Physics 1 06 : Mechanics Lecture 03

1. Physics 106: Mechanics Lecture 03 Wenda Cao NJITPhysics Department

2. Rotational Equilibrium and Rotational Dynamics II • Rotational Kinetic Energy • Moment of Inertia • Torque • Newton 2nd Law for Rotational Motion: Torque and angular acceleration

3. Rotational Kinetic Energy • There is an analogy between the kinetic energies associated with linear motion (K = ½ mv 2) and the kinetic energy associated with rotational motion (KR= ½ Iw2) • Rotational kinetic energy is not a new type of energy, the form is different because it is applied to a rotating object • Units of rotational kinetic energy are Joules (J)

4. Moment of Inertia of Point Mass • For a single particle, the definition of moment of inertia is • m is the mass of the single particle • r is the rotational radius • SI units of moment of inertia are kg.m2 • Moment of inertia and mass of an object are different quantities • It depends on both the quantity of matter and its distribution (through the r2 term)

5. m cm P m L P m cm m L/2 L/2 P m cm m L Moment of Inertia of Point Mass • For a composite particle, the definition of moment of inertia is • mi is the mass of the ith single particle • ri is the rotational radius of ith particle • SI units of moment of inertia are kg.m2

6. Moment of Inertia of Extended Objects • Divided the extended objects into many small volume elements, each of mass Dmi • We can rewrite the expression for Iin terms of Dm • With the small volume segment assumption, • If r is constant, the integral can be evaluated with known geometry, otherwise its variation with position must be known

7. Moment of Inertia for some other common shapes

9. Parallel-Axis Theorem • In the previous examples, the axis of rotation coincided with the axis of symmetry of the object • For an arbitrary axis, the parallel-axis theorem often simplifies calculations • The theorem states I = ICM + MD 2 • I is about any axis parallel to the axis through the center of mass of the object • ICM is about the axis through the center of mass • D is the distance from the center of mass axis to the arbitrary axis

10. h2 = (a/2)2 + (b/2)2 a m m h X cm b h m m P • Rotation axes perpendicular to plane of figure • Masses on the corners of a rectangle, sides a & b • About an axis through the CM: • About an axis “P” through a corner: • Using the Parallel Axis Theorem directly for the same corner axis:

11. Force vs. Torque • Forces cause accelerations • What cause angular accelerations ? • A door is free to rotate about an axis through O • There are three factors that determine the effectiveness of the force in opening the door: • The magnitude of the force • The position of the application of the force • The angle at which the force is applied

12. General Definition of Torque • Let F be a force acting on an object, and let r be a position vector from a rotational center to the point of application of the force. The magnitude of the torque is given by •  = 0° or  = 180 °: torque are equal to zero •  = 90° or  = 270 °:torque attain to the maximum • Torque will have direction • If the turning tendency of the force is counterclockwise, the torque will be positive • If the turning tendency is clockwise, the torque will be negative

13. Net Torque • The force will tend to cause a counterclockwise rotation about O • The force will tend to cause a clockwise rotation about O • St = t1 + t2 = F1d1 – F2d2 • If St 0, starts rotating • If St= 0, rotation rate does not change • Rate of rotation of an object does not change, unless the object is acted on by a net torque

14. Torque on a Rotating Object • Consider a particle of mass m rotating in a circle of radius r under the influence of tangential force • The tangential force provides a tangential acceleration:Ft = mat • Multiply both side by r, then rFt = mrat • Since at= r, we have rFt = mr2 • So, we can rewrite it as  = mr2  = I

15. Torque on a Solid Disk • Consider a solid disk rotating about its axis. • The disk consists of many particles at various distance from the axis of rotation. The torque on each one is given by  = mr2 • The net torque on the disk is given by  = (mr2) • A constant of proportionality is the moment of inertia, I =mr2 = m1r12 + m2r22 + m3r32 + … • So, we can rewrite it as  = I

