Homework Parallel-Axis Theorem And Torque Specifications

 

4/24/2014Chapter 10 Homeworkhttp://session.masteringphysics.com/myct/assignmentPrintView?assignmentID=28458931/32

Chapter 10 Homework

Due: 10:00pm on Wednesday, April 16, 2014

You will receive no credit for items you complete after the assignment is due. Grading Policy

Torque about the z Axis

Learning Goal:

To understand two different techniques for computing the torque on an object due to an applied force.Imagine an object with a pivot point p at the origin of the coordinate system shown . The force vector lies in the

 xy 

plane, and this force of magnitude acts on the object at apoint in the

 xy 

plane. The vector is the position vectorrelative to the pivot point p to the point where is applied.The torque on the object due to the force is equal to thecross product . When, as in this problem, theforce vector and lever arm both lie in the

 xy 

 plane of the paperor computer screen, only the

 z

 component of torque isnonzero.When the torque vector is parallel to the

 z

 axis (), it iseasiest to find the magnitude and sign of the torque, , interms of the angle between the position and force vectorsusing one of two simple methods: the

Tangential Component of the Force

 method or the

Moment Arm of the Force

 method.Note that in this problem, the positive

 z

 direction isperpendicular to the computer screen and points toward you(given by the right-hand rule ), so a positive torque would cause counterclockwise rotation about the

 z

 axis.

Tangential component of the force

Part A

Decompose the force vector into radial (i.e., parallel to ) and tangential (perpendicular to ) components asshown. Find the magnitude of the radial and tangentialcomponents, and . You may assume that isbetween zero and 90 degrees.

Enter your answer as an ordered pair. Express and in terms of and .

Hint 1.

Magnitude of

F     

   ⃗

F     󰁲   

   ⃗

F     

   ⃗

F     

   ⃗=×    

τ   

   ⃗

󰁲   

   ⃗

F     

   ⃗=    

τ   τ   

   ⃗

󰁫   

^  

τ   θ   

×=    

󰁩  

^  

 󰁪   

^  

󰁫   

^  

F     

   ⃗

󰁲   

   ⃗

󰁲   

   ⃗

F     

r  

F     

t  

θ   F     

t  

F     

r  

Fθ   F     

   ⃗

r  

   ⃗⃗

r  

Presentation on theme: "a) Parallel Axis Theorem"— Presentation transcript:

1 a) Parallel Axis Theorem
Physics 211 Lecture 15Today’s Concepts:a) Parallel Axis Theoremb) Torque & Angular Acceleration

2 Your Commentsevery thing was new to me, but not difficult to understandHow many times can Superman run into a wall before he gets a headache?As long as I went through each step, I arrived at a sensible answer. This is all easy to do because you just have to walk through the steps. The one thing I am confused about is direction. So Torque will point along the axis in either direction. I am just confused on how I can see the apply in real life. Like does a force pointing along the axis of rotation do anything?I bet you're going to say this can all be figured out with F = ma huh?My wish for "To Make a Wish Foundation" was to get this put onto today's lecture... and it came true...I discovered that approaching the prelectures with the same zeal as the homework makes me understand things a lot better. Now, onto serious business. You don't have Maiden, which is outrageous, but what about Rush? There is no excuse for not having Rush. Actually, it is a sin. I am expecting Subdivisions on Thursday.Can i get a SHOUT OUT?! so i relaized something tonight...F=MA is that awkward friend that changes his personality depending on who he wants to impress. His been in 1/2mv^2 and now his in TORQUE?!!!liking this, seems cpmlicated but once you think about it its okay. Gotta love the prelectures and garrryyy! cheers, me2

3 Proven Fact: If you don’t prepare for lecture you don’t learn as much…
You're not supposed to understand everything in the prelecture – that’s the whole point of having a lecture.

4 Parallel Axis TheoremSmallest when D = 0

5 Clicker QuestionA solid ball of mass M and radius is connected to a thin rod of mass m and length L as shown. What is the moment of inertia of this system about an axis perpendicular to the other end of the rod?A)B)C)D)RmMLaxis

6 CheckPointA ball of mass 3M at x = 0 is connected to a ball of mass M at x = L by a massless rod. Consider the three rotation axes A, B, and C as shown, all parallel to the y axis.For which rotation axis is the moment of inertia of the object smallest? (It may help you to figure out where the center of mass of the object is.)B3MMCAL/2L/4xyLRotation is smallest around the objects center of mass, and the center of mass for this system is at L/4.

7 Right Hand Rule for finding Directions
I learned that the right hand rule was for B fields about wires and B fields with E fields crossing them. Now you tell me there is another right hand rule. Seriously physics, be more creative. I think that with all the smart people who study you could at least give your rules more diverse names.

8 Clicker QuestionA ball rolls across the floor, and then starts up a ramp as shown below. In what direction does the angular velocity vector point when the ball is rolling up the ramp?A) Into the pageB) Out of the pageC) UpD) Down

9 Clicker QuestionA ball rolls across the floor, and then starts up a ramp as shown below. In what direction does the angular acceleration vector point when the ball is rolling up the ramp?A) Into the pageB) Out of the page

10 Clicker QuestionA ball rolls across the floor, and then starts up a ramp as shown below. In what direction does the angular acceleration vector point when the ball is rolling back down the ramp?A) into the pageB) out of the page

11 TorqueI quite don't understand the concept of torque. So generally, if Torque is greater, what will happen other than the door will closing faster?t = rF sin(q )

12 Torque and F=maWhy in the new equation for torque analogous to Newtons's Second Law deals with angular acceleration only? Are these concepts separate?vqrF

13 Look at Prelecture slide 9 again
It’s all there….

14 =r(Fsinq) =F(rsinq) I know torque = rxF but what exactly is r? q q q q

15 CheckPointIn Case 1, a force F is pushing perpendicular on an object a distance L/2 from the rotation axis. In Case 2 the same force is pushing at an angle of 30 degrees a distance L from the axis.In which case is the torque due to the force about the rotation axis biggest?A) Case 1 B) Case 2 C) SameFL/290oCase 1axisLF30oCase 2axis

16 CheckPointIn which case is the torque due to the force about the rotation axis biggest?A) Case 1 B) Case 2 C) SameFL/290oCase 1axisA) Pushing perpendicular to the door is always best..B) the length of the rotation arm is longer in case 2..LF30oCase 2axisC) FL/2=FLsin30

17 Similarity to 1D motion

18

19 Clicker QuestionStrings are wrapped around the circumference of two solid disks and pulled with identical forces. Disk 1 has a bigger radius, but both have the same moment of inertia.Which disk has the biggest angular acceleration?w2w1A) Disk 1B) Disk 2C) sameFF

20 CheckPointTwo hoops can rotate freely about fixed axles through their centers. The hoops have the same mass, but one has twice the radius of the other. Forces F1 and F2 are applied as shown.How are the magnitudes of the two forces related if the angular acceleration of the two hoops is the same?F1F2Case 1Case 2A) F2 = F1B) F2 = 2F1C) F2 = 4F1

21 CheckPointHow are the magnitudes of the two forces related if the angular acceleration of the two hoops is the same?M, RM, 2RF1F2Case 1Case 2A) F2 = F1B) F2 = 2F1C) F2 = 4F1

22 90-qq

23 Direction is perpendicular to both R and F, given by the right hand rule

24 (i)(ii)(iii)Use (i) & (ii)Use (iii)

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