1. Vertical Circular Motion

2. A demo

3. Motion in a Vertical Circle
Consider the forces on a ball attached to a string as it moves in a vertical loop. T mg
Top of Path
Tension is minimum as weight helps Fc force + T mg +
Left Side
Weight has no effect on T + T mg
Bottom
Maximum tension T, W opposes Fc + T mg
Top Right
Weight causes small decrease in tension T + T mg
Top Right
Weight has no effect on T
As with any object moving in a circle there is a net force acting towards the center of the circle. + T mg
Bottom
The net force is a constant force, therefore the tension must adjust so that the central force remains constant throughout the circle.

4. 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:
at the top?
b) at the bottom?

5. 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.

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

7. Apparent weight will be the normal force at the top:mg + r 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?
Apparent weight will be the normal force at the top:
mg - N =
mv2 r
n = mg -
mv2 r
N= 552 N

8. 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?
Hint: N=0 N mg +y
At Top:
Answer: v =12.2 m/s

9. Spinning Bucket v
m = 80 g, r = 1.2 m, What is the minimum speed v so Rhino stays in the bucket?
Rhino is most likely to lose contact at the top: r
What is the force of the bucket on Rhino at the bottom? +y N mg
What is N on Rhino at top if v = 7 m/s ? N mg +y
Answer: 24.8 (N)
When Rhino loses contact, N = 0

10. Problems in 2 dimensions
Sometimes horizontal or vertical motion problems involve angles

11. 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.

12. FN Fg Fc On a frictionless corner only Fg and FN act on the car.
Using trigonometry tanƟ = Fc / Fg FN Fg Fc

13. 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.

14. Banked Turns Banked turn with no friction
m = 1000 kg, r = 20 m, θ = 20o
What v should car have? r v N mg +y +x θ
N ≠ mgcosθ because there is a component of acceleration in the normal direction

15. Banked Turns with friction
What if car goes faster than 8.4 m/s? Need friction to keep it from sliding up banked turn
Banked Turns with friction
What if car goes slower than 8.4 m/s? Need friction to keep it from sliding down banked turn N mg +y +x fs r v
Static friction acts parallel to and up the banked turn θ +y N mg +x fs
Static friction acts parallel to and down the banked turn
Not easy to solve for vmax or vmin on banked turns with friction

16. The Conical Pendulum
A conical pendulum consists of a mass m revolving in a horizontal circle of radius R at the end of a cord of length L.
T cos q q h T L R T q mg
T sin q
Note: The inward component of tension T sin q gives the needed central force.

17. Angle q and velocity v: T sin q = T cos q = mg L h T mg R mv2 v2 R gR
tan q = v2 gR
Solve two equations to find angle q
mv2 R
T sin q =
T cos q = mg

18. Example : A 2-kg mass swings in a horizontal circle at the end of a cord of length 10 m. What is the constant speed of the mass if the rope makes an angle of 300 with the vertical?
1. Draw & label sketch.
q = 300 q h T L R
2. Recall formula for pendulum.
Find: v = ?
3. To use this formula, we need to find R = ?
R = L sin 300 = (10 m)(0.5)
R = 5 m

19. Example 6(Cont.): Find v for q = 300h T L R
q = 300
R = 5 m
4. Use given info to find the velocity at 300.
R = 5 m
g = 10 m/s2
Solve for v = ?
v = 5.32 m/s

20. 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