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Torque Chap 8 Units: m N 2.

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Presentation on theme: "Torque Chap 8 Units: m N 2."— Presentation transcript:

1 Torque Chap 8 Units: m N 2

2 Lever Arm is the vector from the point of application of a force to the axis of rotation (not necessarily the radius!). F F F

3 Convention: CCW is positive (thumb is up) CW is negative (thumb is down)

4 Which of the forces pictured as acting upon the rod will produce a torque about an axis perpendicular to the plane of the diagram at the left end of the rod? F1 F2 Both. Neither. F2 will produce a torque about an axis at the left end of the rod. F1 has no lever arm with respect to the given axis.

5 The two forces in the diagram have the same magnitude
The two forces in the diagram have the same magnitude. Which orientation will produce the greater torque on the wheel? F1 F2 Both. Neither. F1 provides the larger torque. F2 has a smaller component perpendicular to the radius.

6 A 50-N force is applied at the end of a wrench handle that is 24 cm long. The force is applied in a direction perpendicular to the handle as shown. What is the torque applied to the nut by the wrench? 6 N·m 12 N·m 26 N·m 120 N·m 0.24 m  50 N = 12 N·m

7 What would the torque be if the force were applied half way up the handle instead of at the end?
6 N·m 12 N·m 26 N·m 120 N·m 0.12 m  50 N = 6 N·m

8 What is the torque about the axle of the merry-go-round due to the 50-N force?
+60 N·m -60 N·m +120 N·m -120 N·m -(1.2 m  50 N) = -60 N·m (clockwise)

9 What is the net torque acting on the merry-go-round?
+36 N·m -36 N·m +96 N·m -60 N·m +126 N·m 96 N·m (counterclockwise) 60 N·m (clockwise) = +36 N·m (counterclockwise)

10 Equilibrium

11 We want to balance a 3-N weight against a 5-N weight on a beam
We want to balance a 3-N weight against a 5-N weight on a beam. The 5-N weight is placed 20 cm to the right of a fulcrum. What is the torque produced by the 5-N weight? +1 N·m -1 N·m +4 N·m F = 5 N  = - Fl l = 20 cm = 0.2 m = - (5 N)(0.2 m) = -1 N·m

12 How far do we have to place the 3-N weight from the fulcrum to balance the system?
2 cm 27 cm 33 cm 53 cm F = 3 N l =  / F  = +1 N·m = (+1 N·m) / (3 N) = 0.33 m = 33 cm

13 Quick Quiz A constant net torque is applied to an object. Which one of the following will not be constant? angular acceleration angular velocity moment of inertia center of gravity

14 Center of Mass (CM)

15

16 You have learned have that:
perpendicular r or l is the distance to the center of the rotation!!!!!! But torque is also equal to: Where and is measured in rad/s2 Is called MOMENT OF INERTIA (I), and is measured in kgm2

17 MOMENT OF INERTIA (I), measured in kgm2
Objects with different shapes, will have different I: Objects with the same shape, but spinning around different axis, will have different I: R L

18 The only cases when you can actually calculate the MOMENT OF INERTIA (I) will be:
Case A: When the object is very small, punctual (not an extended object), located at a distance r from the center of the rotation. Ex. a small ball, a piece of putty, etc.

19 The only cases where you can actually calculate the MOMENT OF INERTIA (I) will be:
Case B: Only if you can ignore the mass of the connecting rod:

20 Case C: Only if you can ignore the mass of the connecting rod: for all masses!

21 Only if you can ignore the mass of the connecting rod:
Case D: Only if you can ignore the mass of the connecting rod: 3 2 4 1

22 Otherwise, consult the table in your book.

23 Newton’s 2nd Law for Rotation
Draw free body diagrams of each object Only the cylinder is rotating, so apply St = I a The bucket is falling, but not rotating, so apply SF = ma Remember that a = a r

24 Quick Quiz The two rigid objects shown in the figure below have the same mass, radius, and angular speed. If the same braking torque is applied to each, which takes longer to stop? A B more information is needed

25 Rotational Kinetic Energy (KEr)
Unit: joule (J) From now on, when bodies are TRANSLATING (moving along a straight line) AND ALSO ROTATING (such as a ball rolling down an incline), the Total Mechanic Energy (TME) will be:

26 If Work by external force is zero, then:
Total KE Work done by external force F If Work by external force is zero, then:

27 Quick Quiz Two spheres, one hollow and one solid, are rotating with the same angular speed around an axis through their centers. Both spheres have the same mass and radius. Which sphere, if either, has the higher rotational kinetic energy? the hollow sphere the solid sphere They have the same kinetic energy.

28 Angular Momentum (L) L is measured in kgm2/s or also Js
And, just like we’ve seen for linear momentum (p), the relation between L and torque is:

29 Angular momentum (L) is a vector.
L has the direction of the change in torque. L is in the same axis as torque (parallel or antiparallel). L is in the same direction as angular velocity. L is positive is counterclockwise. L is negative if clockwise.

30 Conservation of Angular Momentum (L)
If Then L is conserved:

31 Conservation of Angular Momentum
Linear momentum is conserved if the net external force acting on the system is zero. Angular momentum (L) is conserved if the net external torque (τ) acting on the system is zero.

32 Conservation of Angular Momentum
How do spinning skaters or divers change their rotational velocities?

33 Conservation of Angular Momentum
Change the rotational inertia (I) in order to change the rotational velocity (ω)

34 Larger I (because R is larger) results in
smaller ω, in order to keep L = I ω constant. Smaller I (because R is smaller) results in larger ω, in order to keep L = I ω constant.

35 Conservation of Angular Momentum
Change the rotational inertia (I) in order to change the rotational velocity (ω)

36 I1 / I2 I2 / I1 I1 / (I1 + I2) I2 / (I1 + I2) Quick Quiz
A horizontal disk with moment of inertia I1 rotates with angular speed ω1 about a vertical frictionless axle. A second horizontal disk, with moment of inertia I2 drops onto the first, initially not rotating but sharing the same axis as the first disk. Because their surfaces are rough, the two disks eventually reach the same angular speed ω. The ratio ω/ω1 is equal to I1 / I2 I2 / I1 I1 / (I1 + I2) I2 / (I1 + I2)


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