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Physics 2210 Fall 2015 smartPhysics 14 Rotational kinematics 15 Parallel Axis Theorem and Torque 11/09/2015

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Page 1: Physics 2210 Fall 2015woolf/2210_Jui/nov9.pdfย ยท Physics 2210 Fall 2015 smartPhysics 14 Rotational kinematics . 15 Parallel Axis Theorem and Torque . ... is the moment-of-inertia

Physics 2210 Fall 2015

smartPhysics 14 Rotational kinematics

15 Parallel Axis Theorem and Torque 11/09/2015

Page 2: Physics 2210 Fall 2015woolf/2210_Jui/nov9.pdfย ยท Physics 2210 Fall 2015 smartPhysics 14 Rotational kinematics . 15 Parallel Axis Theorem and Torque . ... is the moment-of-inertia

Exam 3 Results

Page 3: Physics 2210 Fall 2015woolf/2210_Jui/nov9.pdfย ยท Physics 2210 Fall 2015 smartPhysics 14 Rotational kinematics . 15 Parallel Axis Theorem and Torque . ... is the moment-of-inertia

Unit 14 Main Points 2/2

Page 4: Physics 2210 Fall 2015woolf/2210_Jui/nov9.pdfย ยท Physics 2210 Fall 2015 smartPhysics 14 Rotational kinematics . 15 Parallel Axis Theorem and Torque . ... is the moment-of-inertia

๐ผ =13๐‘€โ„“2

๐ผ =1

12๐‘€โ„“2

This table will be provided on the front page of exam 4 and the final exam

Units of Moment of Inertia kgm2

Page 5: Physics 2210 Fall 2015woolf/2210_Jui/nov9.pdfย ยท Physics 2210 Fall 2015 smartPhysics 14 Rotational kinematics . 15 Parallel Axis Theorem and Torque . ... is the moment-of-inertia

Example 14.3 (1/4) A right, circular cone is made of solid aluminum, with uniform density ฯ=2.70x103 kg/m3. Its base, of radius b=18cm sits on the xy plane, and its axis of symmetry lies along the z-axis. The height, measured from the center of the base to the apex, is h=38cm. Calculate ๐ผ๐‘ง, the moment-of-inertia about the z-axis, of the cone shown. (%i1) /* Break cone into a stack of cylinders of radius r, radial thickness dr and height z. The mass dm of each cylinder is given by rho*L*z*dr where L=2*pi*r is the circumference and in xmaxima we will omit the dr. The moment of inertia of this cylinder is then dm*r^2. We also need to know a as a function of r and we know it decreases linearly from h to 0 for r from 0 to b */

z: h*(1 - r/b);

r

(%o1) h (1 - -)

b

(%i2) L: 2*%pi*r, numer;

(%o2) 6.283185307179586 r

(%i3) m: rho*L*z;

r

(%o3) 6.283185307179586 h r (1 - -) rho

b

... continued

Page 6: Physics 2210 Fall 2015woolf/2210_Jui/nov9.pdfย ยท Physics 2210 Fall 2015 smartPhysics 14 Rotational kinematics . 15 Parallel Axis Theorem and Torque . ... is the moment-of-inertia

Example 14.3 (2/4) Right, circular cone; uniform density ฯ=2.70x103 kg/m3; Base: radius b=18cm, Height, h=38cm. Calculate ๐ผ๐‘ง (%i4) f: m*r^2;

3 r

(%o4) 6.283185307179586 h r (1 - -) rho

b

(%i5) assume(b>0);

(%o5) [b > 0]

(%i6) assume(h>0);

(%o6) [h > 0]

(%i7) integrate(f, r);

5 4

0.3141592653589793 h (4 r - 5 b r ) rho

(%o7) - ----------------------------------------

b

(%i8) Iz: integrate(f, r, 0, b);

4

(%o8) 0.3141592653589793 b h rho

(%i9) Iz, rho=2.7e3, b=0.18, h=0.38;

(%o9) 0.3383664179937264

Answer: Iz = 0.338 kg*m^2

Page 7: Physics 2210 Fall 2015woolf/2210_Jui/nov9.pdfย ยท Physics 2210 Fall 2015 smartPhysics 14 Rotational kinematics . 15 Parallel Axis Theorem and Torque . ... is the moment-of-inertia

