8 Rotational Dynamics Homework: Complete sentence physics answers required. 1, 3, 4, 7, 8, 17, 19, 33, 39, 49, 50, 51, 60, 65, 83, .

Rotational Dynamics: Newton’s 2nd Law for Rotation Clockwise (CW) Counter-clockwise (CCW)

Rotational Inertia & Energy

Mass-Distribution ~ Larger radius Larger Speed Larger Effort   Rotational Inertia ~ R2

Central Axis

Axis on End

Calculated Rot. Inertias, p.273 rotational inertias of solid objects can be calculated need to know: a, b, c, d, e, g, h, j, k. omit: f, i

Torque meter-newton ft-lb

torque lever-arm is the shortest distance from axis to line of the force torque = force x lever-arm 9 9

Torque (t) [m·N] F F = lever-arm

Example Rod is 3m Long.

Newton’s 2nd Law (Rotation)

Concept Review Torque: rotational action Rotational Inertia: resistance to change in rotational motion. Torque = force x lever-arm

Equilibrium Translational: net-force = 0 Rotational: net-torque = 0

The drawing shows a person whose weight is 584N The drawing shows a person whose weight is 584N. Calculate the net force with which the floor pushes on each end of his body.

Rotational Kinetic Energy Rotational K = ½(I)w2. Example: Constant Power Source has 100kg, 20cm radius, solid disk rotating at 7000 rad/s. I = ½MR2 = ½(100kg)(0.2m)2 = 2kgm2. Rot K = ½ (2kgm2)(7000/s)2 = 49 MJ

Rotational Work-Energy Theorem (Work)rot = tDq. Example: torque of 50 mN is applied for one revolution. rotational work = (50Nm)(2prad) = 314 J (Rotational Work)net = DKrot. Krot = ½Iw2.

Angular Momentum (L) analog of translational momentum L = Iw [kgm2/s] Example: Disk R = 1m, M = 1kg, w = 10/s I = ½MR2 = ½(1)(1)2 = 0.5 L = Iw = (0.5kgm2)(10/s) = 5kgm2/s

Conservation of Angular Momentum For an isolated system (Iw)before = (Iw)after Example: Stationary disk M,R is dropped on rotating disk M, R, wi. (½MR2)(wi) = (½MR2 + ½MR2)(wf) (wf) = ½ (wi)

8 Summary All TRANSLATIONAL quantities; speed, velocity, acceleration, force, inertia, energy, and momentum, have ROTATIONAL analogs.

Rotational Inertia ( I ) kg(m)2 Example 3m 2m 4kg 5kg 21 21

Problem 33 Pivot at left joint, Fj = ?, but torque = 0. ccw (Fm)sin15(18) = mg(26) = cw ccw (Fm)sin15(18) = (3)g(26) = cw (Fm) = (3)g(26)/sin15(18) = 160N Note: any point of arm can be considered the pivot (since arm is at rest)

If ball rolls w/o slipping at 4 If ball rolls w/o slipping at 4.0m/s, how large is the height h in the drawing? rolling w/o slipping

#39 Left force = mg = 30g, Right = 25g mg = 30g + 25g m = 55kg ccw mg(xcg) = cw 30g(1.6) (55)g(xcg) = 30g(1.6) (55)(xcg) = 30(1.6) Xcg = (30/55)(1.6)

#60, z-axis Each mass has r2 = 1.52 + 2.52. I = sum mr2 = (2+3+1+4)(1.52 + 2.52)

#65 First with no frictional torque, then with frictional torque as specified in problem. M = 0.2kg, R = 0.15m, m1 = 0.4, m2 = 0.8

#83 Pulley M, R. what torque causes it to reach ang. Speed. 25/s in 3rev? Alpha: use v-squared analog eqn. Torque = Ia = (½MR2)(a)

#89, uniform sphere part Rolling at v = 5m/s, M = 2kg, R = 0.1m K-total = ½mv2 + ½Iw2. = ½(2)(5x5) + ½[(2/5)(2)(0.1x0.1)](5/0.1)2. = 25 + 10 = 35J Roll w/o slipping, no heat created, mech energy is conserved, goes all to Mgh. 35 = Mgh h = 35/Mg = 35(19.6) = 1.79m

#111 Ice skater, approximate isolated system Therefore: (Iw)before = (Iw)after (100)(wi) = (92.5)(wf) (wf) = (100/92.5)(wi) K-rot increases by this factor squared times new rot. Inertia x ½.

