Vertical Circular Motion A demo T8 T8

Vertical Circular Motion

A demo

• http://www.youtube.com/watch?v=mrKLJjx9VT8

Motion in a Vertical CircleConsider the forces on a ball attached to a string as it moves in a vertical loop.As with any object moving in a circle there is a net force acting towards the center of the circle.

+

T

mg

Bottom

Maximum tension T, W opposes Fc

+ T

mg

Top Right

Weight has no effect on

T

+

Tmg

Top Right

Weight causes small decrease in tension T

T

mg

+

Left Side

Weight has no effect on

T

+

T

mg

Bottom

T

mg

Top of PathTension is

minimum as weight helps

Fc force

+

The tension must adjust so that the central force remains constant throughout the circle.

Motion in a Vertical Circle

r

v

v

Resultant force toward

centerFc =

mv2

r

Consider TOP of circle:

AT TOP:

T

mg

T

mg

+

mg + T = mv2

r

T = - mg mv2

r

Vertical Circle; Mass at bottom

r

v

v

Resultant force toward

centerFc =

mv2

r

Consider bottom of circle:

AT Bottom:

Tmg

+

T - mg = mv2

r

T = + mg

mv2

r

T

mg

Example:A 1.7 kg object is swung from the end of a 0.60 m string in a vertical circle. If the time of one revolution is 1.1 s, what is the tension in the string:a) at the top?

b) at the bottom?

Now suppose the mass is spun with just enough speed to keep it moving in a

circular path. What is the tension in the string at the top?

We say that the mass at the peak of the arc is weightless, because the net force working on it is only gravity. This is the

same as an object in total free fall.

0T

Example:

An object is swung in a vertical circle with a radius of 0.75 m. What is the minimum speed of the object at the top of the motion for the object to remain in circular motion?

The Loop-the-Loop and Ferris Wheel

Same as string, N replaces T

AT TOP:

N

mg

+

AT BOTTOM:

Nmg

+

N = - mg

mv2

r

N= + mg

mv2

r

r

v

v

Example What is the apparent weight of a 60-kg person as she moves through the highest point when r = 45 m and the speed at that point is 6 m/s?

n

mg

+

r

v

v

mg - N = mv2

rn = mg -

mv2

r

Apparent weight will be the normal force at the

top:

22 (60kg)(6m/s)

60 kg(10 m/s )45 m

N N= 540 NN= 540 N

Loop the loopWhat is minimum height to release the ball so that it stays in the loop?

Vertical Circular Motion – Ferris Wheel

m = 50 kg, R = 15 m, At what v would the rider lose contact with the seat at the top?

N

mg

+y

2

2

/

0 50 (10 /15)

N mg mv r

v

At Top:

Answer: v =12.2 m/s

Hint: N=0

Spinning Bucketm = 80 g, r = 1.2 m, What is the minimum speed v so Rhino stays in the bucket?

Fy mac mv 2

r

N mg

+y

mg N mv 2

r

mgmv 2

r

v gr 9.81.2 3.43m /s

Rhino is most likely to lose contact at the top:

N mgmv 2

r

N 0.8(9.8 3.432

1.2) 15.7 (N)

When Rhino loses contact, N = 0

r

v

What is N on Rhino at top if v = 7 m/s ?

Answer: 24.8 (N)

What is the force of the bucket (Normal force) on Rhino at the bottom?

N

mg

+y

Fy mac mv 2

r

N mgmv 2

r

Banked Curves:

When cars travel at high speeds on highways, they do not rely solely on friction to keep the cars from sliding off the road. A greater centripetal force can exist if the turn is banked.

Consider a car traveling at a constant speed around a frictionless banked corner.

On a frictionless corner only Fg and FN act on the car.

Note that in this case FN is larger Fg because it both:

(1) Balances Fg

(2) Provides Fc towards the center of the circle

The sum of FN and Fg must equal Fc .FN

Fg

Fc

Example

Calculate the angle at which a frictionless curve must be banked if a car is to round it safely at a speed of 22 m/s if its radius is 475 m.

Banked Turns

θ

N

mg

+y

+x

Banked turn with no friction m = 1000 kg, r = 20 m, θ = 20o

What v should car have?

r

v

Fy 0

N cos mg0

N mg

cos

N ≠ mgcosθ because there is a component of acceleration in the normal direction

Fx mac mv 2

r

mgsincos

mv 2

r

N sin mv 2

r

v gr tan

v 9.820tan(20) 8.4 m /s

Banked Turns with frictionWhat if car goes faster than 8.4 m/s? Need friction to keep it from sliding up banked turn

Static friction acts parallel to and down the banked turn

θ

r

vN

mg

+y

+x

fs

What if car goes slower than 8.4 m/s? Need friction to keep it from sliding down banked turn

+yN

mg+x

fs

Static friction acts parallel to and up the banked turn

Not easy to solve for vmax or vmin on banked turns with friction

Example

A 0.25 kg toy plane is attached to a string so that it flies in a horizontal circle with a radius of 0.80 m. The string makes a 28o angle to the vertical. What is its period of rotation?

28o