1 / 24

Understanding Center of Mass and Torque in Rotational Dynamics

Explore principles of center of mass, rolling motion, torque, equilibrium dynamics, and conservation of angular momentum in rotational systems of particles. Learn to analyze and calculate rotational dynamics using examples and demonstrations.

Download Presentation

Understanding Center of Mass and Torque in Rotational Dynamics

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Lecture 17 Goals: • Chapter 12 • Define center of mass • Analyze rolling motion • Introduce and analyze torque • Understand the equilibrium dynamics of an extended object in response to forces • Employ “conservation of angular momentum” concept Assignment: • HW7 due tomorrow • Wednesday, Exam Review

  2. + + m2 m1 m1 m2 A special point for rotationSystem of Particles: Center of Mass (CM) • A supported object will rotate about its center of mass. • Center of mass: Where the system is balanced ! • Building a mobile is an exercise in finding centers of mass. mobile

  3. System of Particles: Center of Mass • How do we describe the “position” of a system made up of many parts ? • Define the Center of Mass (average position): • For a collection of N individual point like particles whose masses and positions we know: RCM m2 m1 r2 r1 y x (In this case, N = 2)

  4. RCM = (12,6) (12,12) 2m m m (0,0) (24,0) Sample calculation: m at ( 0, 0) 2m at (12,12) m at (24, 0) • Consider the following mass distribution: XCM = (m x 0 + 2m x 12 + m x 24 )/4m meters YCM = (m x 0 + 2m x 12 + m x 0 )/4m meters XCM= 12 meters YCM= 6 meters

  5. p p VCM p  p p p p Connection with motion... • So a rigid object that has rotation and translation rotates about its center of mass! • And Newton’s Laws apply to the center of mass For a point p rotating:

  6. Work & Kinetic Energy: • Recall the Work Kinetic-Energy Theorem: K = WNET • This applies to both rotational as well as linear motion. • What if there is rolling?

  7. M M M M M h M who is 1st? q M Demo Example :A race rolling down an incline • Two cylinders with identical radii and total masses roll down an inclined plane. • The 1st has more of the mass concentrated at the center while the 2nd has more mass concentrated at the rim. • Which gets down first? Two cylinders with radius R and mass m Mass 1 Mass 2 They both arrive at same time

  8. M M M M M h M v ? q M Same Example :Rolling, without slipping, Motion • A solid disk is about to roll down an inclined plane. • What is its speed at the bottom of the plane ?

  9. Rolling without slipping motion • Again consider a cylinder rolling at a constant speed. 2VCM CM VCM

  10. Motion • Again consider a cylinder rolling at a constant speed. Both with |VTang| = |VCM | Rotation only VTang = wR Sliding only 2VCM VCM CM CM CM VCM If acceleration acenter of mass = - aR

  11. Disk has radius R M M M M M h M v ? q M Example :Rolling Motion • A solid cylinder is about to roll down an inclined plane. What is its speed at the bottom of the plane ? • Use Work-Energy theorem Mgh = ½ Mv2 + ½ ICMw2 and v =wR Mgh = ½ Mv2 + ½ (½ M R2)(v/R)2 = ¾ Mv2 v = 2(gh/3)½

  12. How do we reconcile force, angular velocity and angular acceleration?

  13. Angular motion can be described by vectors • With rotation the distribution of mass matters. Actual result depends on the distance from the axis of rotation. • Hence, only the axis of rotation remains fixed in reference to rotation. We find that angular motions may be quantified by defining a vector along the axis of rotation. • We can employ the right hand rule to find the vector direction

  14. The Angular Velocity Vector • The magnitude of the angular velocity vector is ω. • The angular velocity vector points along the axis of rotation in the direction given by the right-hand rule as illustrated above. • As w increased the vector lengthens

  15. From force to spin (i.e., w) ? A force applied at a distance from the rotation axis gives a torque a FTangential NET = |r| |FTang| ≡|r||F| sin q F q Fradial r r FTangential Fradial =|FTang| sin q • If a force points at the axis of rotation the wheel won’t turn • Thus, only the tangential component of the force matters • With torque the position & angle of the force matters

  16. a FTangential F Fradial r Rotational Dynamics: What makes it spin? A force applied at a distance from the rotation axis NET = |r| |FTang| ≡|r||F| sin q • Torque is the rotational equivalent of force Torque has units of kg m2/s2 = (kg m/s2) m = N m NET = r FTang = r m aTang = r m r a = (m r2) a For every little part of the wheel

  17. a FTangential F Frandial r For a point massNET= m r2a The further a mass is away from this axis the greater the inertia (resistance) to rotation (as we saw on Wednesday) NET = Ia • This is the rotational version of FNET = ma • Moment of inertia, I≡Simi ri2 , is the rotational equivalent of mass. • If I is big, more torque is required to achieve a given angular acceleration.

  18. Rotational Dynamics: What makes it spin? A force applied at a distance from the rotation axis gives a torque a FTangential NET = |r| |FTang| ≡|r||F| sin q F Fradial r • A constant torque gives constant angularacceleration if and only if the mass distribution and the axis of rotation remain constant.

  19. Torque, like w, is a vector quantity • Magnitude is given by (1) |r| |F| sin q (2) |Ftangential | |r| (3) |F| |rperpendicular to line of action | • Direction is parallel to the axis of rotation with respect to the “right hand rule” • And for a rigid object= I a r sin q line of action F cos(90°-q) = FTang. r a 90°-q q F F F Fradial r r r

  20. Example :Rolling Motion • Newton’s Laws: M N f x dir Mg q Notice rotation CW (i.e. negative) when ax is positive! Combining 3rd and 4th expressions gives f = Max / 2 Top expression gives Max + f = 3/2 M ax = Mg sin q So ax =2/3 Mg sin q

  21. Statics Equilibrium is established when In 3D this implies SIX expressions (x, y & z)

  22. Example • Two children (60 kg and 30 kg) sit on a horizontal teeter-totter. The larger child is 1.0 m from the pivot point while the smaller child is trying to figure out where to sit so that the teeter-totter remains motionless. The teeter-totter is a uniform bar of 30 kg its moment of inertia about the support point is 30 kg m2. • Assuming you can treat both children as point like particles, what is the initial angular acceleration of the teeter-totter when the large child lifts up their legs off the ground (the smaller child can’t reach)? • For the static case:

  23. N 60 kg 30 kg 0.5 m 1 m 300 N 300 N 600 N 30 kg Example: Soln. • Draw a Free Body diagram (assume g = 10 m/s2) • 0 = 300 d + 300 x 0.5 + N x 0 – 600 x 1.0 0= 2d + 1 – 4 d = 1.5 m from pivot point

  24. Recap Assignment: • HW7 due tomorrow • Wednesday: review session

More Related