Rotation Torque Moment of Inertia

Transcription

1 Rotation Torque Moment of Inertia Lana Sheridan De Anza College Nov 17, 2017

2 Last time rotational quantities rotational kinematics torque

3 Quick review of Vector Expressions Let a, b, and c be (non-null) vectors. Let l, m, and n be non-zero scalars. Could this possibly be a valid equation? a = b c (A) yes (B) no

4 Quick review of Vector Expressions Let a, b, and c be (non-null) vectors. Let l, m, and n be non-zero scalars. Could this possibly be a valid equation? a = b c (A) yes (B) no

5 Quick review of Vector Expressions Let a, b, and c be vectors. Let m and n be non-zero scalars. Could this possibly be a valid equation? m = b c (A) yes (B) no

6 Quick review of Vector Expressions Let a, b, and c be vectors. Let m and n be non-zero scalars. Could this possibly be a valid equation? m = b c (A) yes (B) no

7 Quick review of Vector Expressions Let a, b, and c be vectors. Let m and n be non-zero scalars. Suppose a 0 and b 0 and a n b for any value of n. Could this possibly be a true equation? a b = b a (A) yes (B) no

8 Quick review of Vector Expressions Let a, b, and c be vectors. Let m and n be non-zero scalars. Suppose a 0 and b 0 and a n b for any value of n. Could this possibly be a true equation? a b = b a (A) yes (B) no

9 Overview torque example moment of inertia calculating moments of inertia

10 Torque Torque is a measure of force-causing-rotation. It is not a force, but it is related. It depends on a force vector and its point of application relative to an axis of rotation. Torque is given by: τ = r F That is: the cross product between a vector r, the displacement of the point of application of the force from the axis of rotation, and an the force vector F

11 Torque O The component F sin f tends to rotate the wrench about an axis through O. d r S r f F sin f Figure 10.7 The force F S has a f S F F cos f Line of action greater rotating tendency about an axis through τ = r O as F = F increases rf sin φand ˆn as the moment arm d increases. where φ is the angle between r and F, and ˆn is the unit vector perpendicular to r and F, as determined by the right-hand rule Torqu In our study of tra we studied the cau What is the cause Imagine trying ular to the door s hinges. You will ac force near the doo When a force is to rotate about th axis is measured b but we will consid Chapter 11. Consider the wr perpendicular to t

12 Torque Diagram also illustrates two points of view about torque: or O The component F sin f tends to rotate the wrench about an axis through O. d r S r f F sin f f S F F cos f Line of action Figure 10.7 The force S F has a greater rotating τ = r(f tendency sin φ) about ˆn an axis through O as F increases and as the moment arm d increases. τ = F (r sin φ) ˆn 10.4 Torque In our study of translation we studied the cause of c What is the cause of chang Imagine trying to rotat ular to the door surface n hinges. You will achieve a force near the doorknob t When a force is exerted to rotate about that axis. axis is measured by a quan but we will consider only Chapter 11. Consider the wrench in perpendicular to the page In the diagram, the distance d = r sin φ and is called the moment arm or lever arm of the torque.

13 Torque Torque: Torque: No torque: when a force is applied CM above the center of mass. The system rotates clockwise when a force is applied a above the center of mass. CM The system rotates counterclockwise acm when a force is applied below the center of mass. The a system rotates counterclockwise when a force is applied below CM the center of mass. The system rotates counterclockwise when a force is applied below b the center of mass. CM The CM system moves in the b direction of the force without rotating when a force is applied bat the center of mass. The system moves in the direction of the force without rotating when a force is applied The at the system center moves of mass. in the direction CM of the force without rotating when a force is applied at the center of mass. c CM out the rotating x axis and (see lies Fig. somewhe 9.13c) is this applied procedure. at the center of ma out The rotating center (see of mass Fig. 9.13c) of the this the x procedure. axis and lies somewhe For example, if x 1 5 0, x 2 center The center of mass of lies mass closer of the to the the x center axis and of mass lies somewher lies midw For We example, can extend if x 1 this 5 0, conc x 2 5 center dimensions. of mass The lies x closer coordin to For the example, center of mass if x 5 lies 0, midw x 2 5 center We can of mass extend lies this closer conce to the dimensions. x CM ; m 1x 1 1 m 2 x 2 1 center of The mass mx 1 lies 1coordina mmidw 2 1 We can extend this conce dimensions. The x coordina x CM ; m 1x 1 1 m 2 x 2 1 m where x i is the mx coordinate 1 1 m 2 1 m the sum runs x CM ; m over 1x 1 1 mall 2 x n 2 1par m similarly defined by the equ m 1 1 m 2 1 m where x i is the x coordinate the sum runs over all n y CM part ; where similarly x i is defined the x coordinate by the equo the The sum center runs over of mass all n can partb similarly The components defined by of the this y CM equ vec ; Therefore, The center of mass ycan CM ; b S The components r CM 5of x CM this i^ 1vect y C The Therefore, center of mass can be

14 Question B A A torque is supplied by applying a force at point A. To produce the same torque, the force applied at point B must be: (A) greater (B) less (C) the same 1 Image from Harbor Freight Tools,

