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Physics 101: Lecture 14, Pg 1 Torque Looking back l Newton’s second law. l The mathematics of circular motion. Looking ahead l Apply rigid-body model to extended object. l Understand the rotation of a rigid body around a fixed axis.

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Physics 101: Lecture 14, Pg 2 Torque Definition Rotational effect of force. Tells how effective force is at twisting or rotating an object. ¿ = +- r F tangential = r F sin Á è Units N m è Sign, CCW rotation is positive è Depends on 3 properties: 1) Magnitude of force F. 2) Its distance r from pivot point. 3) Angle Á.

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Physics 101: Lecture 14, Pg 3 Concept Quiz The picture below shows three different ways of using a wrench to loosen a stuck nut. Assume the applied force F is the same in each case. In which of the cases is the torque on the nut the biggest? A. Case 1 B. Case 2 C. Case 3 CORRECT 10

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Physics 101: Lecture 14, Pg 4 Concept Quiz The picture below shows three different ways of using a wrench to loosen a stuck nut. Assume the applied force F is the same in each case. In which of the cases is the torque on the nut the smallest? A. Case 1 B. Case 2 C. Case 3 CORRECT 12

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Physics 101: Lecture 14, Pg 5 Torque Example and Quiz A person raises one leg to an angle of 30 degrees. An ankle weight (89 N) attached a distance of 0.84 m from her hip. What is to torque due to this weight? 1) Draw Diagram 2) = F r sin F r sin(90 – 30) If she raises her leg higher, the torque due to the weight will A) Increase B) Same C) Decrease 30 F=89 N r = 0.84 m = 65 N m 17

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Physics 101: Lecture 14, Pg 6 Equilibrium Quiz l A rod is lying on a table and has two equal but opposite forces acting on it. What is the net force on the rod? A) UpB) DownC) Zero l Will the rod move? A) Yes B) No F F y x Y direction: F y = ma y +F – F = 0 Yes, it rotates! 21

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Physics 101: Lecture 14, Pg 7Equilibrium l Conditions for Equilibrium F = 0 Translational EQ = 0 Rotational EQ »Can choose any axis of rotation…. Choose Wisely! 25

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Physics 101: Lecture 14, Pg 8 Equilibrium Example A meter stick is suspended at the center. If a 1 kg weight is placed at x=0. Where do you need to place a 2 kg weight to balance it? A) x = 25B) x=50C) x=75D) x=100 E) 1 kg can’t balance a 2 kg weight. 25 Balance Demo 9.8 N 17.6 N 50 cm d = (0.5) – (17.6)d = 0 d = 25

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Physics 101: Lecture 14, Pg 9 Static Equilibrium and Center of Mass l Gravitational Force Weight = mg è Acts as force at center of mass è Torque about pivot due to gravity: ¿ = Wrsin Á =mgd è Object not in static equilibrium Center of mass pivot d W=mg

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Physics 101: Lecture 14, Pg 10 Center of mass pivot d W=mg Torque about pivot 0 Center of mass pivot Torque about pivot = 0 A method to find center of mass of an irregular object Not in equilibrium Equilibrium Static Equilibrium

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Physics 101: Lecture 14, Pg 11 Equilibrium Quiz and Example The picture below shows two people lifting a heavy log. Which of the two people is supporting the greatest weight? 1. The person on the left is supporting the greatest weight 2. The person on the right is supporting the greatest weight 3. They are supporting the same weight CORRECT 34 mg FLFL FRFR L/2 L Look at torque about center: F R L – F L L/2 = 0 F R = ½ F L

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Physics 101: Lecture 14, Pg 12Example A 75 kg painter stands at the center of a 50 kg 3 meter plank. The supports are 1 meter in from each edge. Calculate the force on support A. A B 40 1 meter FAFA FBFB Mg mg 1) Draw FBD 2) F = 0 3) Choose pivot 4) = 0 F A + F B – mg – Mg = 0 -F A (1) sin(90)+ F B (0) sin(90) + mg (0.5)sin(90) + Mg(0.5) sin(90) = 0 F A = 0.5 mg Mg = Newtons 1 meter 0.5meter

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Physics 101: Lecture 14, Pg 13 Concept Quiz If the painter moves to the right, the force exerted by support A A) IncreasesB) UnchangedC) Decreases A B 43 1 meter FAFA FBFB Mg mg ¿ M decreases ¿ A also decreases

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Physics 101: Lecture 14, Pg 14Example How far to the right of support B can the painter stand before the plank tips? A B 47 1 meter Just before board tips, force from A becomes zero FBFB Mg mg 1) Draw FBD 2) F = 0 3) Choose pivot 4) = 0 F B – mg – Mg = 0 F B (0) sin(90) + mg (0.5)sin(90) – Mg(x) sin(90) = m = x M 0.5meter x

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Physics 101: Lecture 14, Pg 15 Summary l Torque = Force that causes rotation = F r sin l Equilibrium F = 0 = 0 »Can choose any axis.

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Physics 101: Lecture 14, Pg 16 Torque and rotational motion Torque is the quantity that changes rotational motion. For rotations, it is the object that replaces force as the cause of angular acceleration.

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Physics 101: Lecture 14, Pg 17 Torque and Rotational Motion Linear motion: F = ma (Newton’s 2 nd law linear motion) If force is tangential to a circular path: F=F t and a = a t =r ® Rotational motion: ¿ = F t r = m a t r=(mr 2 ) ® =I ® (Newton’s 2 nd law rotational motion) Extended object: I = mr 2 r F t

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Physics 101: Lecture 14, Pg 18 Moment of inertia I = i m i r i 2 Net torque leads to angular acceleration, but we need the moment of inertia to tell us how large an acceleration results. Moment of inertia in rotational problems takes the place that mass has in translational problems. Depends upon the mass and location of the mass from the axis of rotation. Large moment of inertia requires large torque to accelerate object (just like large mass requires large force)

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Physics 101: Lecture 14, Pg 19 Consider a rod of uniform density with an axis of rotation through its center and an identical rod with the axis of rotation through one end. Which has the larger moment of inertia? A IC > IE B IC < IE C IC = IE Concept Quiz

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Physics 101: Lecture 14, Pg 20 Consider two masses, each of size 2m at the ends of a light rod of length L with the axis of rotation through the center of the rod. The rod is doubled in length and the masses are halved. What happens to I? A: I A > I B B: I A < I B C: I A = I B Concept Quiz

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Physics 101: Lecture 14, Pg 21 Consider a solid disk with an axis of rotation through the center (perpendicular to the diagram). Two holes are cut out near the center and the material is placed near the rim. Which has the larger I? A I A > I B B I A < I B C I A = I B Concept Test

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Physics 101: Lecture 14, Pg 22 A mass m hangs from string wrapped around a pulley of radius R. The pulley has a moment of inertia I and its pivot is frictionless. Because of gravity the mass falls and the pulley rotates. The magnitude of the torque on the pulley is.. A: greater than mgR B: equal to mgR C: less than mgR (Hint: Is the tension in the string = mg?) Concept Test

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Physics 101: Lecture 14, Pg 23 Acceleration of the mass m? ¿ = T s R= I cyl ® ® = T_sR/ (0.5MR^2)= 2T s /MR Tension of the string from linear motion of block: T s -mg=ma y a y = a t = - ® R ® = -a y /R = -(T s -mg)/(mR) T s = -mR ® +mg ® = 2T s /MR = 2(-mR ® +mg)R/MR = 2(mR ® +mg)/M

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