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Lecture 17 (Catch up)

Lecture 17 (Catch up). Goals:. Chapter 12 Introduce and analyze torque U nderstand the equilibrium dynamics of an extended object in response to forces Employ “conservation of angular momentum” concept. Assignment: HW8 due March 17 th Thursday, Exam Review.

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Lecture 17 (Catch up)

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  1. Lecture 17 (Catch up) Goals: • Chapter 12 • Introduce and analyze torque • Understand the equilibrium dynamics of an extended object in response to forces • Employ “conservation of angular momentum” concept Assignment: • HW8 due March 17th • Thursday, Exam Review

  2. Rotational Motion: Statics and Dynamics • Forces are still necessary but the outcome depends on the location from the axis of rotation • This is in contrast to the translational motion and acceleration of the center of mass. Here the position of these forces doesn’t matter (doesn’t alter the physics we see) • However: For rotational statics & dynamics: we must reference the specific position of the force relative to an axis of rotation. It may be necessary to consider more than one rotation axis! • Vectors remain the key tool for visualizing Newton’s Laws

  3. 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

  4. 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

  5. So: What makes it spin (causes 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

  6. 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

  7. 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 Thursday) 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.

  8. 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.

  9. 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

  10. Exercise Torque Magnitude • Case 1 • Case 2 • Same • In which of the cases shown below is the torque provided by the applied force about the rotation axis biggest? In both cases the magnitude and direction of the applied force is the same. • Remember torque requires F, rand sin q or the tangential force component times perpendicular distance L F F L axis case 1 case 2

  11. L m Example: Rotating Rod Again • A uniform rod of length L=0.5 m and mass m=1 kg is free to rotate on a frictionless pin passing through one end as in the Figure. The rod is released from rest in the horizontal position. • What is the initial angular acceleration a?

  12. L m Example: Rotating Rod • A uniform rod of length L=0.5 m and mass m=1 kg is free … What is its initial angular acceleration ? 1. For forces you need to locate the Center of Mass CM is at L/2 ( halfway ) and put in the Force on a FBD 2. The hinge changes everything! S F = 0 occurs only at the hinge buttz = I az = - r F sin 90° at the center of mass and IEndaz = - (L/2) mg and solve foraz mg

  13. F  R dr =Rd d Work (in rotational motion) • Consider the work done by a force F acting on an object constrained to move around a fixed axis. For an infinitesimal angular displacement d:where dr =R d dW = FTangential dr dW = (FTangential R)d • dW =  d (and with a constant torque) • We can integrate this to find: W =  = t (qf-qi) • Analog of W = F •r • W will be negativeif  and have opposite sign ! axis of rotation

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

  15. 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:

  16. 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

  17. Classic Example: Will the ladder slip? Not if

  18. Classic Example: Will the ladder slip?

  19. EXAMPLE 12.17 Will the ladder slip?

  20. EXAMPLE 12.17 Will the ladder slip?

  21. Angular Momentum: • We have shown that for a system of particles, momentum is conserved if • What is the rotational equivalent of this (rotational “mass” times rotational velocity)? angular momentum is conserved if

  22. v1 m2 j  m1 r2 r1 i v2 r3 v3 m3 Angular momentum of a rigid body about a fixed axis: • Consider a rigid distribution of point particles rotating in the x-y plane around the z axis, as shown below. The total angular momentum around the origin is the sum of the angular momentum of each particle: • Even if no connecting rod we can define an Lz ( ri and vi, are perpendicular) Using vi =  ri, we get

  23. z z F Example: Two Disks • A disk of mass M and radius R rotates around the z axis with angular velocity 0. A second identical disk, initially not rotating, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity F. 0

  24. z z 0 F Example: Two Disks • A disk of mass M and radius R rotates around the z axis with initial angular velocity 0. A second identical disk, at rest, is dropped on top of the first. There is friction between the disks, and eventually they rotate together with angular velocity F. No External Torque so Lz is constant Li = Lf I wii = I wf½ mR2w0 = ½ 2mR2wf

  25. Example: Throwing ball from stool • A student sits on a stool, initially at rest, but which is free to rotate. The moment of inertia of the student plus the stool is I. They throw a heavy ball of mass M with speed v such that its velocity vector moves a distance d from the axis of rotation. • What is the angular speed F of the student-stool system after they throw the ball ? M v F d I I Top view: before after

  26. Example: Throwing ball from stool • What is the angular speed F of the student-stool system after they throw the ball ? • Process: (1) Define system (2) Identify Conditions (1) System: student, stool and ball (No Ext. torque, L is constant) (2) Momentum is conserved Linit = 0 = Lfinal = -mvd + I wf M v F d I I Top view: before after

  27. Angular Momentum as a Fundamental Quantity • The concept of angular momentum is also valid on a submicroscopic scale • Angular momentum has been used in the development of modern theories of atomic, molecular and nuclear physics • In these systems, the angular momentum has been found to be a fundamental quantity • Fundamental here means that it is an intrinsic property of these objects

  28. Lecture 17 Assignment: • HW7 due March 17th • Thursday: Review session

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