Physics 2210 Fall 2015

smartPhysics 14 Rotational kinematics

15 Parallel Axis Theorem and Torque 11/09/2015

Exam 3 Results

Unit 14 Main Points 2/2

𝐼 =13𝑀ℓ2

𝐼 =1

12𝑀ℓ2

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

Units of Moment of Inertia kgm2

Example 14.3 (1/4) A right, circular cone is made of solid aluminum, with uniform density ρ=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 𝐼𝑧, 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

Example 14.3 (2/4) Right, circular cone; uniform density ρ=2.70x103 kg/m3; Base: radius b=18cm, Height, h=38cm. Calculate 𝐼𝑧 (%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

Example 14.3 (3/4) Right, circular cone; uniform density ρ=2.70x103 kg/m3; Base: radius b=18cm, Height, h=38cm. Calculate 𝐼𝑧 (%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

Example 14.3 (4/4) Right, circular cone; uniform density ρ=2.70x103 kg/m3; Base: radius b=18cm, Height, h=38cm. Calculate 𝐼𝑧 (%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

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

center-of-mass. • Moment-of-inertia is usually tabulated/listed about a symmetry

axis through the center-of-mass

𝐼 =13𝑀ℓ2

𝐼 =1

12𝑀ℓ2

• 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.

𝑥 𝑧

𝑦

CM CM

CM

CM CM

CM

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

𝐾 = 𝐾∗ + 𝐾𝐶𝐶 Where 𝐾′ is the kinetic energy of the body ABOUT the CM (i.e. in the CM frame)

𝐾∗ = 12𝐼𝐶𝐶𝜔

2 𝐼𝐶𝐶 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 𝑅𝐶𝐶 (perpendicular distance 𝑟 from rotation axis to the CM):

𝑉𝐶𝐶 = 𝑅𝐶𝐶𝜔

𝐾𝐶𝐶 = 12𝑀𝑉𝐶𝐶

2 = 12𝑀𝑅𝐶𝐶

2𝜔2 Adding the two terms together we have

𝐾 = 𝐾∗ + 𝐾𝐶𝐶 = 12𝐼𝐶𝐶𝜔

2 + 12𝑀𝑅𝐶𝐶

2𝜔2 = 12 𝐼𝐶𝐶 + 𝑀𝑅𝐶𝐶2 𝜔2

By definition: 𝐾 = 1

2𝐼𝜔2

Where "𝐼” is by this definition the moment of inertia about the ACTUAL axis of rotation.

→ 𝐼 = 𝐼𝐶𝐶 + 𝑀𝑅𝐶𝐶2

Example 15.1:

In this case: 𝐷 = 𝑅𝐶𝐶 = ℓ 2⁄ , 𝐼𝐸𝐸𝐸 = 𝐼𝐶𝐶 + 𝑀𝑅𝐶𝐶2

=1

12𝑀ℓ2 + 𝑀

ℓ2

2

=1

12 +14 𝑀ℓ2 =

13𝑀ℓ

2

Unit 15

𝐼 =1

12𝑀ℓ2 𝐼 =

13𝑀ℓ2

1 of 3

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

𝑋𝐶𝐶

=3𝑀 ∙ 0 + 𝑀. 𝐿3𝑀 + 𝑀

= 𝐿 4⁄

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

𝑟

�⃗�

�⃗� 𝜃

NOTE: sin(180°−θ) = sinθ

Angular Velocity as a (pseudo-) Vector

𝜑

𝜑

Cross Product: (a) Magnitude A second product of multiplying TWO vectors is used extensively for describing rotational dynamics: The product here is a “vector”

𝐴 × 𝐵 known as “cross product”, “vector product”, “exterior Product”

Basic Definition: (a) Magnitude of 𝐴 × 𝐵: 𝐴 × 𝐵 = 𝐴 𝐵 sin𝜃

Where𝜃𝐴,𝐵 is the smaller (< 𝜋 radians) angle between 𝐴 and 𝐵.

You can think of 𝐴 × 𝐵 as the product of 𝐴 with 𝐵⊥𝐴, the component of 𝐵 perpendicular to 𝐴.

Alternatively, you can think of 𝐴 × 𝐵 as the area of the parallelogram formed by vectors 𝐴 and 𝐵.

𝐵 𝐴 𝜃�⃗�,𝐵

𝐵⊥𝐴 = 𝐵 sin 𝜃�⃗�,𝐵

𝐴 × 𝐵 = 𝐴 𝐵⊥𝐴

𝐵 𝐴 𝜃�⃗�,𝐵

𝐴 × 𝐵 = 𝐴 𝐵 sin𝜃

𝐵 sin 𝜃�⃗�,𝐵

Cross Product: (b) Direction The cross/vector product 𝐴 × 𝐵 is perpendicular to both 𝐴 and 𝐵.

i.e. 𝐴 × 𝐵 is perpendicular to the parallelogram formed by vectors 𝐴 and 𝐵, which in this case in the plane of this page.