15 Question B A A torque is supplied by applying a force at point A. To produce the same torque, the force applied at point B must be: (A) greater (B) less (C) the same 1 Image from Harbor Freight Tools,

16 the rotation axis to the line of action of ary line extending out both ends of the Net Torque line extending from the tail of S F in Fig. m the right triangle in Figure 10.7 that hat d 5 r sin f. The quantity d is called S F rotation axis. Object that can rotate about Moment anarm axis at O: that tends to cause rotation of the f, the component perpendicular to a int of application of the force. The horiof action passes through O, has no tenassing through O. From the definition as F increases and as d increases, which e push at the doorknob rather than at a apply our push as closely perpendicular 908. Pushing sideways on the doorknob ect as in Figure 10.8, each tends to pro- In this example, S F2 tends to rotate the ounterclockwise. We use the convention force is positive if the turning tendency ive if the turning tendency is clockwise. ulting from S F1, which has a moment arm e from S F2 is negative and equal to 2F 2 d 2. h O is F 1 d 1 2 F 2 d 2 ce. Forces can cause a change in translaond law. Forces can also cause a change ss of the forces in causing this change rces and the moment arms of the forces, has units of force times length newton reported in these units. Do not confuse O d 2 S F 2 d 1 S F 1 Figure 10.8 The force S F 1 tends to rotate the object counterclockwise about an axis through O, and S F 2 tends to rotate it clockwise. There are two forces acting, but the two torques produced, τ 1 and τ 2 point in opposite directions. τ 1 would produce a counterclockwise rotation τ 2 would produce a clockwise rotation

17 m the right triangle in Figure 10.7 that hat d 5 r sin f. The quantity d is called SNet Torque F that tends to cause rotation of the f, the component perpendicular to a int of application of the force. The horiof action passes through O, has no tenassing through O. From the definition as F increases and as d increases, which e push at the doorknob rather than at a apply our push as closely perpendicular 908. Pushing sideways on the doorknob Moment arm O d 2 d 1 S F 1 ect as in Figure 10.8, each tends to pro- In this example, S F2 tends to rotate the ounterclockwise. We use the convention force is positive if the turning tendency ive if the turning tendency is clockwise. ulting from S F1, which has a moment arm e from S F2 is negative and equal to 2F 2 d 2. h O is F 1 d 1 2 F 2 d 2 ce. Forces can cause a change in translaond law. Forces can also cause a change ss of the forces in causing this change rces and the moment arms of the forces, has units of force times length newton reported in these units. Do not confuse ts but are very different concepts. osen a stubborn screw from a piece of you find a screwdriver for which the S F 2 Figure 10.8 The force S F 1 tends to rotate the object counterclockwise about an axis through O, and S F 2 tends to rotate it clockwise. The net torque is the sum of the torques acting on the object: τ net = i In this case, with ˆn pointing out of the slide: τ i τ net = τ 1 + τ 2 = (F 1 d 1 F 2 d 2 )ˆn

18 om F 1, which has a moment arm 2 is negative and equal to 2F 2 d 2. 1 to rotate the object counterclockwise about an axis through O, and Example NetS Torque on a Cylinder F 2 d 2 F 2 tends to rotate it clockwise. A one-piece cylinder is shaped as shown, with a core section protruding from the larger drum. The cylinder is free to rotate about the central z axis shown in the drawing. A rope wrapped around the drum, which has radius R 1, exerts a force T 1 to the right on the cylinder. A rope wrapped around the core, which has radius R 2, exerts a force T 2 downward on the cylinder. s can cause a change in transla-. Forces can also cause a change forces in causing this change the moment arms of the forces, s of force times length newton d in these units. Do not confuse e very different concepts. What is the net torque acting on the cylinder about the rotation axis (which is the z axis)? tubborn screw from a piece of a screwdriver for which the rying to loosen a stubborn l, should you find a wrench for y S T 1 nder.9, with a core section protrudrotate about the central z axis e drum, which has radius R 1, pe wrapped around the core, the cylinder. z S T 2 R 2 O R 1 Figure 10.9 (Example 10.3) A x

19 d in these units. Do not confuse e very different concepts. Example Net Torque on a Cylinder tubborn screw from a piece of a screwdriver for which the rying to loosen a stubborn l, should you find a wrench for nder.9, with a core section protrudrotate about the central z axis e drum, which has radius R 1, pe wrapped around the core, the cylinder. z S T 2 y R 2 O out the rotation axis (which is Figure 10.9 (Example 10.3) A solid cylinder pivoted about the z axis through O. The moment arm of S T1 is about the continued rotation axis. R 1, and the moment arm of S T2 is R 2. R 1 First: Find an expression for the net torque acting on the cylinder Second: Let T 1 = 5.0 N, R 1 = 1.0 m, T 2 = 15 N, and R 2 = 0.50 m. What is the net torque? Which way is the rotation? S T 1 x

20 Rotational Version of Newton s Second Law Tangential components of forces give rise to torques. They also cause tangential accelerations. Consider the tangential component of the net force, F net,t : from Newton s second law. F net,t = m a t τ net = r F net = r F net,t ˆn Now let s specifically consider the case of a single particle, mass m, at a fixed radius r.

