Rotational Motion

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Rotational Motion

How can a car be accelerating if it is traveling at a constant speed?

If it is changing direction

Centripetal Acceleration always points toward the center of the circle.

a_c = centripetal accel. (m/s²) v = velocity (m/s)

r = radius of circle (m)

v

a

= r

If there is an acceleration for a car

moving in a circle, there must also be a…

FORCE

F

= ma

F

= mv

a

= v

r

Centripetal Force always points toward

r

the center of the circle.

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Remember that the force in F

= ma

stands for the NET FORCE acting on an object.

So…the F

in F

=ma

stands for the NET FORCE acting on an object that is making it move in a circle.

Centripetal Force is a NET FORCE!!!

Applying a force to an object perpendicular to the direction of its motion causes the object to change direction but not speed.

A roller coaster car is moving around a loop, but the second half of the track is missing.

•Which path (A,B, or C) will the car travel? Explain.

 B - inertia would want to keep moving it in a straight line

A B C

F_N The centripetal force is

the net force causing it to move in a circle.

F_net = ma

F_net = ma_c F_W F_W+ F_N =

mv

r

Centripetal force

If all the forces are pointed down,

why doesn’t the car fall down? INERTIA

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A roller-coaster car has a mass of 500 kg when fully loaded with passengers.

A

B

10 m 15 m

a) If the car has a speed of 20 m/s at point A, what is the force of the track on the car (Normal Force) at this point?

At Pt. A  2 Forces  Weight (down) & Normal Force (up) So F_c = F_net = F_N – F_G = mv²

r

F_N = mv² = (500)(20)(20) + (500)(10) =

r 10 25,000 N

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b) What is the maximum speed the car can have at B in order for only gravity to hold it on the track?

At Pt. B  2 Forces  Weight (down) & Normal Force (down), A

B

10 m 15 m

So F_c = F_G = mv² r

v² = F_G r = (500)(10) x (15) = 150 = m 500

but the question only wants to consider gravity (Weight)

12.2 m/s

1) A 0.5 kg ball is swung on a string in a

horizontal circle of radius 2 m and at a velocity of 6 m/s. a) Draw a free body diagram

T

F_W

The ball is not changing speed, only changing direction, so this is a centripetal acceleration.

F_net = ma F_c = ma_c

b) What is the tension in the string?

Tension is the NET

FORCE causing the ball to move in a circle.

T

w

r m v T



N m) 9

(2

m/s) kg) (6

(0.5

T 



** Always call toward the center of the

circle as positive

d) What’s the centripetal force acting on the ball?

F_c = ma_c

F_c = (0.5 kg)(18 m/s²) F_c = 9 N

a_c = v² r

= (6 m/s)²

2 m = 18 m/s²

Toward center of the circle

c) What is the ball’s centripetal acceleration?

2) A ball (mass = 2kg) is swung in a vertical

circle at a speed of 5 m/s at a radius of 0.5 m.

a. What is the centripetal acceleration?

a_c = v² r

= (5 m/s)²

0.5 m = 50 m/s²

Toward center of the circle

b. What is the tension in the string at the top of the path? (Draw a free-body diagram)

F_net = ma

T + F_W = ma_c

R m v

w T



 F

) m/s kg)(10

2 m) (

(.5

m/s) kg) (5

(2

T

2





T

r w m v

T

2

 F



N 80



F_W

r mg m v

2





c. What is the tension in the string at the bottom of the path? (Draw a free-body diagram)

F_net = ma T - F_W = ma_c

R m v

w T



 F

N m) 20

(.5

m/s) kg) (5

(2

T 



T

R w m v

T

2

 F



N 120



F_W

3a) Draw the all forces that cause circular motion in the following situations of an

amusement park ride that rotates horizontally.

Tension

b) What is its acceleration if the ride has a radius of 10

meters, is moving at 12 m/s, and the car has a mass of 200 kg? (still in horizontal circle) a_c = v²

r

= (12 m/s)²

10 m = 14.4 m/s², toward the center of the circle

c) What is the centripetal force if it is moving at the same speed?

F_c = T = ma _c = (200 kg)(14.4 m/s²) = 2880 N

d) How fast would the car have to be moving to have an acceleration 3 times that of gravity?

a_c = v² r

v² = a_c x r = (3 x 10 m/s²) (10 m) v = 17.3 m/s

4. Monkey Joe’s cousin (m= 85 kg) tries to cross a river by

swinging from a 10 m long vine. His speed at the bottom of the swing is 8 m/s. He does not know that the vine has a breaking strength of 1000 N. Does he make it safely

across the river without the vine breaking?

T

F_W F_net = ma

T - F_W = ma_c

r m v w

T



 F

r w m v

T

2

 F



N m) 850

(10

m/s) kg) (8

(85 T

2



  1394 N

No, he won’t make it.

The vine will break because the tension (1394 N) is greater than the breaking strength (1000 N).

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5) A 1000 kg car is going in a circle at a constant speed of 24 m/s, and the coefficient of static friction between the tires and road is 0.8.

a) Draw a free body diagram.

b) If the car is turning, what force is causing it to move in a circle? What is the

direction of this force?