Question remains: Is 𝐴 × 𝐵 into the page or out of the page? Answer: determine the pointing direction of 𝐴 × 𝐵 with the Right Hand Rule:

In this CASE: 𝑨 × 𝑩 points OUT of the page

𝐵 𝐴 𝜃�⃗�,𝐵

𝐵 sin 𝜃�⃗�,𝐵

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

𝑟 𝑟 𝑟 𝑟: Vector from the rotation axis perpendicularly to the point of application of the force

We mean here the magnitude of the torque

Torque

𝜏 = 𝑟×�⃗�

90°

𝑭

30°

𝝉𝟏 =𝑳𝟐 ∙ 𝑭 ∙ 𝐬𝐬𝐬𝟗𝟗𝟗 =

𝟏𝟐𝑳𝑭 𝝉𝟐 = 𝑳 ∙ 𝑭 ∙ 𝐬𝐬𝐬𝟑𝟗𝟗 =

𝟏𝟐𝑳𝑭

𝜏1 𝜏2

Torque and Angular Acceleration 1/3 • We apply a force �⃗� of constant

magnitude 𝐹 on a point 𝑟 from the origin (rotation axis): it acts at a distance 𝑟 from the rotation axis, but at an angle of 𝜑 relative to 𝑟

• Only the tangential (to the circle of motion) component of the force, 𝐹𝑡 , does work:

𝐹𝑡 = 𝐹 sin𝜑

�⃗�

𝑟 𝐹𝑡

𝐹𝑟

𝜑

𝑆 = 𝑟𝜃

𝑥 𝑧

𝑦

• As the body rotates through an angle 𝜃 we maintain the relative orientation of the moving 𝑟 to �⃗�.

• The force now acted through a distance of 𝑆 = 𝑟𝜃 and has done work: 𝑊 = 𝐹𝑡𝑆 = 𝑟𝜃𝐹 sin𝜑

• Applying work-kinetic- energy theorem: ∆𝐾 = 𝑊 = 𝑟𝐹 sin𝜑 𝜃 • Differentiating with respect to time: we then get (noting 𝑟𝐹 sin𝜑 is constant):

𝑑𝐾𝑑𝑑

= 𝑟𝐹 sin𝜑𝑑𝜃𝑑𝑑

= 𝑟𝐹 sin𝜑 𝜔

�⃗�

Torque and Angular Acceleration 2/3 • The quantity (which we are

keeping constant here) 𝜏 ≡ 𝑟𝐹 sin𝜑 is called “torque”

• unit of torque: N⋅m • Torque is the rotational analog

to “force”

• Note that in the case shown, �⃗� makes a positive angle 𝜑 (CCW) from 𝑟 and so 𝜏 is positive. If 𝜑 is negative then 𝜏 is negative.

�⃗�

𝑟 𝐹𝑡

𝐹𝑟

𝜑

𝑆 = 𝑟𝜃

𝑥 𝑧

𝑦

• Rotational kinetic energy is given by 𝐾 = 12𝐼𝜔

2. Its time derivative is then 𝑑𝐾𝑑𝑑

=12𝐼𝑑𝑑𝑑

𝜔2 =12𝐼 ∙ 2𝜔 ∙

𝑑𝜔𝑑𝑑

= 𝐼𝜔𝐼

So we have 𝐼𝜔𝐼 = 𝑟𝐹 sin𝜑 𝜔 = τ𝜔

→ 𝐼𝐼 = 𝜏 → 𝐼 =𝜏𝐼

Which is like Newton’s 2nd Law for rotation.

Compare to

𝒂 =𝑭𝒎

Torque and Angular Acceleration 3/3 Important notes: • The angle 𝜑 is measured from

the pointing directions of the vector 𝑟 to that of the force vector �⃗�.

• When measuring the angle between two vectors, they should be drawn tail-to-tail

�⃗�

𝑟 𝐹𝑡

𝐹𝑟

𝜑

𝑆 = 𝑟𝜃

𝑥 𝑧

𝑦

• The vector 𝑟 is the vector drawn perpendicularly from the rotation axis to the point of application of the force.

• Angle 𝜑 is positive if rotation from(the direction of) 𝑟 to �⃗� is counter-clock-wise (CCW): it means that the force/torque tends to push the body to rotate in the positive (CCW) direction.

• The diagram to the right here shows th 𝜑 < 0 case • smartPhysics uses 𝜃 for both this angle and the angular position: it’s confusing

𝑟

�⃗�

𝜑 < 0

𝑟 �⃗�

𝜑 > 0

Unit 15

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