbooks, physics websites, your Sketchbooks. 2.  Show your thinking through calculations, words and sketches. Writing just the solution is worth 0 points. 3.  Team leaders should coordinate the work done by team members. But all team members must be

able to solve every question eventually. So teach each other.

4.  Work hard. Make good use of your time. 5.  It’s perfectly OK to be confused and lost at first. I expect you to struggle with the problems; that is how

you build your intelligence and knowledge.

6.  You will find clarity and correct solutions. It’s only a matter of effort and time.

A) A car is traveling around a curve at a steady 4 m/s. Which vector shows the direction of the car’s acceleration? Why? A. B. C. D. E. The acceleration is zero. B) A toy car moves around a circular track at constant speed. It suddenly doubles its speed — a change of a factor of 2. As a result, the centripetal acceleration changes by a factor of: A. 1/4 B. 1/2 C. No change since the radius doesn’t change. D. 2 E. 4 Use calculations to support your answer.

2Blueberry Muffin Nov 29-30, 2016 Mr. Han's Physics

C) A metal ball at the end of a rope is being swung in a perfectly horizontal circle at a constant speed. The ball is accelerating because (select one and explain):

A. The speed is changing. B. The direction is changing. C. The speed and the direction are changing. D. The ball is not accelerating, because it's moving at a constant speed. E. The ball is not accelerating, because its movement is perpendicular to the force of gravity.

Explain your answer with words and sketches. D) If the ball is accelerating, what is the direction of the acceleration? (select one and explain)

A. Tangent to the circle, in the direction of the ball’s tangential velocity. B. Toward the center of the circle. C. Towards the direction opposite to the force of tension on the rope. D. The ball is not accelerating. E. The ball is not accelerating due to the speed of its swing, but it is accelerating downward (towards ground) because of a net force due to gravity.

Explain your answer with words and sketches.


E) A ball at the end of a rope is being swung in a horizontal circle. What force is producing the centripetal acceleration of the ball?

A. Gravity B. Air resistance C. Normal force D. Tension in the rope E. Friction between the ground and person's feet.

Explain your answer with words and sketches. F) What is the direction of the net force on the ball?

A. Tangent to the circle B. Toward the center of the circle C. There is no net force. D. Toward the ground. E. Toward the direction opposite to the tension in the rope.



G) If you were a cosmonaut in space orbiting the Earth, you would feel weightless because:

A. Gravity would be the only force acting upon you. B. The net force on you would be zero. C. The centrifugal force balances the gravitational force. D. The force due to gravity is extremely small (but not zero) so you would feel weightless. E. There’s no gravity in outer space.

Explain your answer with words and sketches. H) You are curled up in a ball on the ground next to a red panda which is similarly curled up in a ball. The distance from your respective centers is 0.50 meters. Calculate the gravitational force between you. Assume that you have a mass of 40 kg and the red panda a mass of 10 kg. Show your calculations and a sketches.


H) If Planet Susie has twice the mass of Planet Patrick, then the gravitational force Planet Susie exerts on Planet Patrick, compared to the gravitational force Planet Patrick exerts on Planet Susie, is:

A. One quarter as much B. One half as much C. The same D. Twice as much E. Four times as much

Explain your answer with words and sketches. I) If the distance separating Planet Susie from Planet Patrick were to increase to twice the original distance, then how woul the gravitational force Planet Susie exerts on Planet Patrick change? It would be:




RmmGFgravity 2 21=

J) If your mass is 150 kg, how much would you weigh (in N) on the Moon? And what % is this of your weight on Earth? If g on Earth is 9.8 m/s2, show the calculation for finding g on the Moon. Don't just look up the answer; instead show how you would use this equation to calculate the answer. K) If your mass is 50 kg, on which planet would you weigh the most and why? R = planet radius M = planet mass Explain your answer with calculations, words and sketches.


L) Moons A and B both orbit around a planet. If Moon B's mass is twice that of Moon A and Moon B's distance from the planet's center is 50% greater than that of Moon A, which Moon orbits faster (moves with greater tangential velocity) around the planet? Explain your answer with calculations, words and sketches.

