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Physics Exam 3: Energy, Angular Momentum, and Collisions, Exams of Classical Mechanics

Jaypee University of Engineering & TechnologyClassical Mechanics

Solutions for exam 3 of physics 218, which covers topics such as energy, angular momentum, and collisions. It includes problem-solving steps for various questions related to these topics.

Typology: Exams

2012/2013

Uploaded on 03/06/2013

gannesh
gannesh 🇮🇳

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1!
Physics 218 Exam 3
Fall$2010,$Sections$5212524$
$
Do$not$fill$out$the$information$below$until$
instructed$to$do$so!$
$
Name$:_________SOLUTIONS_______!
Signature$ :______________________________!
Student$ID$:______________________________!
E2mail$:______________________________!
Section$#$:______________________________$
!
Rules$of$the$exam:$
1. You!have!the!full!class!period!to!complete!the!exam.!
2. When!calculating!numerical!values,!be!sure!to!keep!track!of!units.!!
3. You!may!use!this!exam!or!come!up!front!for!scratch!paper.!!
4. Be!sure!to!put$a$box$around$your$final$answers!and!clearly!indicate!your!work!to!your!grader.!
5. Clearly!erase!any!unwanted!marks.!No!credit!will!be!given!if!we!can’t!figure!out!which!answer!you!are!
choosing,!or!which!answer!you!want!us!to!consider.!
6. Partial!credit!can!be!given!only!if!your!work!is!clearly!explained!and!labeled.!
7. !All$work$must$be$shown$to$get$credit$for$the$answer$marked.$If$the$answer$marked$does$not$
obviously$follow$from$the$shown$work,$even$if$the$answer$is$correct,$you$will$not$get$credit$for$
the$answer.!
Put!your!initials!here!after!reading!the!above!instructions:__________!
!
!
!
!
!
!
!
!
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pf4
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Download Physics Exam 3: Energy, Angular Momentum, and Collisions and more Exams Classical Mechanics in PDF only on Docsity!

Physics 218 Exam 3

Fall 2010, Sections 521 524

Do not fill out the information below until

instructed to do so!

Name :_________SOLUTIONS_______ Signature :______________________________ Student ID :______________________________ E mail :______________________________ Section # :______________________________ Rules of the exam:

  1. You have the full class period to complete the exam.
  2. When calculating numerical values, be sure to keep track of units.
  3. You may use this exam or come up front for scratch paper.
  4. Be sure to put a box around your final answers and clearly indicate your work to your grader.
  5. Clearly erase any unwanted marks. No credit will be given if we can’t figure out which answer you are choosing, or which answer you want us to consider.
  6. Partial credit can be given only if your work is clearly explained and labeled.
  7. All work must be shown to get credit for the answer marked. If the answer marked does not obviously follow from the shown work, even if the answer is correct, you will not get credit for the answer. Put your initials here after reading the above instructions:__________

Table to be filled by the graders Part Score Part 1 ( 25 ) Part 2 ( 25 ) Part 3 (25) Part 4 (25) Bonus (5) Exam Total

Part 2: (25p) Energy and Kinematics Question 2.1.1: A disk of mass m and radius R is spinning with angular velocity ω. The disk is spinning and slipping with respect to the surface of the ground. Initially the disk is just spinning but not moving in the horizontal direction. There is friction between the disk and the table with kinetic coefficient μ K. Question 2.1.2: (5p) Draw a coordinate system, the orientation in which the disk rotates, and a free body diagram of the disk. Question 2.1.3: (10p) Find the angular acceleration of the disk. Is it constant ?, is it parallel or antiparallel to the angular velocity? If you write a torque specify the point from where the torque is computed. € From the center of the disk the torque is τ = + Rf = Id α ⇒ α = Rf Id = R μ K N Id = R μ K mg Id = 2 R μ K mg mR^2 = 2 μ K g R which is clearly constant as the all the components are constant. In the coordinate system depicted above the angular velocity is negative, while the angular acceleration found is positive, so the angular acceleration is antiparallel to the angular velocity. As expected as the object is deaccelerating. Question 2.1.4: ( 10 p) After an N number of revolutions the disk is found rotating without slipping. Using the work-­‐energy theorem find the angular velocity at which the disk is now rotating. Assume the distance travelled in the horizontal direction is zero. € The work - energy theorem states that : Ki + Ri + Wnon − cons = K (^) f + Rf In our case we have Ki = 0 , Ri = 1 2 I ω 2 and Rf = 1 2 I ω (^) f^2 , K (^) f = 1 2 mv (^) f^2 = 1 2 mR^2 ω (^2) f After N revolutions the work of the torque is Wnon − cons = − τ 2 π N = − R μ K mg 2 π N where the minus sign comes from the fact that the torque was going against the angle displacement. 1 2 I ω 2 − R μ K mg 2 π N = 1 2 I ω (^) f^2 + 1 2 mR^2 ω (^2) f^ ⇒ I ω 2 − 2 R μ K mg 2 π N = ω (^) f^2 ( I + mR^2 ) ⇒ ω (^) f = I ω 2 − 4 R μ K mg π N ( I + mR^2 ) = mR^2 ω 2 2 −^4 R^ μ K^ mg^ π N ( mR 2 2 +^ mR (^2) ) = R ω 2 − 8 μ K g π N 3 R = g R m Ground R

Y

Z X

f

N

w

Part 3: (25p) Angular Momentum Problem 3.1: A tandem rotor helicopter has two sets of blades powered by the same motor that rotate in opposite direction with angular velocity of magnitude ω b. The moment of inertia of each set of helicopter blades with respect to each set’s center is Ib and the moment of inertia of the helicopter around its center of mass is Ih ,. The helicopter is initially hovering at high altitude without moving in the horizontal or vertical position and without rotation of the body of the helicopter. Ignore any air resistance. Question 3.1.1: (5p) Find the total angular momentum of the helicopter, and show how you reached that number. The total angular momentum is € L = Ib ω (^) b − Ib ω (^) b = 0 as the body is not rotating and the blades are rotating in opposite direction. Question 3.1.2: (10p) If a malfunction suddenly stops one set of blades to a screeching halt, find the angular velocity the body of the helicopter develops around its center of mass. € The angular momentum has to be conserved as there are no external forces. Li = 0 = Lf = Ib ω (^) b + Ih ω (^) h ⇒ ω (^) h = − Ib ω (^) b Ih Question 3.1.3: (10p) To stop the spinning, the helicopter posses a jet engine that can provide a constant torque τjet on the body of the helicopter and with respect to the center of mass in the direction of the angular velocity of the remaining set. Find the time it takes to bring the broken helicopter to the status of no-­‐body rotation. € Basically find the time the torque τjet reduces the angular velocity from ω h to zero. The torque τ (^) jet produces a angular acceleration of α = τjet Ih ω (^) f = 0 = ω h + α t ⇒ t = − ω h α

Ib ω (^) b Ih Ih τ (^) jet

Ib ω b τ (^) jet Ib Ib Ih CM