Example 14.3 (3/4) Right, circular cone; uniform density ฯ=2.70x103 kg/m3; Base: radius b=18cm, Height, h=38cm. Calculate ๐ผ๐‘ง (%i1) /* Alternate solution: break up cone into a stack of horizontal disks each of thickness dz, radius r (function of z). We start by writing r as a function of z */ r: b*(1 - z/h); z (%o1) b (1 - -) h (%i2) /* mass of disk is dm=rho*A*dz, we omit dz, A=pi*r^2 */ A: %pi*r^2, numer; 2 z 2 (%o2) 3.141592653589793 b (1 - -) h m: rho*A; 2 z 2 (%o3) 3.141592653589793 b rho (1 - -) h (%i4) /* moment of inertia of each disk is dm/2*r^2 */ f: m/2*r^2; 4 z 4 (%o4) 1.570796326794896 b rho (1 - -) h ... continued

Page 8: Physics 2210 Fall 2015woolf/2210_Jui/nov9.pdfย ยท Physics 2210 Fall 2015 smartPhysics 14 Rotational kinematics . 15 Parallel Axis Theorem and Torque . ... is the moment-of-inertia

Example 14.3 (4/4) Right, circular cone; uniform density ฯ=2.70x103 kg/m3; Base: radius b=18cm, Height, h=38cm. Calculate ๐ผ๐‘ง (%i5) integrate(f, z); 5 4 3 2 4 z z 2 z 2 z (%o5) 1.570796326794896 b rho (---- - -- + ---- - ---- + z) 4 3 2 h 5 h h h (%i6) integrate(f, z, 0, h); 4 (%o6) 0.3141592653589793 b h rho Which is the same answer as before

... continued

Page 9: Physics 2210 Fall 2015woolf/2210_Jui/nov9.pdfย ยท Physics 2210 Fall 2015 smartPhysics 14 Rotational kinematics . 15 Parallel Axis Theorem and Torque . ... is the moment-of-inertia

Unit 15: Parallel Axis Theorem โ€ข Objects rotate naturally (without external forces) around their

center-of-mass. โ€ข Moment-of-inertia is usually tabulated/listed about a symmetry

axis through the center-of-mass

๐ผ =13๐‘€โ„“2

๐ผ =1

12๐‘€โ„“2

Page 10: Physics 2210 Fall 2015woolf/2210_Jui/nov9.pdfย ยท Physics 2210 Fall 2015 smartPhysics 14 Rotational kinematics . 15 Parallel Axis Theorem and Torque . ... is the moment-of-inertia

โ€ข Rotation of a rigid body about arbitrary axis can be separated into A. rotation of the CM about the axis, and B. rotation of the object about the CM. These occur at the SAME angular velocity for a rigid body (SPECIAL CASE) When you stand still on a carousel, as your CM goes through ONE rotation around the axis of the carousel , you also execute exactly ONE rotation about your CM.

๐‘ฅ ๐‘ง

๐‘ฆ

CM CM

CM

CM CM

CM

Page 11: Physics 2210 Fall 2015woolf/2210_Jui/nov9.pdfย ยท Physics 2210 Fall 2015 smartPhysics 14 Rotational kinematics . 15 Parallel Axis Theorem and Torque . ... is the moment-of-inertia

Parallel Axis Theorem continued So the kinetic energy of the body can be written in two parts

๐พ = ๐พโˆ— + ๐พ๐ถ๐ถ Where ๐พโ€ฒ is the kinetic energy of the body ABOUT the CM (i.e. in the CM frame)

๐พโˆ— = 12๐ผ๐ถ๐ถ๐œ”

2 ๐ผ๐ถ๐ถ is the moment-of-inertia of the object about an (imaginary) axis parallel to the actual rotation axis, that goes through its CM) The CM is executing circular motion at radius ๐‘…๐ถ๐ถ (perpendicular distance ๐‘Ÿ from rotation axis to the CM):

๐‘‰๐ถ๐ถ = ๐‘…๐ถ๐ถ๐œ”

๐พ๐ถ๐ถ = 12๐‘€๐‘‰๐ถ๐ถ

2 = 12๐‘€๐‘…๐ถ๐ถ

2๐œ”2 Adding the two terms together we have

๐พ = ๐พโˆ— + ๐พ๐ถ๐ถ = 12๐ผ๐ถ๐ถ๐œ”

2 + 12๐‘€๐‘…๐ถ๐ถ

2๐œ”2 = 12 ๐ผ๐ถ๐ถ + ๐‘€๐‘…๐ถ๐ถ2 ๐œ”2

By definition: ๐พ = 1

2๐ผ๐œ”2

Where "๐ผโ€ is by this definition the moment of inertia about the ACTUAL axis of rotation.