16. Newton’s Second Law for a Rotating Object • When a rigid object is subject to a net torque (≠0), it undergoes an angular acceleration • The angular acceleration is directly proportional to the net torque • The angular acceleration is inversely proportional to the moment of inertia of the object • The relationship is analogous to

18. Example 1: second law for rotation When she is launched from a springboard, a diver's angular speed about her center of mass changes from zero to 6.20 rad/s in 220 ms. Her rotational inertia about her center of mass is constant at 12.0 kg·m2. During the launch, what are the magnitudes of (a) her average angular acceleration and (b) the average external torque on her from the board? a) Use: or b) Use:

19. +y N L1 L2 a1 = - aL1 m1g fulcrum m2g Constraints: a2 = + aL2 Example 2: afor an unbalanced bar • Bar is massless and originally horizontal • Rotation axis at fulcrum point  N has zero torque • Find angular acceleration of bar and the linear acceleration of m1 just after you let go Use: Using specific numbers: where: Let m1 = m2= m L1=20 cm, L2 = 80 cm What happened to sin(q) in moment arm? net torque Clockwise Accelerates UP total I about pivot

20. q L1 N m1g L2 fulcrum m2g Newton 2nd Law in Rotation • Suppose everything is as it was in the preceding example, but the bar is NOT horizontal. Assume both masses are equal. Which of the following is the correct equation for the angular acceleration?

21. Note: can have Fnet .eq. 0 but tnet.ne. 0 Strategy to use the Newton 2nd Law Many components in the system means several (N) unknowns…. … need an equal number of independent equations Draw or sketch system. Adopt coordinates, name the variables, indicate rotation axes, list the known and unknown quantities, … • Draw free body diagrams of key parts. Show forces at their points of application. find torques about a (common) axis • May need to apply Second Law twice to each part • Translation: • Rotation: • Make sure there are enough (N) equations; there may be constraint equations (extra conditions connecting unknowns) • Simplify and solve the set of (simultaneous) equations. • Interpret the final formulas. Do they make intuitive sense? Refer back to the sketches and original problem • Calculate numerical results, and sanity check anwers (e.g., right order of magnitude?)

22. Rotating Rod • A uniform rod of length L and mass M is attached at one end to a frictionless pivot and is free to rotate about the pivot in the vertical plane as in Figure. The rod is released from rest in the horizontal position. What are the initial angular acceleration of the rod and the initial translational acceleration of its right end?

23. Rotating Rod

24. The Falling Object • A solid, frictionless cylindrical reel of mass M = 2.5 kg and radius R = 0.2 m is used to draw water from a well. A bucket of mass m = 1.2 kg is attached to a cord that is wrapped around the cylinder. • (a) Find the tension T in the cord and acceleration a of the object. • (b) If the object starts from rest at the top of the well and falls for 3.0 s before hitting the water, how far does it fall ?

25. Example, Newton’s Second Law for Rotation • Draw free body diagrams of each object • Only the cylinder is rotating, so apply St = I a • The bucket is falling, but not rotating, so apply SF = m a • Remember that a = a r and solve the resulting equations

26. r a T y mg mg N T Mg • Cord wrapped around disk, hanging weight • Cord does not slip or stretch  constraint • Disk’s rotational inertia slows accelerations • Let m = 1.2 kg, M = 2.5 kg, r =0.2 m For mass m: Unknowns: T, a support force at axis “O” has zero torque FBD for disk, with axis at “o”: Unknowns: a, a from “no slipping” assumption So far: 2 Equations, 3 unknowns Need a constraint: Substitute and solve:

27. r a T y mg mg • Cord wrapped around disk, hanging weight • Cord does not slip or stretch  constraint • Disk’s rotational inertia slows accelerations • Let m = 1.2 kg, M = 2.5 kg, r =0.2 m For mass m: Unknowns: T, a support force at axis “O” has zero torque