8

A

B

Blueberry Muffin Nov 29-30, 2016 Mr. Han's Physics

M) Velocity x Time = Distance, so Time = Distance / Velocity. For a satellite in orbit at radius, r, (orbit radius is the distance between the satellite to the center of the planet it orbits) with speed, v, (tangential speed) the period, T, is the time it takes to complete one full orbit:

T = 2 π r / v ( 2 π r is the circumference of the orbit with radius r. ) Replace the v above with the equation for the speed (tangential velocity) of an object in circular obit:

v tangential = √ (G m(planet) / r(object to planet center)) yields the period of a satellite: Elon Musk's SpaceX launched a communication satellite into orbit which appears to “hover” over one point on the earth’s equator. Such a satellite is said to be in a geostationary orbit. What is the radius of such a satellite's orbit?


N)

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The diameter of a lazy susan is 12.0 cm. When it spins at 54 revolutions per minute, what is the speed of a point at the outside edge of the lazy susan, 6.0 cm from its center?


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O) In the following three examples of Uniform Circular Motion (UCM), each object travels at a constant tangential velocity in a circular orbit. Because they move in UCM, they all have Centripetal Acceleration (ac ) directed toward the center of their orbits. This ac is due to the fact that their tangential velocity (vt) constantly changes direction, toward the center of their orbits.

ac = vt2 / r where r is the radius of the orbit. If there is acceleration, then there must be a net force.

Just as F = ma, Centripetal Force (Fc) = mac = mvt2 / r In each of the following examples of UCM, explain what keeps the respective objects moving in their orbits and thus is responsible for creating the Centripetal Forces (Fc). What information would you need to determine the tension in the string for the first item below? Use words and sketches to explain your answers.

1.  Tennis ball being swung on string in a horizontal orbit. 2.  Car turning in a circle. 3.  Moon orbiting the Earth.


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P) A coin of 1.0 kg has been set revolving friction-less on a granite table in a circle of radius 2.0 meters at a constant velocity around a hole in the table in the middle of the circle. The revolving coin is tied to a string that travels through the hole and on the other end this string is connected to a wooden block of 5.0 kg. The wooden block stays at a the same height below the table because it is in equilibrium with the revolving coin. 1.  What is the tangential speed of the coin as it revolves? 2.  What cause the Centripetal acceleration of the revolving coin? 3.  What is the magnitude of the Centripetal Force on the revolving coin? 4.  What is the tension force the string experiences as the coin revolves?

Show your calculations.


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Q) A Smart car of 900 kg traveling at 1 m/s can barely make a tight circular turn of radius 5 m without skidding sideways. If the driver using the same car, tires and road wants to travel at 1.4 m/s while making a circular turn without skidding sideways, what would be the smallest possible radius of this turn? Show your calculations and make a sketch to explain your answer.


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R) You are swinging a tennis ball of mass 0.10 kg on a string at a constant speed in a circle of radius 1 meter in the vertical direction. What is the slowest speed you can swing the ball and still make a circle with your swing? At this speed, what is the tension on the string at the bottom of circle? Show your calculations and make a sketch to explain your answer.


Blueberry Muffin Nov 29-30, 2016 Mr. Han's Physics 15

1) Use the following equation to calculate the mass of the Earth in kilograms. Show your work in detail, including correct units for all quantities. Anyone can find the mass of the Earth on Google. So this problem is about using this equation.


2) Use the following equation to calculate the tangential velocity in m/s of the Earth as it orbits around the Sun. Show your work in detail, including correct units for all quantities. Anyone can find this fact on Google. So this problem is about using this equation. G = 6.7 x 10 -11 Nm2/kg2 m1 = mass of the Sun, m2 = mass of the Earth


3) Use the following general equation for potential energy due to gravity, U, and conservation of mechanical energy (PEi + KEi = PEf + KEf) to calculate the escape velocity for Earth. This is the velocity with which a projectile must be launched from Earth's surface to escape completely Earth's gravitational pull. Show your work in detail, including correct units for all quantities. G = 6.7 x 10 -11 Nm2/kg2 , M = mass of Earth, m = mass of projectile, r = distance between projectile and center of Earth.