 The force of friction; towards the center of the circle.

d) What is the radius of the tightest circle that the car could make on a level parking lot so that it would not skid?

c) Find the force of static friction.

F_s,max = μ_sF _N = 0.8 x (1000 kg)(10 m/s²) F_s,max = 8000 N

F

= mv

r

r = mv

F

r = mv

= (1000 kg) (24 m/s)

F

8000 N = 72 m

6) Rank in order, from largest to smallest, the

centripetal accelerations a_ato a_e of particles a to e.

1. a_b > a_e > a_a > a_d > a_c 2. a_b = a_e > a_a = a_c > a_d 3. a_b > a_a = a_c = a_e > a_d 4. a_b > a_a = a_a > a_e > a_d 5. a_b > a_e > a_a = a_c > a_d

7) A ball on a string is swung in a vertical

circle. The string happens to break when it is parallel to the ground and the ball is moving up. Which trajectory does the ball follow?

1) a 2) b 3) c 4) d

1. n > w. 2. n = w. 3. n < w.

4. We can’t tell about n without knowing v. 8) A car is rolling over the top of a hill at speed v. At this instant,

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Every particle in the universe attracts every other particle.

F_G = Gravitational Force (N)

G = constant of Universal Gravitation = 6.673 x 10^-11 N•m²/kg²

r = distance between the two objects (m) m = mass of each object (kg)

F

= Gm

m

r

Monkey Joe has a mass of 30 kg and Monkey Jane has a mass of 25 kg. If they are

standing 2 m apart, what is the force of attraction between them?

F

F

= Gm

m

r

F

=

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The moon’s orbit around Earth

There is a gravitational force between the Earth and moon (responsible for tides)

F_G

This force makes it move in a circle…so it is a centripetal force.

1. one quarter as big.

2. half as big.

3. twice as big.

4. four times as big.

5. the same size.

The figure shows a binary star system. The mass of star 2 is twice the mass of star 1.

Compared to F_{2 on 1}, the magnitude of the force F_{1 on 2} is

1) If the force between you and the person sitting next to you is 3 x 10^-3 N, and each of you has a mass of 100 kg, how far apart are you sitting?

2) If the distance between you and your partner doubles, what will be the force of attraction between you.

Practice Questions

0.0149 m

7.5 x 10^-4 N

(reduced by ¼)

A satellite orbits the earth with constant speed at height above the surface equal to the earth’s radius. The magnitude of the satellite’s acceleration is

1. ¼g_earth 2. ½g_earth 3. g_earth 4. 2g_earth 5. 4g_earth

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The ability of a force to rotate a body about some axis.

 = Fd

 = Torque (N•m) (greek “tau”) F = Force (N)

d = lever arm (m)

A lever arm (d) is the perpendicular distance from the axis of rotation (pivot point) to a line drawn along the direction of force.

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The units of Torque are N•m, but do not confuse this with work (Energy) which has the same units (1 Joule = 1 N

•m)

Torque is a vector and work is a scalar

Units of Torque

 = Fd

 = Newton • meter

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F

d₁ F d₂

Which situation will create more torque?

(i.e. which will more effectively rotate the bar?)

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F

d

F F

Notice that F// does not cause the bar to rotate.

 = d F

What if the force is at an angle?

Rank in order, from largest to smallest, the five torques. The rods all have the same

length and are pivoted at the dot.

1.

2.

3.

4.

5.

A student holds a meter stick straight out with one or more masses dangling from it. Rank in order, from most difficult to least difficult, how hard it will be for the student to keep the meter stick from

rotating.

1. c > d > b > a 2. b = c = d > a 3. c > b > d > a 4. b > d > c > a 5. c > d > a = b

How much torque is created when a 50 N force is

applied perpendicularly 0.5 m away from the pivot point?

 = Fd

 = (50 N)(0.5 m)

 = 25 N

What size force would create a torque of 100 N•m on a door if it is applied 0.75 m from the hinge?

F _  d

m 0.75

m N

100 

 = 133 N

Position where all of the weight of an object can be considered to be

concentrated.

• Always draw the weight of an object at the center of mass

For an object in projectile motion, the center of mass follows a parabolic path.

Center of Mass or Center of Gravity

w

The hammer

rotates about its center of mass as it moves through

the air. As the rest of the hammer

spins, the center of mass moves

along the path of the projectile.

A uniform 40.0 N board supports two children weighing 500 N & 350 N. The support (called a fulcrum) is under the center of gravity of the board. The 500 N child is 1.50 m from the fulcrum.

a) Draw this situation

(b) Where should the 350 N child stand so that the system will be in equilibrium?

(500 N)(1.50 m) = (350 N)(d)

d = 2.14 m from the center

1.50 m

d = ? m F = 500 N

F = 350 N

The torques ( = Fd) of both children should be equal to each other but in the opposite direction so that _net = 0.

(c) Determine the upward force, F_N, exerted on the board by the fulcrum.

F_net = ma

F_net = F_N – 500 N – 350 N – 40.0 N = ma

F_N = 890 N

**There is no acceleration so a = 0.

F_N – 500 N – 350 N – 40.0 N = 0