โ†’ ๐ผ = ๐ผ๐ถ๐ถ + ๐‘€๐‘…๐ถ๐ถ2

Page 12: Physics 2210 Fall 2015woolf/2210_Jui/nov9.pdfย ยท Physics 2210 Fall 2015 smartPhysics 14 Rotational kinematics . 15 Parallel Axis Theorem and Torque . ... is the moment-of-inertia

Example 15.1:

In this case: ๐ท = ๐‘…๐ถ๐ถ = โ„“ 2โ„ , ๐ผ๐ธ๐ธ๐ธ = ๐ผ๐ถ๐ถ + ๐‘€๐‘…๐ถ๐ถ2

=1

12๐‘€โ„“2 + ๐‘€

โ„“2

2

=1

12 +14 ๐‘€โ„“2 =

13๐‘€โ„“

2

Unit 15

๐ผ =1

12๐‘€โ„“2 ๐ผ =

13๐‘€โ„“2

1 of 3

Page 13: Physics 2210 Fall 2015woolf/2210_Jui/nov9.pdfย ยท Physics 2210 Fall 2015 smartPhysics 14 Rotational kinematics . 15 Parallel Axis Theorem and Torque . ... is the moment-of-inertia

Poll 11-09-01 A ball of mass 3M at x=0 is connected to a ball of mass M at x=L by a massless rod. Consider the three rotation axes A, B and C as shown, all parallel to the y axis.

For which rotation axis is the moment of inertia of the object smallest? (It may help you to figure out where the center of mass of the object is.)

A. A B. B C. C

๐‘‹๐ถ๐ถ

=3๐‘€ โˆ™ 0 + ๐‘€. ๐ฟ3๐‘€ + ๐‘€

= ๐ฟ 4โ„

Page 14: Physics 2210 Fall 2015woolf/2210_Jui/nov9.pdfย ยท Physics 2210 Fall 2015 smartPhysics 14 Rotational kinematics . 15 Parallel Axis Theorem and Torque . ... is the moment-of-inertia

Unit 15 2 of 3

For a more in-depth look at the Cross Product See the introduction to Cross Products in the Khan Academy https://www.khanacademy.org/math/linear-algebra/vectors_and_spaces/dot_cross_products/v/linear-algebra-cross-product-introduction

Vector/Cross Product

๐‘Ÿ

๏ฟฝโƒ—๏ฟฝ

๏ฟฝโƒ—๏ฟฝ ๐œƒ

NOTE: sin(180ยฐโˆ’ฮธ) = sinฮธ

Page 15: Physics 2210 Fall 2015woolf/2210_Jui/nov9.pdfย ยท Physics 2210 Fall 2015 smartPhysics 14 Rotational kinematics . 15 Parallel Axis Theorem and Torque . ... is the moment-of-inertia

Angular Velocity as a (pseudo-) Vector

๐œ‘

๐œ‘

Page 16: Physics 2210 Fall 2015woolf/2210_Jui/nov9.pdfย ยท Physics 2210 Fall 2015 smartPhysics 14 Rotational kinematics . 15 Parallel Axis Theorem and Torque . ... is the moment-of-inertia

Cross Product: (a) Magnitude A second product of multiplying TWO vectors is used extensively for describing rotational dynamics: The product here is a โ€œvectorโ€

๐ด ร— ๐ต known as โ€œcross productโ€, โ€œvector productโ€, โ€œexterior Productโ€

Basic Definition: (a) Magnitude of ๐ด ร— ๐ต: ๐ด ร— ๐ต = ๐ด ๐ต sin๐œƒ

Where๐œƒ๐ด,๐ต is the smaller (< ๐œ‹ radians) angle between ๐ด and ๐ต.

You can think of ๐ด ร— ๐ต as the product of ๐ด with ๐ตโŠฅ๐ด, the component of ๐ต perpendicular to ๐ด.

Alternatively, you can think of ๐ด ร— ๐ต as the area of the parallelogram formed by vectors ๐ด and ๐ต.