Example: Thin rod formulas.

Angular Momentum Symbol: L Unit: kg·m2/s L = mvr = m(rw)r = mr2w = Iw. v is perpendicular to axis r is perpendicular distance from axis to line containing v.

Angular Momentum Symbol: L Unit: kg·m2/s L = mvr = m(rw)r = mr2w = Iw. v is perpendicular to axis r is perpendicular distance from axis to line containing v.

13) Consider a bus designed to obtain its motive power from a large rotating flywheel (1400. kg of diameter 1.5 m) that is periodically brought up to its maximum speed of 3600. rpm by an electric motor at the terminal. If the bus requires an average power of 12. kilowatts, how long will it operate between recharges? Answer: 39. minutes Diff: 2 Var: 1 Page Ref: Sec. 8.4

6) A 82. 0 kg painter stands on a long horizontal board 1 6) A 82.0 kg painter stands on a long horizontal board 1.55 m from one end. The 15.5 kg board is 5.50 m long. The board is supported at each end. (a) What is the total force provided by both supports? (b) With what force does the support, closest to the painter, push upward?

28) A 4.0 kg mass is hung from a string which is wrapped around a cylindrical pulley (a cylindrical shell). If the mass accelerates downward at 4.90 m/s2, what is the mass of the pulley? A) 10.0 kg B) 4.0 kg C) 8.0 kg D) 2.0 kg E) 6.0 kg

(a) 2.0 kg-m2 (b) 2.3 kg-m2 (c) 31. rpm 19) A solid disk with diameter 2.00 meters and mass 4.0 kg freely rotates about a vertical axis at 36. rpm. A 0.50 kg hunk of bubblegum is dropped onto the disk and sticks to the disk at a distance d = 80. cm from the axis of rotation. (a) What was the moment of inertia before the gum fell? (b) What was the moment of inertia after the gum stuck? (c) What is the angular velocity after the gum fell onto the disk? (a) 2.0 kg-m2 (b) 2.3 kg-m2 (c) 31. rpm

1. A pair of forces with equal magnitudes and opposite directions is acts as shown. Calculate the torque on the wrench.

3. The drawing shows the top view of two doors 3. The drawing shows the top view of two doors. The doors are uniform and identical. The mass of each door is M and width as shown below is L. How do their rotational accelerations compare?

A Ring, a Solid-Disk, and a Solid-Sphere are released from rest from the top of an incline. Each has the same mass and radius. Which will reach the bottom first?

5. The device shown below is spinning with rotational rate wi when the movable rods are out. Each moveable rod has length L and mass M. The central rod is length 2L and mass 2M. Calculate the factor by which the angular velocity is increased by pulling up the arms as shown.

Rotational Review (angles in radians) + 4 kinematic equations

L = Iw Angular Momentum Calculation Example: Solid Disk M = 2kg R = 25cm Spins about its center-of-mass at 35 rev/s

4. A one-meter-stick has a mass of 480grams. a) Calculate its rotational inertia about an axis perpendicular to the stick and through one of its ends. b) Calculate its rotational inertia about an axis perpendicular to the stick and through its center-of-mass. c) Calculate its angular momentum if spinning on axis (b) at a rate of 57rad/s.

Conservation of Angular Momentum Example: 50 grams of putty shot at 3m/s at end of 200 gram thin 80cm long rod free to rotate about its center. Li = mvr = (0.050kg)(3m/s)(0.4m) Lf = Iw = {(1/12)(0.200kg)(0.8m)2 + (0.050kg)(0.4m)2}(w) final rotational speed of rod&putty =