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
This table will be provided on the front page of exam 4 and the final exam Units of Moment of Inertia kgm 2 I = 1 12 Ml2 I = 1 3 Ml2
Example 14.3 (1/4) A right, circular cone is made of solid aluminum, with uniform density ρ=2.70x10 3 kg/m 3. 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 I z, 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); (%o1) h (1 - -) (%i2) L: 2*%pi*r, numer; (%o2) (%i3) m: rho*l*z; (%o3)... continued r b 6.283185307179586 r 6.283185307179586 h r (1 - -) rho r b
Example 14.3 (2/4) Right, circular cone; uniform density ρ=2.70x10 3 kg/m 3 ; Base: radius b=18cm, Height, h=38cm. Calculate I z (%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=2.70x10 3 kg/m 3 ; Base: radius b=18cm, Height, h=38cm. Calculate I z (%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=2.70x10 3 kg/m 3 ; Base: radius b=18cm, Height, h=38cm. Calculate I z (%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 = 1 12 Ml2 I = 1 3 Ml2
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. y CM CM CM z CM x CM CM
Parallel Axis Theorem continued So the kinetic energy of the body can be written in two parts K = K + K CC Where K is the kinetic energy of the body ABOUT the CM (i.e. in the CM frame) K = 1 2 I CCω 2 I CC 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 R CC (perpendicular distance r from rotation axis to the CM): V CC = R CC ω K CC = 1 2 MV CC 2 = 1 2 MR CC 2 ω 2 Adding the two terms together we have K = K + K CC = 1 2 I CCω 2 + 1 2 MR CC 2 ω 2 = 1 2 I CC + MR 2 CC ω 2 By definition: K = 1 2 Iω2 Where "I is by this definition the moment of inertia about the ACTUAL axis of rotation. 2 I = I CC + MR CC
Unit 15 1 of 3 Example 15.1: I = 1 12 Ml2 I = 1 3 Ml2 In this case: D = R CC = l 2, 2 I EEE = I CC + MR CC = 1 12 Ml2 + M l 2 = 1 12 + 1 4 Ml2 = 1 3 Ml2 2
Poll 11-09-01 = X CC 3M 0 + M. L 3M + M = L 4 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
Unit 15 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/linearalgebra/vectors_and_spaces/dot_cross_prod ucts/v/linear-algebra-cross-productintroduction 2 of 3 Vector/Cross Product θ r F F 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 sin θ Whereθ A,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 B θ A,B θ A,B A B = A B sin θ B A = B sin θ A,B A A B = A B A B sin θ A,B A
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 θ A,B B sin θ A,B A
Torque τ = r F r Poll 11-09-02 τ 1 τ 2 r : Vector from the rotation axis 90 perpendicularly 30 to the point of application of the force τ 1 = L 2 F sss 999 = 1 2 LL τ 2 = L F sss 333 = 1 2 LL 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? We mean here the magnitude of the torque A. Case 1 B. Case 2 C. Same r F r
Torque and Angular Acceleration 1/3 We apply a force F 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, F t, does work: F t = F sin φ S = rθ As the body rotates through an angle θ we maintain the relative orientation of the moving r to F. The force now acted through a distance of S = rθ and has done work: W = F t S = rθf sin φ Applying work-kinetic- energy theorem: K = W = rr sin φ θ Differentiating with respect to time: we then get (noting rr sin φ is constant): dd dd = dθ rr sin φ dd = z y rr sin φ ω r F F F r φ F t x
Torque and Angular Acceleration 2/3 y The quantity (which we are keeping constant here) τ rr sin φ is called torque unit of torque: N m S = rθ Torque is the rotational analog to force F Note that in the case shown, F makes a positive angle φ (CCW) r φ from r and so τ is positive. If φ is negative then τ is negative. z F r Rotational kinetic energy is given by K = 1 2 Iω2. Its time derivative is then dd dd = 1 2 I d dd ω2 = 1 dd I 2ω 2 dd = III So we have III = rr sin φ ω = τω Iα = τ α = τ I Which is like Newton s 2 nd Law for rotation. Compare to a = F m F t x
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 F. φ > 0 When measuring the angle between two vectors, they should be drawn tail-to-tail r F z y S = rθ 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 F is counter-clock-wise (CCW): it means that the force/torque tends to push the body to rotate in the positive (CCW) direction. r The diagram to the right here shows th φ < 0 case φ < 0 smartphysics uses θ for both this angle and the angular position: it s confusing F r F F r φ F t x
Unit 15 3 of 3