๐ต ๐ด ๐œƒ๏ฟฝโƒ—๏ฟฝ,๐ต

๐ตโŠฅ๐ด = ๐ต sin ๐œƒ๏ฟฝโƒ—๏ฟฝ,๐ต

๐ด ร— ๐ต = ๐ด ๐ตโŠฅ๐ด

๐ต ๐ด ๐œƒ๏ฟฝโƒ—๏ฟฝ,๐ต

๐ด ร— ๐ต = ๐ด ๐ต sin๐œƒ

๐ต sin ๐œƒ๏ฟฝโƒ—๏ฟฝ,๐ต

Page 17: Physics 2210 Fall 2015woolf/2210_Jui/nov9.pdfย ยท Physics 2210 Fall 2015 smartPhysics 14 Rotational kinematics . 15 Parallel Axis Theorem and Torque . ... is the moment-of-inertia

Cross Product: (b) Direction The cross/vector product ๐ด ร— ๐ต is perpendicular to both ๐ด and ๐ต.

i.e. ๐ด ร— ๐ต is perpendicular to the parallelogram formed by vectors ๐ด and ๐ต, which in this case in the plane of this page.

Question remains: Is ๐ด ร— ๐ต into the page or out of the page? Answer: determine the pointing direction of ๐ด ร— ๐ต with the Right Hand Rule:

In this CASE: ๐‘จ ร— ๐‘ฉ points OUT of the page

๐ต ๐ด ๐œƒ๏ฟฝโƒ—๏ฟฝ,๐ต

๐ต sin ๐œƒ๏ฟฝโƒ—๏ฟฝ,๐ต

Page 18: Physics 2210 Fall 2015woolf/2210_Jui/nov9.pdfย ยท Physics 2210 Fall 2015 smartPhysics 14 Rotational kinematics . 15 Parallel Axis Theorem and Torque . ... is the moment-of-inertia

Poll 11-09-02

In Case 1, a force F is pushing perpendicular on an object a distance L/2 from the rotation axis. In Case 2 the same force is pushing at an angle of 30 degrees a distance L from the axis.

In which case is the torque due to the force about the rotation axis biggest?

A. Case 1 B. Case 2 C. Same

๐‘Ÿ ๐‘Ÿ ๐‘Ÿ ๐‘Ÿ: Vector from the rotation axis perpendicularly to the point of application of the force

We mean here the magnitude of the torque

Torque

๐œ = ๐‘Ÿร—๏ฟฝโƒ—๏ฟฝ

90ยฐ

๐‘ญ

30ยฐ

๐‰๐Ÿ =๐‘ณ๐Ÿ โˆ™ ๐‘ญ โˆ™ ๐ฌ๐ฌ๐ฌ๐Ÿ—๐Ÿ—๐Ÿ— =

๐Ÿ๐Ÿ๐‘ณ๐‘ญ ๐‰๐Ÿ = ๐‘ณ โˆ™ ๐‘ญ โˆ™ ๐ฌ๐ฌ๐ฌ๐Ÿ‘๐Ÿ—๐Ÿ— =

๐Ÿ๐Ÿ๐‘ณ๐‘ญ

๐œ1 ๐œ2

Page 19: Physics 2210 Fall 2015woolf/2210_Jui/nov9.pdfย ยท Physics 2210 Fall 2015 smartPhysics 14 Rotational kinematics . 15 Parallel Axis Theorem and Torque . ... is the moment-of-inertia

Torque and Angular Acceleration 1/3 โ€ข We apply a force ๏ฟฝโƒ—๏ฟฝ of constant

magnitude ๐น on a point ๐‘Ÿ from the origin (rotation axis): it acts at a distance ๐‘Ÿ from the rotation axis, but at an angle of ๐œ‘ relative to ๐‘Ÿ

โ€ข Only the tangential (to the circle of motion) component of the force, ๐น๐‘ก , does work:

๐น๐‘ก = ๐น sin๐œ‘

๏ฟฝโƒ—๏ฟฝ

๐‘Ÿ ๐น๐‘ก

๐น๐‘Ÿ

๐œ‘

๐‘† = ๐‘Ÿ๐œƒ

๐‘ฅ ๐‘ง

๐‘ฆ

โ€ข As the body rotates through an angle ๐œƒ we maintain the relative orientation of the moving ๐‘Ÿ to ๏ฟฝโƒ—๏ฟฝ.

