# Definition. is a measure of how much a force acting on an object causes that object to rotate, symbol is, (Greek letter tau)

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1 Torque

2 Definition is a measure of how much a force acting on an object causes that object to rotate, symbol is, (Greek letter tau) = r F = rfsin, r = distance from pivot to force, F is the applied force and is the angle between r and F

3 SI unit of torque is N m, this is not the same as a joule Torque is a vector, rfsin is a vector crossproduct

4 Equal Torques = rfsin = 20 lb ft

5 The magnitude of the torque depends on the magnitude of the applied force, the distance between the force and the pivot and the direction the force acts Only the component of the force to the beam can cause rotation Pivot this force is to the beam and will cause the greatest torque

6

7 Centre of Mass (gravity) The point at which the mass of an object seems to act The centre of mass of a homogenous symmetric object is on an axis of symmetry An object is in a stable position when the centre of gravity is directly above or below the lowest support point.

8 No torque will result when a vertical line passes through the centre of gravity and the point of support

9 Locating Centre of Mass A long, thin beam has 3 masses distributed along it as shown. Locate the centre of mass relative to the left edge of the beam. 5.0 m 1.0 m 2.5 m 4.0 kg 2.0 kg 2.0 kg

10 x 1.0 m cm Solution m x m x m x m m m 2.5 m m 4.0 kg 2.0 kg 2.0 kg

11 Net Torque = 0, stable

12 Net torque =0, unstable

13 Net torque 0, Unstable

14

15

16 Example A 3.0 m long uniform beam has a weight of 60 N and is attached to a wall. Determine the torque acting on the attachment point.

17 Solution The weight of the beam acts as if all the mass is concentrated at the centre of the beam. r = 1.5 m F g = 60 N

18 Calculation = rfsin = 1.5 m x 60N x sin 90 o = 90 N m

19

20 Static Equilibrium No acceleration and no rotation F = 0, = 0

21 Example A uniform beam with a weight of 300 N is pivoted at its centre of gravity. A 350 N weight is placed 1.5 m from the pivot. How far from the pivot should a 550 N weight be placed so that the beam does not rotate? 350 N 550 N 1.5 m

22 Solution Since the beam is supported at the centre of gravity, the beam would be balanced without the placement of the extra weights. 350 N 550 N 1.5 m

23 Solution 350 N N = N = N 1.5 m x 350 N = r x 550 N r = 0.95 m 350 N 550 N = 90 o 1.5 m r

24 Determine the upward force exerted by the biceps muscle on the forearm. Example

25 Solution book + arm + muscle = 0, = 90 o -(40 x 0.38) + -(25 x 0.16) + muscle = 0

26 Solution N m N m + muscle = 0 muscle = 19.2 N m

27 Solution muscle = 19.2 N m muscle = 4.8 x 10 2 N UP

28 Example A 65 kg person is ¾ of the way up the 4.0 m ladder as shown in the diagram. What are the magnitude and direction of the torque about the base of the ladder at P produced by the person? A B C D Magnitude of torque Direction 9.8 x 10 2 N m clockwise 9.8 x 10 2 N m counter clockwise 1.7 x 10 3 N m clockwise 1.7 x 10 3 N m counter clockwise

29 Solution = rfsin = 3.0 m x 65 kg x 10 m/s 2 x sin 30 o = 3.0 m x 65 kg x 10 m/s 2 x ½ = 3.0 m x 325 N = 975 N m

30 Example A 4.00 m long ladder weighing 380 N is leaning against a wall. The ladder makes a 60 o angle with the ground and there is no friction between the ladder and the wall. Determine the force of friction between the ladder and the floor.

31 Solution FBD! F N from wall on ladder F N from ground on ladder F g = 380 N F friction from ground on ladder

32 Solution F = 0 F N from wall + F friction from ground = 0 x-direction F N from wall on ladder F N from wall = - F friction from ground F friction from ground on ladder

33 Solution y-direction F = 0 F N from ground + F g = 0 F N from ground = 380 N up F N from ground on ladder F g = 380 N

34 Solution if the wall wasn t there, the ladder would pivot about the point of contact with the ground. = 0 = 60 o 30 o

35 Solution 0 = (4.00 m)(f N wall )(sin 60 o )+(-2.0m)(380N)(sin30 o ) 0 = m x F N wall N m 380 N m = m x F N wall = 60 o F N wall = 110 N Since F N wall = F friction F friction = 110 N 30 o

36 Example A uniform 3.5 m beam of negligible mass, hinged at P, supports a hanging block as shown. If the tension F T in the horizontal cord is 150 N, what is the mass of the hanging block?

37 Solution = 0 cord + mass = m x 150 N x sin 37 o m x F g x sin 53 o = N m = m x F g F g = N m = 9.2 kg 53 o 37 o

38 Torque & Rotational Inertia A net force causes a mass to accelerate (translational motion) A net torque causes a mass to rotate, which is accelerated motion

39 Angular Acceleration I net I I is the moment of inertia or rotational inertia (units are kg m 2 ) I = mr 2 Units of are radians/s 2

40 A radian is the angle subtended by an arc that is equal to the radius of the circle Definition of Radian

41 Arc length Length of arc, s, = r

42 Two cylinders have the same mass but different radii. Which one has the higher moment of inertia?

43 Consider two objects: a hoop and a solid disc. If they start from the same position on an inclined ramp, which one would make it to the bottom of the ramp first? Example

44 the solid disc would beat the hoop to the bottom of the ramp! mr 2 is the rotational inertia of the object: the disc has more of its mass at a greater radius more inertia

45 I disc = ½ mr 2 I hoop = mr 2

46 Example A 15 N force acts on a 4.0 kg wheel 33 cm in radius. There is a frictional torque of 1.1 N m at the axle. Determine the angular acceleration of the wheel.

47 Example I = mr 2 = kg m 2 rf sin 1.1N m 2 I N m 8.8 rad s 2

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