physics 2210 fall 2015woolf/2210_jui/nov9.pdfย ยท physics 2210 fall 2015 smartphysics 14 rotational...
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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
๐ผ =13๐โ2
๐ผ =1
12๐โ2
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 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 ๐ผ๐ง, 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 ฯ=2.70x103 kg/m3; Base: radius b=18cm, Height, h=38cm. Calculate ๐ผ๐ง (%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 ฯ=2.70x103 kg/m3; Base: radius b=18cm, Height, h=38cm. Calculate ๐ผ๐ง (%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 ฯ=2.70x103 kg/m3; Base: radius b=18cm, Height, h=38cm. Calculate ๐ผ๐ง (%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 โข Objects rotate naturally (without external forces) around their
center-of-mass. โข Moment-of-inertia is usually tabulated/listed about a symmetry
axis through the center-of-mass
๐ผ =13๐โ2
๐ผ =1
12๐โ2
โข 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.
๐ฅ ๐ง
๐ฆ
CM CM
CM
CM CM
CM
Parallel Axis Theorem continued So the kinetic energy of the body can be written in two parts
๐พ = ๐พโ + ๐พ๐ถ๐ถ Where ๐พโฒ is the kinetic energy of the body ABOUT the CM (i.e. in the CM frame)
๐พโ = 12๐ผ๐ถ๐ถ๐
2 ๐ผ๐ถ๐ถ 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 ๐ถ๐ถ (perpendicular distance ๐ from rotation axis to the CM):
๐๐ถ๐ถ = ๐ ๐ถ๐ถ๐
๐พ๐ถ๐ถ = 12๐๐๐ถ๐ถ
2 = 12๐๐ ๐ถ๐ถ
2๐2 Adding the two terms together we have
๐พ = ๐พโ + ๐พ๐ถ๐ถ = 12๐ผ๐ถ๐ถ๐
2 + 12๐๐ ๐ถ๐ถ
2๐2 = 12 ๐ผ๐ถ๐ถ + ๐๐ ๐ถ๐ถ2 ๐2
By definition: ๐พ = 1
2๐ผ๐2
Where "๐ผโ is by this definition the moment of inertia about the ACTUAL axis of rotation.
โ ๐ผ = ๐ผ๐ถ๐ถ + ๐๐ ๐ถ๐ถ2
Example 15.1:
In this case: ๐ท = ๐ ๐ถ๐ถ = โ 2โ , ๐ผ๐ธ๐ธ๐ธ = ๐ผ๐ถ๐ถ + ๐๐ ๐ถ๐ถ2
=1
12๐โ2 + ๐
โ2
2
=1
12 +14 ๐โ2 =
13๐โ
2
Unit 15
๐ผ =1
12๐โ2 ๐ผ =
13๐โ2
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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
๐๐ถ๐ถ
=3๐ โ 0 + ๐. ๐ฟ3๐ + ๐
= ๐ฟ 4โ
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
๏ฟฝโ๏ฟฝ
๏ฟฝโ๏ฟฝ ๐
NOTE: sin(180ยฐโฮธ) = sinฮธ
Angular Velocity as a (pseudo-) Vector
๐
๐
Cross Product: (a) Magnitude A second product of multiplying TWO vectors is used extensively for describing rotational dynamics: The product here is a โvectorโ
๐ด ร ๐ต known as โcross productโ, โvector productโ, โexterior Productโ
Basic Definition: (a) Magnitude of ๐ด ร ๐ต: ๐ด ร ๐ต = ๐ด ๐ต sin๐
Where๐๐ด,๐ต is the smaller (< ๐ radians) angle between ๐ด and ๐ต.
You can think of ๐ด ร ๐ต as the product of ๐ด with ๐ตโฅ๐ด, the component of ๐ต perpendicular to ๐ด.
Alternatively, you can think of ๐ด ร ๐ต as the area of the parallelogram formed by vectors ๐ด and ๐ต.
๐ต ๐ด ๐๏ฟฝโ๏ฟฝ,๐ต
๐ตโฅ๐ด = ๐ต sin ๐๏ฟฝโ๏ฟฝ,๐ต
๐ด ร ๐ต = ๐ด ๐ตโฅ๐ด
๐ต ๐ด ๐๏ฟฝโ๏ฟฝ,๐ต
๐ด ร ๐ต = ๐ด ๐ต sin๐
๐ต sin ๐๏ฟฝโ๏ฟฝ,๐ต
Cross Product: (b) Direction The cross/vector product ๐ด ร ๐ต is perpendicular to both ๐ด and ๐ต.
i.e. ๐ด ร ๐ต is perpendicular to the parallelogram formed by vectors ๐ด and ๐ต, which in this case in the plane of this page.