โ€ข The force now acted through a distance of ๐‘† = ๐‘Ÿ๐œƒ and has done work: ๐‘Š = ๐น๐‘ก๐‘† = ๐‘Ÿ๐œƒ๐น sin๐œ‘

โ€ข Applying work-kinetic- energy theorem: โˆ†๐พ = ๐‘Š = ๐‘Ÿ๐น sin๐œ‘ ๐œƒ โ€ข Differentiating with respect to time: we then get (noting ๐‘Ÿ๐น sin๐œ‘ is constant):

๐‘‘๐พ๐‘‘๐‘‘

= ๐‘Ÿ๐น sin๐œ‘๐‘‘๐œƒ๐‘‘๐‘‘

= ๐‘Ÿ๐น sin๐œ‘ ๐œ”

๏ฟฝโƒ—๏ฟฝ

Page 20: Physics 2210 Fall 2015woolf/2210_Jui/nov9.pdfย ยท Physics 2210 Fall 2015 smartPhysics 14 Rotational kinematics . 15 Parallel Axis Theorem and Torque . ... is the moment-of-inertia

Torque and Angular Acceleration 2/3 โ€ข The quantity (which we are

keeping constant here) ๐œ โ‰ก ๐‘Ÿ๐น sin๐œ‘ is called โ€œtorqueโ€

โ€ข unit of torque: Nโ‹…m โ€ข Torque is the rotational analog

to โ€œforceโ€

โ€ข Note that in the case shown, ๏ฟฝโƒ—๏ฟฝ makes a positive angle ๐œ‘ (CCW) from ๐‘Ÿ and so ๐œ is positive. If ๐œ‘ is negative then ๐œ is negative.

๏ฟฝโƒ—๏ฟฝ

๐‘Ÿ ๐น๐‘ก

๐น๐‘Ÿ

๐œ‘

๐‘† = ๐‘Ÿ๐œƒ

๐‘ฅ ๐‘ง

๐‘ฆ

โ€ข Rotational kinetic energy is given by ๐พ = 12๐ผ๐œ”

2. Its time derivative is then ๐‘‘๐พ๐‘‘๐‘‘

=12๐ผ๐‘‘๐‘‘๐‘‘

๐œ”2 =12๐ผ โˆ™ 2๐œ” โˆ™

๐‘‘๐œ”๐‘‘๐‘‘

= ๐ผ๐œ”๐ผ

So we have ๐ผ๐œ”๐ผ = ๐‘Ÿ๐น sin๐œ‘ ๐œ” = ฯ„๐œ”

โ†’ ๐ผ๐ผ = ๐œ โ†’ ๐ผ =๐œ๐ผ

Which is like Newtonโ€™s 2nd Law for rotation.

Compare to

๐’‚ =๐‘ญ๐’Ž

Page 21: Physics 2210 Fall 2015woolf/2210_Jui/nov9.pdfย ยท Physics 2210 Fall 2015 smartPhysics 14 Rotational kinematics . 15 Parallel Axis Theorem and Torque . ... is the moment-of-inertia

Torque and Angular Acceleration 3/3 Important notes: โ€ข The angle ๐œ‘ is measured from

the pointing directions of the vector ๐‘Ÿ to that of the force vector ๏ฟฝโƒ—๏ฟฝ.

โ€ข When measuring the angle between two vectors, they should be drawn tail-to-tail

๏ฟฝโƒ—๏ฟฝ

๐‘Ÿ ๐น๐‘ก

๐น๐‘Ÿ

๐œ‘

๐‘† = ๐‘Ÿ๐œƒ

๐‘ฅ ๐‘ง

๐‘ฆ

โ€ข The vector ๐‘Ÿ is the vector drawn perpendicularly from the rotation axis to the point of application of the force.

โ€ข Angle ๐œ‘ is positive if rotation from(the direction of) ๐‘Ÿ to ๏ฟฝโƒ—๏ฟฝ is counter-clock-wise (CCW): it means that the force/torque tends to push the body to rotate in the positive (CCW) direction.

โ€ข The diagram to the right here shows th ๐œ‘ < 0 case โ€ข smartPhysics uses ๐œƒ for both this angle and the angular position: itโ€™s confusing

๐‘Ÿ

๏ฟฝโƒ—๏ฟฝ

๐œ‘ < 0

๐‘Ÿ ๏ฟฝโƒ—๏ฟฝ

๐œ‘ > 0

Page 22: Physics 2210 Fall 2015woolf/2210_Jui/nov9.pdfย ยท Physics 2210 Fall 2015 smartPhysics 14 Rotational kinematics . 15 Parallel Axis Theorem and Torque . ... is the moment-of-inertia

Unit 15

3 of 3