Final Exam During class (1 - 3:55 pm) on 6/27, Mon Room: 412 FMH (classroom)
Bring scientific calculators No smart phone callltculators are allowed . Exam covers everything learned in this course.
Review session: Thursday during class
1 How to find torque
F From figure, θ F T m τ = rFTp== rFsinθ r F r
rp axis Line of action
If we know components G G rrr= (0)( x ,,y 0) FFF= (0)( x ,,y 0) G τ =−(0,0,rFx yyx rF )
Reminder: Rotational variables are vectors, having direction
The directions are given by the right-hand rule:
Fingers of right hand curl along the angular direction (See Fig.)
Then, the direction of thumb is thdihe direct ion o fhf the angu lar quantity.
(a) A (b) B C (c) C (d) All forces have the same torque (e) Not enough information
G 5.1.G A particle located at the position vector rmm = (1 , 2 ) has a force F = (2NN ,3 ) acting on it. The torque in N.m about the origin is?
3 Gravitational force on extended objects
Torque on extended object by gravitational force Æ Assume that the total gravitational force effectively acts at the center of mass.
Gravitational potential energy of extended object Æ M g H, where H is the height of the center of mass and M is the total mass.
Axis of rotation
HiHorizonta l unif orm rod of length L & mass M Find the torque by gravitational force.
A. LMg B. (L/2)Mg C. 2LMg D. (3/2)LMg E. None of the above 1 Find the angular acceleration. I = ML2 end, rod 3
4 Example: Torque and Angular Acceleration of a Wheel
•Cord wrapped around disk, hanging weight • Cord does not slip or stretch •Let m = 1.2 kg, M = 2.5 kg, r =0.2 m • Find acceleration of mass m, r find angular acceleration α for disk, tension, and torque on the a disk 1 Formula: I = Mr2 2 mg
Example: A uniform circular disk of radius r and mass M is pulled by constant horizontal force F applied to the center of mass, and is rolling without slipping. M=2 kg, r=0.5 m, I_(cm,disk)=(1/2)MR^2, F=5N. a) Find the angular acceleration. b) Find the static friction force.
cm r CCW = + F P
iClicker: If there’s no friction from the surface on the disk, and if the disk is initially not rotating , would it rotate about center of mass by applied force?
A) Yes B) No
5 Kinetic energy for rotation
Kinetic energy for rotation If the rigid body rotates with respect to a moving axis through the center of mass, such as rolling ω
KK=+rot K cm vcm 1 KI= ω 2 rot2 cm 1 KMv= 2 cm2 cm For rolling without slipping ΔsR=Δθ
vRcm = ω
6 For any rotational or translational motion,
UMgravity= gh cm
Δ=EWmech non− conservative
is conserved. If Wnon− conservative = 0 , then Emech
Example of energy conservation
Axis of rotation Initial
Final Horizontal uniform rod of length L & mass M is released from rest.
Find its angular speed at the lowest point, assuming no friction between axis of rotation and the rod.
7 Example: Use energy conservation to find the speed of the bowling ball as it rolls w/o slipping to the bottom of the ramp Given: h=2m Formula: For a solid sphere I = 2 MR2 cm 5 Hint:
Rotation accelerates if there is friction between the sphere and the ramp Friction force produces the net torque and angular acceleration. There is no mechanical energy change because the contact point is always at rest relative to the surface, so no work is done against friction
iClicker Q: A solid sphere and a spherical shell of the same radius r and same mass M roll to the bottom of a ramp without slipping from the same height h.
True or false? : “The two have the same speed at the bottom.”
A) True B) False. Shell is faster. C) False. Solid sphere is faster. D) Not enough information.
I_(cm, spherical shell) = (2/3) MR^2 I_(cm, solid sphere)=(2/5) MR^2
8 Angular momentum – concepts & definition
- Linear momentum: p = mv
linear rotational inertia m I speed v ω rigid body linear p=mv L=Iω angular momentum momentum
Angular momentum of a bowling ball 6.1. A bowling ball is rotating as shown about its mass center axis. Find it’s angular momentum about that axis, in kg.m2/s
A) 4 B) ½ C) 7 D) 2 E) ¼
ω = 4d/4 rad/s M = 5 kg r = ½ m I = 2/5 MR2 L = Iω
9 Angular momentum of a point particle 2 L== Iωω mr = mvTpTp r = mvrsin θ = mvr = p r = pr
vT = ω r O G r r p vT G θ v
Note: L = 0 if v is parallel to r (radially in or out)
LI==ωθ mvrmvrTp =sin = mvr
= rpTp== rp sinθ r p
Angular momentum of a point particle
O G L = rp== rpsinθ r p r r T p p pT G θ p
If we know components G G ppp= (0)( ,,0) rrr= (0)( xy,,0) x y G L =−(0,0,rpx yyx rp )
10 Angular momentum for car
5.2. A car of mass 1000 kg moves with a speed of 50 m/s on a circular track of radius 100 m. What is the magnitude of its angular momentum (in kg • m2/s) relative to the center of the race track (point “P”) ?
A) 5.0 × 102 A
6 B) 5.0 × 10 B 4 C) 2.5 × 10 P D) 2.5 × 106 E) 5.0 × 103
5.3. What would the anggpular momentum about point “P” be if the car leaves the track at “A” and ends up at point “B” with the same velocity ? A) Same as above B) Different from above
C) Not Enough Information L = r ⊥ p = p ⊥r = pr sin( θ)
Net angular momentum of particles
11 Example: calculating angular momentum for particles
PP10602-23*: Two objects are moving as shown in the figure . What is their total angular momentum about point O?
ΔΔω L τα==II = ΔLt=Δτ net net ΔΔtt
Conservation of angular momentum :
Angular momentum, L is consreved if τ net = 0 Conservation of net angular momentum for a system
Net angular momentum, Lnet is consreved if τ net,ext = 0
12 A puck on a frictionless air hockey table has a mass of 5.0 g and is attached to a cord passing through a hole in the surface as in the figure. The puck is revolving at a distance 2.0 m from the hole with an angular velocity of 3.0 rad/s. The cord is then pulled from below, shortening the radius to 1 .0 m . The new angular velocity (in rad/s) is ______.
Demonstration: Spinning Professor: Web link
τ net, ext =⇒0 ⇒ L = constant
Iiiω = I fω f
Moment of inertia changes Another link
13 Internal torques do not change total angular momentum...... G it is redistributed within the isolatedG system τnet = 0 about z - axis ⇒ Lsys = constant Demonstration: Weblink, more Spacecraft maneuvers by Internal torques not 0 spinning the flywheel
they reverse L wh the craft count er-rotttates
G G G G G L = 0 = L + L Lboy,final = 2Lwheel tot craft flywheel
6.3. Two astronauts each having mass M are connected by a mass-less rope of length d. They are isolated in space, orbiting their center of mass at identical speeds v. By pulling on the rope, one of them shortens the distance between them to d/2. What are the new angular momentum L’ and speed v’?
A) L’ =mvd/2= mvd/2, vv=’=v/2 v/2 B) L’ = mvd, v’ = 2v C) L’ = 2mvd, v’ = v D) L’ = 2mv’d, v’ = v/2 E) L’ = mvd, v’= v/2
L = Iω
L = mv Tr
14 Example: A merry-go-round problem
A 40-kg child running at 4.0 m/s jumps tangentially onto a stationary circular merry-go-round platform whose radius is 2.0 m and whose moment of inertia is 20 kg-m2.
Find the angular velocity of the platform after the child has jumped on.