Blueberry Muffin Nov. 29/30, 2016 Period: Names: 4) AP Physics problem: Show how the following general equation for potential energy due to gravity, U, reduces to the more familiar ΔPEg = mgh when h is the height above the surface of the Earth and h is small compared to the radius of the Earth. Show your work in detail, including correct units for all quantities. G = 6.7 x 10 -11 Nm2/kg2 , M = mass of Earth, m = mass of object with ΔPEg

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ANSWERKEY


A) A car is traveling around a curve at a steady 4 m/s. Which vector shows the direction of the car’s acceleration? Why? A. B. C. D. E. The acceleration is zero. By definition, any object moving in a circular motion is subjected to an acceleration that is directed toward the center of that circle. This acceleration is called the Centripetal acceleration. B) A toy car moves around a circular track at constant speed. It suddenly doubles its speed — a change of a factor of 2. As a result, the centripetal acceleration changes by a factor of: A. 1/4 B. 1/2 C. No change since the radius doesn’t change. D. 2 E. 4 Use calculations to support your answer. Fcentripetal = m acentripetal = m v2tangential / r So as v doubles, the square of v becomes (2v)2 which is 4v2 so the increase is 4 times more.


1 point maximum

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C) A metal ball at the end of a rope is being swung in a perfectly horizontal circle at a constant speed. The ball is accelerating because (select one and explain):

A. The speed is changing. B. The direction is changing. C. The speed and the direction are changing. D. The ball is not accelerating, because it's moving at a constant speed. E. The ball is not accelerating, because its movement is perpendicular to the force of gravity.

Explain your answer with words and sketches. Centripetal acceleration exists even when the object travels at a constant speed as long as the object moves in a circular motion. The fact that the object keeps changing direction (keeps turning) results in that object having an acceleration – the Centripetal acceleration. D) If the ball is accelerating, what is the direction of the acceleration? (select one and explain)

A. Tangent to the circle, in the direction of the ball’s tangential velocity. B. Toward the center of the circle. C. Towards the direction opposite to the force of tension on the rope. D. The ball is not accelerating. E. The ball is not accelerating due to the speed of its swing, but it is accelerating downward (towards ground) because of a net force due to gravity.

Explain your answer with words and sketches. See answer to question A


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1 point maximum

E) A ball at the end of a rope is being swung in a horizontal circle. What force is producing the centripetal acceleration of the ball?

A. Gravity B. Air resistance C. Normal force D. Tension in the rope E. Friction between the ground and person's feet.

Explain your answer with words and sketches. F) What is the direction of the net force on the ball?

A. Tangent to the circle B. Toward the center of the circle C. There is no net force. D. Toward the ground. E. Toward the direction opposite to the tension in the rope.



1 point maximum

1 point maximum

G) If you were a cosmonaut in space orbiting the Earth, you would feel weightless because:

A. Gravity would be the only force acting upon you. B. The net force on you would be zero. C. The centrifugal force balances the gravitational force. D. The force due to gravity is extremely small (but not zero) so you would feel weightless. E. There’s no gravity in outer space.



1 point maximum

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H) You are curled up in a ball on the ground next to a red panda which is similarly curled up in a ball. The distance from your respective centers is 0.50 meters. Calculate the gravitational force between you. Assume that you have a mass of 40 kg and the red panda a mass of 10 kg. ANSWER KEY

F = (6.7 x 10 -11) (40) (10) / (0.50)2

F = 5.4 x 10 -6 / (0.25) F = 1.1 x 10 -7 N (approx weight of 1 long strand of human hair)

RmmGF redpandayoundayouonredpa 2=


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H) If Planet Susie has twice the mass of Planet Patrick, then the gravitational force Planet Susie exerts on Planet Patrick, compared to the gravitational force Planet Patrick exerts on Planet Susie, is:


Explain your answer with words and sketches. The force of gravity is due to the interaction of two objects, m1 and m2. Those two objects experience the same force of gravity according to Newton’s Third Law. Moreover, m1 and m2 are interchangeable in the F gravity equation. I) If the distance separating Planet Susie from Planet Patrick were to increase to twice the original distance, then how would the gravitational force Planet Susie exerts on Planet Patrick change? It would be:


Explain your answer with words and sketches. As “R” doubles in the F gravity equation, R2 quadruples and making the fraction one fourth as large.