Question remains: Is ๐ด ร ๐ต into the page or out of the page? Answer: determine the pointing direction of ๐ด ร ๐ต with the Right Hand Rule:
In this CASE: ๐จ ร ๐ฉ points OUT of the page
๐ต ๐ด ๐๏ฟฝโ๏ฟฝ,๐ต
๐ต sin ๐๏ฟฝโ๏ฟฝ,๐ต
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 ๐ ๐ ๐: Vector from the rotation axis perpendicularly to the point of application of the force
We mean here the magnitude of the torque
Torque
๐ = ๐ร๏ฟฝโ๏ฟฝ
90ยฐ
๐ญ
30ยฐ
๐๐ =๐ณ๐ โ ๐ญ โ ๐ฌ๐ฌ๐ฌ๐๐๐ =
๐๐๐ณ๐ญ ๐๐ = ๐ณ โ ๐ญ โ ๐ฌ๐ฌ๐ฌ๐๐๐ =
๐๐๐ณ๐ญ
๐1 ๐2
Torque and Angular Acceleration 1/3 โข We apply a force ๏ฟฝโ๏ฟฝ of constant
magnitude ๐น on a point ๐ from the origin (rotation axis): it acts at a distance ๐ from the rotation axis, but at an angle of ๐ relative to ๐
โข Only the tangential (to the circle of motion) component of the force, ๐น๐ก , does work:
๐น๐ก = ๐น sin๐
๏ฟฝโ๏ฟฝ
๐ ๐น๐ก
๐น๐
๐
๐ = ๐๐
๐ฅ ๐ง
๐ฆ
โข As the body rotates through an angle ๐ we maintain the relative orientation of the moving ๐ to ๏ฟฝโ๏ฟฝ.
โข The force now acted through a distance of ๐ = ๐๐ and has done work: ๐ = ๐น๐ก๐ = ๐๐๐น sin๐
โข Applying work-kinetic- energy theorem: โ๐พ = ๐ = ๐๐น sin๐ ๐ โข Differentiating with respect to time: we then get (noting ๐๐น sin๐ is constant):
๐๐พ๐๐
= ๐๐น sin๐๐๐๐๐
= ๐๐น sin๐ ๐
๏ฟฝโ๏ฟฝ
Torque and Angular Acceleration 2/3 โข The quantity (which we are
keeping constant here) ๐ โก ๐๐น sin๐ is called โtorqueโ
โข unit of torque: Nโ m โข Torque is the rotational analog
to โforceโ
โข Note that in the case shown, ๏ฟฝโ๏ฟฝ makes a positive angle ๐ (CCW) from ๐ and so ๐ is positive. If ๐ is negative then ๐ is negative.
๏ฟฝโ๏ฟฝ
๐ ๐น๐ก
๐น๐
๐
๐ = ๐๐
๐ฅ ๐ง
๐ฆ
โข Rotational kinetic energy is given by ๐พ = 12๐ผ๐
2. Its time derivative is then ๐๐พ๐๐
=12๐ผ๐๐๐
๐2 =12๐ผ โ 2๐ โ
๐๐๐๐
= ๐ผ๐๐ผ
So we have ๐ผ๐๐ผ = ๐๐น sin๐ ๐ = ฯ๐
โ ๐ผ๐ผ = ๐ โ ๐ผ =๐๐ผ
Which is like Newtonโs 2nd Law for rotation.
Compare to
๐ =๐ญ๐
Torque and Angular Acceleration 3/3 Important notes: โข The angle ๐ is measured from
the pointing directions of the vector ๐ to that of the force vector ๏ฟฝโ๏ฟฝ.
โข When measuring the angle between two vectors, they should be drawn tail-to-tail
๏ฟฝโ๏ฟฝ
๐ ๐น๐ก
๐น๐
๐
๐ = ๐๐
๐ฅ ๐ง
๐ฆ
โข The vector ๐ is the vector drawn perpendicularly from the rotation axis to the point of application of the force.
โข Angle ๐ is positive if rotation from(the direction of) ๐ to ๏ฟฝโ๏ฟฝ is counter-clock-wise (CCW): it means that the force/torque tends to push the body to rotate in the positive (CCW) direction.
โข The diagram to the right here shows th ๐ < 0 case โข smartPhysics uses ๐ for both this angle and the angular position: itโs confusing
๐
๏ฟฝโ๏ฟฝ
๐ < 0
๐ ๏ฟฝโ๏ฟฝ
๐ > 0
Unit 15
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