1. Rotational Mechanics

2. Torque The ability of a force to rotate an object around some axis
Produces rotation
Produced when a force is applied with leverage
Longer the lever = more torque
More force = more torque
Ex: Door

3. More Torque τ = Fd Torque = Force x lever arm Units: N x m = Nm
Lever arm: distance from the turning axis to the point of contact
Force is always perpendicular to lever arm.
Torque = Force x lever arm
τ = Fd
Units: N x m = Nm

4. Torque also depends on the angle between force and lever arm
τ = Fd (sin θ)
Perpendicular = most torque
Increase angle = less torque
Increase angle more = least torque

5. Sign of Torque Counterclockwise rotation = (+) torque

6. Problems
A force of 50N is applied perpendicular to the door, .5m from the hinges. How much torque is produced?
What if the force were applied .1 m from the hinges?
If you cannot exert enough torque with a wrench to turn a stubborn bolt, would more torque be produced if you fastened a length of rope to the wrench handle?

7. Example Problem
Ned tightens a bolt in his car engine by exerting 12 N of force on his wrench at a distance of 0.40 m from the fulcrum. How much torque must Ned produce to turn the bolt?

8. Sample Problem 8A
A basketball is being pushed by two players during tip-off. One player exerts a downward force of 11 N at a distance of 7.0 cm from the axis of rotation. The second player applies an upward force of 15 N at a perpendicular distance of 14 cm from the axis of rotation. Find the net torque acting on the ball.

9. Balanced Torques
Ex: on a seesaw, balance can be achieved if clockwise torque = counter clockwise torque
If a meter stick is balanced in the center A 20 N block is attached to the 80cm mark, and another block is attached to the 10cm mark. What is the weight of the second block?

10. Balancing Torque Problem
Mabel and Maude are seesawing on the school playground and decide to see if they can move to the correct location to make the seesaw balance. Mabel weighs 400. N and she sits 2.00 m from the fulcrum of the seesaw. Where should 450. N Maude sit to balance the seesaw?

11. Center of Mass Average position of the mass of an object
Object will rotate around this point is gravity is the only force acting on it.
When working problems – all mass can be considered to be concentrated here
For regular shapes – CoM is at the center of the object
Center of Gravity – average position of weight of an object
An object will fall over if the center of gravity extends beyond the area of support.
Falling over
C o G outside the object
Usually, the center of mass and center of gravity are at the same point.

12. Rotational Inertia
The resistance of an object to changes in its rotational motion.
AKA: moment of inertia
An object rotating about an axis tends to keep rotating about that axis.

13. Moment of Inertia
Torque is required to change the rotational state of motion of an object.
Rotational inertia depends on mass and on the distribution of the mass.
The greater the distance between the axis and the majority of the mass, the greater the rotational inertia.
Ex: Bat

14. Formulas for Rotational Inertia

15. Rolling
Which will roll down an incline with greater acceleration, a hollow cylinder or a solid cylinder if both have the same mass and diameter?
Solid cylinder
Why?
Smaller cylinder has less rotational inertia.

16. Rotational Equilibrium
Equilibrium requires zero net force and zero net torque
Translational Equilibrium
ΣF = 0
Rotational Equilibrium
ΣΤ = 0

17. Sample Problem 8B
A uniform 5.00 m long horizontal beam that weighs 315 N is attached to a wall by a pin connection that allows the beam to rotate. Its far end is supported by a cable that makes an angle of 53o with the horizontal, and a 545 N person is standing 1.50 m from the pin. Find the force in the cable, FT, and the force exerted on the beam by the wall, R, if the beam is in equilibrium.

18. Newton’s 2nd Law of Rotation
Tnet = Iα
Net torque = moment of inertia x angular accel.
A bicycle tire of radius .330 m and mass 1.50 kg is rotating at 98.7 rad/s. What torque is needed to stop the tire in 2.00 s?

20. Sample Problem 8C
A student tosses a dart using only the rotation of her forearm to accelerate the dart. The forearm rotates in a vertical plane about an axis at the elbow joint. The forearm and dart have a combined moment of inertia of kg x m2 about the axis, and the length of the forearm is 0.26 m. If the dart has a tangential acceleration of 45 m/s2 just before it is released, what is the net torque on the arm and dart?

21. Angular Momentum Angular momentum = inertia of rotation
Angular momentum = rotational inertia x rotational velocity
Angular momentum L =  x 
Angular momentum always acts outward along the axis
Angular momentum = mvr

22. Law of Conservation of Angular Momentum
If no unbalanced external torque acts on a rotating system, the angular momentum of that system is constant.

23. Example
A merry-go-round rotates at the rate of 0.30 rad/s w/an 80. kg man standing at a point 2.0 m from the axis of rotation. If the man walks to a point 1.0 m form the center, what will be the new angular speed?
Merry-go-round = solid cylinder, m = 6.5 x 102 kg, r=2.0 m

24. Sample Problem 8D
A 65 kg student is spinning on a merry-go-round that has a mass of 5.25 x 102 kg and a radius of 2.00 m. She walks from the edge of the merry-go-round toward the center. If the angular speed of the merry-go-round is initially 0.20 rad/s, what is the angular speed when the student reaches a point 0.50 m from the center.

25. Homework, yea!!!! Practice problems
8D: 3, 4, & 5
Stop complaining, it’s only 6 problems

26. Rotational Kinetic Energy
Just as with an object moving in a straight line, a rotating object also possess kinetic energy
KErot = ½ I ω2
This causes the conservation of energy equation to change to
(PE + KEtrans +KErot)i = (PE + KEtrans +KErot)f
We call this Mechanical Energy
ME = KEtrans + KErot + Peg
ME = ½ mv2 + ½ Iω2 + mgh

27. Example
A 1.5 kg bicycle tire of radius .33m starts from rest and rolls down a hill 14.8m high. What is the translational speed when it reaches the bottom of the hill?

29. Sample Problem 8E
A solid ball with a mass of 4.10 kg and a radius of m starts from rest at a height of 2.00 m and rolls down a 30.0o slope. What is the translational speed of the ball when it leaves the incline?

30. Homework
Practice 8E (all)

31. Simple Machines
Machine – any device that transmits or modifies force, usually by changing the force applied to an object.
Types of simple machines
Lever Inclined Plane
Wheel & Axle Pulley
Wedge Screw

33. Mechanical Advantage MA compares output force to input force
MA = Fout/Fin
Fout = resistance
Fin = effort
The MA can also be compared to the distance moved – like in a lever or incline
MA = din/dout
din = effort
dout = resistance

34. In the absence of friction, energy is conserved in a machine
The force applied may be decreased, if the distance it is applied is increased.
The product of the 2 remains constant
In reality, no machine exists without friction
We can calculate the efficiency of a machine by..
Eff = Wout/Win
Wout = resistance
Win = effort

35. Example
You are pushing a block up a ramp that is 3.0 m high and 6.0 m long. It is a 30.o ramp. The block has a mass of 5.0 kg. It is takes you 37 N of force to push it up the incline.
Draw this picture.
What is the Effort Force?
What is the Resistance Force?
What is the Effort Distance?
What is the Resistance Distance?
What is the Mechanical Advantage?
What is the Efficiency of the machine?

36. Homework
Section Review (all)

37. Let’s try these Starting on page 306:
#12, 14, 18, 21, 26, 31, 35, & 45