RmmGFgravity 2 21=

RmmGFgravity 2 21=

J) If your mass is 150 kg, how much would you weigh (in N) on the Moon? What % is this of your weight on Earth? If g on Earth is 9.8 m/s2, show the calculation for finding g on the Moon. Don't just look up the answer; instead show how you would use this equation to calculate the answer. ANSWER KEY

Fg = m1 g = G m1 m2 / R2

Fg (Moon) = m(you) g(Moon) = G m(you) m(Moon) / R(radius of Moon)2

g = G m(Moon) / R(radius of Moon)2

g = (6.7 x 10 -11) (7.3 x 10 22) / (1.7 x 10 6)2

g = 1.7 x m/s

Fg (you on moon) = 150 kg ( 1.7 x m/s2 ) = 255 N This is your weight.

Fg (you on Earth) = 150 kg ( 9.8 x m/s2 ) = 1470 N

Fg (you on moon) is 17% of Fg (you on Earth) 26

RmmGFgravity 2 21=



K) If your mass is 50 kg, on which planet would you weigh the most and why? R = planet radius M = planet mass Explain your answer with calculations, words and sketches. Your weight on a planet is another name for F gravity between you and that planet. So look for the planet which produces the greatest F gravity. Looking at the F gravity equation above, you can see that as mass increases from M to 2M to 3M, radius increases from R2 to (2R)2 which is 4R2 to (3R)2 which is 9R2 .

M 2M 3M R2 4R2 9R2

Which fraction is greatest?

RmmGFgravity 2 21=

L) Moons A and B both orbit around a planet. If Moon B's mass is twice that of Moon A, and Moon B's distance from the planet's center is 50% greater than that of Moon A, which Moon orbits faster (moves with greater tangential velocity) around the planet? Explain your answer with calculations, words and sketches. ANSWER KEY F = ma Fcentripetal = m acentripetal = m v2tangential / r Fcentripetal = Fgravity thus m(Moon)v2tangential / R(Moon to planet center) = G m(Moon) m(planet) / R(Moon to planet center)2 m(Moon)v2tangential / R(Moon to planet center) = G m(Moon) m(planet) / R(Moon to planet center)2 m(Moon)v tangential / R(Moon to planet center) = √ (G m(planet) / R(Moon to planet center)) So a moon's tangential velocity (orbit speed) depends on only the mass of the planet around which it orbits and the distance between the center of the moon and the planet. So the mass of the moon is irrelevant. Therefore, planet A orbits faster due to its smaller distance to the center of the planet.

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A

B

RmmGFgravity 2 21=


M) Velocity x Time = Distance, so Time = Distance / Velocity. For a satellite in orbit at radius, r, (orbit radius is the distance between the satellite to the center of the planet it orbits) with speed, v, (tangential speed) the period, T, is the time it takes to complete one full orbit: T = 2 π r / v ( 2 π r is the circumference of the orbit with radius r. ) Plugging in the equation for the speed (tangential velocity) of an object in circular obit:

v tangential = √ (G m(planet) / r(object to planet center)) yields the period of a satellite: Elon Musk's SpaceX launched a communication satellite into orbit which appears to “hover” over one point on the earth’s equator. Such a satellite is said to be in a geostationary orbit. What is the radius of such a satellite's orbit? ANSWER KEY Geostationary orbit means the object rotates once per 24 hours. So T is 24 hours. Convert 24 hours to 8.64 x 104 seconds


srad

60)2(54 π

ω =

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N) ANSWER KEY The diameter of a lazy susan is12.0 cm. When it spins at 54 revolutions per minute, pm, what is the velocity of a point at the outside edge of the lazy susan, 6.0 cm from its center? Angular velocity ω = 54 revolutions per second; each revolution is 2 π rad; so ω = 5.6 rad/s r = 0.06 m v = ω r v = (5.6 rad/s) (0.06 m) = 0.34 m/s Details: 5.6 rad x C x 2 π r = 5.6 (r) = 5.6 (0.06 m) / 1 s = 0.34 m/s 1 s 2 π rad C 1 s

C = circumference of circle.

There are 2 π rad per revolution and 1 C per revolution.

Also, the linear distance of C is 2 π r.

So use these two conversion factors.


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O) In the following three examples of Uniform Circular Motion (UCM), each object travels at a constant tangential velocity in a circular orbit. Because they move in UCM, they all have Centripetal Acceleration (ac ) directed toward the center of their orbits. This ac is due to the fact that their tangential velocity (vt) constantly changes direction, toward the center of their orbits.

ac = vt2 / r where r is the radius of the orbit. If there is acceleration, then there must be a net force.