21 The tangential force on the 10.5 Analysis M Rotational Versionparticle of results Newton s in a torque on Second the Law particle about an axis through the center of the circle. A single particle, mass m, at a fixed radius r. S F t r S F r Figure A particle rotating For such a particle, in F a circle under the influence of a net,t = ma t tangential net force g S Ft. A radial net force g S Fr also must be present to maintain τ net the = circular rf net,t motion. ˆn = r m a t ˆn m = r m (αr) = (mr 2 ) α In Chapter 5, we learned t object and that the acceler basis of the particle unde is Newton s second law. In second law: the angular ac proportional to the net to complex case of rigid-obje case of a particle moving i ence of an external force. Consider a particle of m of a tangential net force g The radial net force cause etal acceleration. The tang The magnitude of the net dicular to the page throug Because the tangential ac the relationship a t 5 ra (E

22 Rotational Version of Newton s Second Law (mr 2 ) is just some constant for this particle and this axis of rotation. Let this constant be (scalar) I = mr 2.

23 Rotational Version of Newton s Second Law (mr 2 ) is just some constant for this particle and this axis of rotation. Let this constant be (scalar) I = mr 2. Scalar I is not impulse! This is just an unfortunate notation coincidence.

24 Rotational Version of Newton s Second Law (mr 2 ) is just some constant for this particle and this axis of rotation. Let this constant be (scalar) I = mr 2. Scalar I is not impulse! This is just an unfortunate notation coincidence. I is called the moment of inertia of this system, for this particular axis of rotation.

25 Rotational Version of Newton s Second Law (mr 2 ) is just some constant for this particle and this axis of rotation. Let this constant be (scalar) I = mr 2. Scalar I is not impulse! This is just an unfortunate notation coincidence. I is called the moment of inertia of this system, for this particular axis of rotation. Replacing the constant quantity in our expression for τ net : τ net = Iα

26 Rotational Version of Newton s Second Law Compare! τ net = Iα F net = ma Now the moment of inertia, I, stands in for the inertial mass, m.

27 Rotational Version of Newton s Second Law Compare! τ net = Iα F net = ma Now the moment of inertia, I, stands in for the inertial mass, m. The moment of inertia measures the rotational inertia of an object, just as mass is a measure of inertia.

28 Moment of Inertia We just found that for a single particle, mass m, radius r, I = mr 2 However, this will not be the moment of inertia for an extended object with mass distributed over varying distances from the rotational axis. For that case, the torque on each individual mass m i will be: τ i = m i r 2 i α And we sum over these torques to get the net torque. So, for a collection of particles, masses m i at radiuses r i : I = i m i r 2 i

29 Moment of Inertia And we sum over these torques to get the net torque. So, for a collection of particles, masses m i at radiuses r i : I = i m i r 2 i For a continuous distribution of mass, we must integrate over each small mass m: I = r 2 dm

30 Moment of Inertia For a continuous distribution of mass, we must integrate over each small mass m: I = r 2 dm For an object of density ρ: I = ρ r 2 dv If ρ varies with position, it must stay inside the integral.

31 Moment of Inertia Important caveat: Moment of inertia depends on the object s mass, shape, and the axis of rotation. A single object will have different moments of inertia for different axes of rotation.

32 Moment of Inertia Important caveat: Moment of inertia depends on the object s mass, shape, and the axis of rotation. A single object will have different moments of inertia for different axes of rotation. Also notice that these integrals and sums are similar to the expression for the center-of-mass, but for I we have r 2 and we do not divide by the total mass. Units: kg m 2

33 Moment of Inertia For a collection of particles, masses m i at radiuses r i : I = i m i r 2 i For a continuous distribution of mass, we must integrate over each small mass m: I = r 2 dm And for an object of density ρ: I = ρ r 2 dv If ρ varies with position, it must stay inside the integral.

34 Example: Calculating Moment of Inertia Calculate the moment of inertia of two equal point-like masses connected by a light rod, length l, rotating about the center of mass. m m l

35 Example: Calculating Moment of Inertia Calculate the moment of inertia of two equal point-like masses connected by a light rod, length l, rotating about the center of mass. m m Let the CM the origin. It will be in the center of the rod. l

36 Example: Calculating Moment of Inertia m m l Moment of inertia: I = i m i x 2 i ( = m 2) l 2 + m ( ) l 2 2

37 Example: Calculating Moment of Inertia m m l Moment of inertia: I = i m i x 2 i ( = m 2) l 2 + m = ml2 2 ( ) l 2 2

38 Summary moments of inertia Atwood machine with massive pulley 3rd Test Monday, Nov 27. (Uncollected) Homework Serway & Jewett, Ch 10, onward from page 288. Probs: 27, 29, 35, 39, 43 Look at examples 10.6, 10.13, Ch 10, onward from page 288. Probs: 45, 47, 51, 53????