Physics 121, Sections 9, 10, 11, and 12 Lecture 18 Today's Topics Homework 7: due Friday Nov 4@ 6: 00 PM Chap7:#3,11,20,21,25,27,30,40, 46,47,52,and68 Chapter 8 Rotation Moment of inertia Rolling motion Physics 121: Lecture 18, Pg 1
Physics 121: Lecture 18, Pg 1 Physics 121, Sections 9, 10, 11, and 12 Lecture 18 Today’s Topics: Homework 7: due Friday Nov. 4 @ 6:00 PM. Chap. 7: # 3, 11, 20, 21, 25, 27, 30, 40, 46, 47, 52, and 68. Chapter 8: Rotation Moment of inertia Rolling motion
Summary (with comparison to 1-D kinematics) Angular Linear a= constant a= constant V= Vo+at 0=00+Oot+ ot+at And for a point at a distance R from the rotation axis X=RO V=OR a= ar Physics 121: Lecture 18, Pg 2
Physics 121: Lecture 18, Pg 2 Summary (with comparison to 1-D kinematics) Angular Linear = constant = 0 +0 + 1 2 2 t t a = constant v = v + at 0 x = x + v t + at 0 0 1 2 2 And for a point at a distance R from the rotation axis: x = R v = R a = R
Rotation Kinetic Energy The kinetic energy of a rotating system looks similar to that of a point particle Point particle Rotating System KE=-mv KE=-I v is"linear velocity o is angular velocity m is the mass I is the moment of inertia about the rotation axis Physics 121: Lecture 18, Pg 3
Physics 121: Lecture 18, Pg 3 Rotation & Kinetic Energy... The kinetic energy of a rotating system looks similar to that of a point particle: Point Particle Rotating System 2 I 2 1 KE = I = m r i i i 2 2 2 1 KE = mv v is “linear” velocity m is the mass. is angular velocity I is the moment of inertia about the rotation axis
Moment of inertia So KE=-Io where I Notice that the moment of inertia /depends on the distribution of mass in the system The further the mass is from the rotation axis, the bigger the moment of inertia For a given object, the moment of inertia will depend on where we choose the rotation axis(unlike the center of mass We will see that in rotational dynamics, the moment of inertia appears in the same way that mass m does when we study linear dynamics Physics 121: Lecture 18, Pg 4
Physics 121: Lecture 18, Pg 4 Moment of Inertia Notice that the moment of inertia I depends on the distribution of mass in the system. The further the mass is from the rotation axis, the bigger the moment of inertia. For a given object, the moment of inertia will depend on where we choose the rotation axis (unlike the center of mass). We will see that in rotational dynamics, the moment of inertia I appears in the same way that mass m does when we study linear dynamics ! 2 I 2 1 KE = I = m r i i i 2 So where
Calculating Moment of Inertia We have shown that for N discrete point masses distributed about a fixed axis the moment of inertia is Where ris the distance from the mass to the axis of rotation Example: Calculate the moment of inertia of four point masses (m)on the corners of a square whose sides have length L, about a perpendicular axis through the center of the square m m m m Physics 121: Lecture 18, Pg 5
Physics 121: Lecture 18, Pg 5 Calculating Moment of Inertia We have shown that for N discrete point masses distributed about a fixed axis, the moment of inertia is: I = = m r i i i N 2 1 where r is the distance from the mass to the axis of rotation. Example: Calculate the moment of inertia of four point masses (m) on the corners of a square whose sides have length L, about a perpendicular axis through the center of the square: m m m m L
Calculating Moment of Inertia The squared distance from each point mass to the axis is Using the Pythagorean Theorem s0/=m=m2+m+m2+m=4m 1= 2mL2 L2 m● m m m Physics 121: Lecture 18, Pg 6
Physics 121: Lecture 18, Pg 6 Calculating Moment of Inertia... The squared distance from each point mass to the axis is: m m m m L r L/2 2 L 2 L r 2 2 2 2 = = 2 L 4m 2 L m 2 L m 2 L m 2 L I m r m 2 2 2 2 2 N i 1 2 = i i = + + + = = so I = 2mL2 Using the Pythagorean Theorem
Calculating Moment of Inertia Now calculate /for the same object about an axis through the center, parallel to the plane(as shown) =∑mr=m+m+m+m==4m I= ml2 m m m● Physics 121: Lecture 18, Pg 7
Physics 121: Lecture 18, Pg 7 Calculating Moment of Inertia... Now calculate I for the same object about an axis through the center, parallel to the plane (as shown): m m m m L r 4 L 4m 4 L m 4 L m 4 L m 4 L I m r m 2 2 2 2 2 N i 1 2 = i i = + + + = = I = mL2
Calculating Moment of Inertia Finally, calculate / for the same object about an axis along one side(as shown) =∑mr2=ml2+m2+m02+m0 1= 2mL2 m m Physics 121: Lecture 18, Pg 8
Physics 121: Lecture 18, Pg 8 Calculating Moment of Inertia... Finally, calculate I for the same object about an axis along one side (as shown): m m m m L r 2 2 2 2 N i 1 2 I = mi r i = mL + mL + m0 + m0 = I = 2mL2
Calculating Moment of Inertia. For a single object, I clearly depends on the rotation axis 1= 2mL2 ml 1= 2mL2 m m● m Physics 121: Lecture 18, Pg 9
Physics 121: Lecture 18, Pg 9 Calculating Moment of Inertia... For a single object, I clearly depends on the rotation axis !! L I = 2mL2 I = mL2 m m m m I = 2mL2
Lecture 18. Act 1 Moment of Inertia a triangular shape is made from identical balls and identical rigid, massless rods as shown the moment of inertia about the a, b, and c axes is la, Ib, and Ic respectively Which of the following is correct (a)12>1b>1c (b)la>Ic>Ib (c)b>a>Ic abc Physics 121: Lecture 18, Pg 10
Physics 121: Lecture 18, Pg 10 Lecture 18, Act 1 Moment of Inertia A triangular shape is made from identical balls and identical rigid, massless rods as shown. The moment of inertia about the a, b, and c axes is Ia , Ib , and Ic respectively. Which of the following is correct: (a) Ia > Ib > Ic (b) Ia > Ic > Ib (c) Ib > Ia > Ic a b c