Just as F = ma, Centripetal Force (Fc) = mac = mvt2 / r In each of the following examples of UCM, explain what keeps the respective objects moving in their orbits and thus is responsible for creating the Centripetal Forces (Fc). What information would you need to determine the tension in the string for the first item below? Use words and sketches to explain your answers.

1.  Tennis ball being swung on string in a horizontal orbit. String tension. 2.  Car turning in a circle. Tire friction against road. 3.  Moon orbiting the Earth. Gravitational force.

To determine the tension in the string, you need to know that Tension = Centripetal Force (Fc) = mac = mvt2 / r So to find tension, you need to know m, vt and r


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P) A coin of 1.0 kg has been set revolving friction-less on a granite table in a circle of radius 2.0 meters at a constant velocity around a hole in the table in the middle of the circle. The revolving coin is tied to a string that travels through the hole and on the other end this string is connected to a wooden block of 5.0 kg. The wooden block stays at a the same height below the table because it is in equilibrium with the revolving coin. 1.  What is the tangential speed of the coin as it revolves? 2.  What causes the Centripetal acceleration of the revolving coin? 3.  What is the magnitude of the Centripetal Force on the revolving coin? 4.  What is the tension force the string experiences as the coin revolves?

ANSWER KEY

String Tension = weight of block =

m2g = 5.0 kg x 9.8 m/s2 = 49 N

String Tension = FCentripetal

FCentripetal causes the coin m1 to move in a circle

FCentripetal = m1 v2 / r

v2 = FCentripetal r / m1

v2 = 49 N x 2 m / 1.0 kg = 98 m2/s2

v = 9.9 m/s


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Q) A Smart car of 900 kg traveling at 1 m/s can barely make a tight circular turn of radius 5 m without skidding sideways. If the driver using the same car, tires and road wants to travel at 1.4 m/s while making a circular turn without skidding sideways, what is the smallest radius of this turn? ANSWER KEY FCentripetal causes the car to move in a circle and is due to the car tires exerting a frictional force against the road. When the car starts skidding sideways while turning, the frictional force is no longer able to provide the necessary FC to keep the car in the circular turn. Since the drive is using the same car, tires and road for both turns, the maximum frictional force generated by the tires is the same for both turns. This maximum frictional force = FCentripetal

For the first turn, FC = m v12 / r1 For the second turn, FC = m v22 / r2

FC is the same for both turns because it depends on the car tires and road, which remain the same for both turns.

m v12 / r1 = m v22 / r2

v12 / r1 = v22 / r2

r2 = ( v22 / v12 ) r1

r2 = ( 1.42 / 1.02 ) 5 = 9.8 m This is the radius of the tightest turn the car can make if it turns in a circle at 1.4 m/s


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R) You are swinging a tennis ball of mass 0.10 kg on a string at a constant speed in a circle of radius 1 meter in the vertical direction. What is the slowest speed you can swing the ball and still make a circle with your swing? At this speed, what is the tension on the string at the bottom of circle? ANSWER KEY At all moments during circular swing, the ball traveling at tangential velocity, v, requires FC = m v2 / r to keep the ball in a circle. The source of this FC is the tension in the string and/or gravity acting upon the ball. Since FC = m v2 / r the object's velocity is the lowest when FC is also the lowest. In the top half of the circle, gravity (mg) works together with string tension (T) to create FC. In the bottom half, gravity works against string tension to create FC. So in the top half, T + mg = FC and in the bottom half, T - mg = FC When the ball travels the slowest tangential velocity possible to still make a full circle, then T = 0. This occurs at the top of the swing. If T < 0, then the string would no longer remain taut and thus the ball would no longer follow a circular path. When T = 0 at the top of the swing, 0 + mg = FC so 0.1 kg x 9.8 m/s2 = FC The ball velocity which requires this amount of FC is given by FC = m v2 / r

So 0.1 kg x 9.8 m/s2 = FC = (0.1 kg) x (v2 ) / 1.0 m v2 = 9.8 , v = 3.1 m/s

At the circle's bottom, T - mg = FC or T = FC + mg and FC = m v2 / r

T = (0.1 kg x (3.1 m/s)2 / 1.0 m) + (0.1 kg x 9.8 m/s2 ) = 1.9 N

FC mg

FC

mg

T


blueberry muffin nov. 29/30, 2016 period: names...blueberry muffin nov 29-30, 2016 mr. han's physics...

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