Final Exam Q6: Finding Problem with Torque Equation

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The discussion focuses on solving physics problems related to torque, average velocity, and motion equations. Key points include the importance of understanding the angle in torque calculations, specifically the angle between the radius vector and the force vector. Participants express confusion over average velocity calculations, emphasizing the need to consider displacement over time rather than simply averaging speeds. The conversation also touches on using kinematic equations to solve for time and velocity in various scenarios, highlighting the significance of horizontal and vertical components in motion. Overall, the thread serves as a collaborative effort to clarify complex physics concepts and problem-solving strategies.
  • #31
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/fa05

Question 21

So i drew out free body diarm, i got the tension for previous which was 24N
OK.

-T - mgsin30 + UkMg = ma

-24 - (X)(9.8)sin 30 + (.2)(X)(9.8) = (X)(-5)

I end up getting 11.65
Careful with your signs: The tension and acceleration point up the ramp; friction and weight point down.
 
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  • #32
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/fa05

Question 26

So what i did was

(1.5 * 10^5)(-4) = (1.5*10^6)(X) but that don't make sense

I know since it bounces off it will be negative
Think about it qualitatively, don't just reach for a formula. You know that the total momentum (which is a vector) will be the same for any kind of collision. But which collision gives more momentum to the iceberg? Hint: How does the momentum of the ship (post collision) differ in each case?
 
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  • #33
Doc Al said:
Think about it qualitatively, don't just reach for a formula. You know that the total momentum (which is a vector) will be the same for any kind of collision. But which collision gives more momentum to the iceberg? Hint: How does the momentum of the ship (post collision) differ in each case?
ya i understood it was gona be less than but i wana know in case he asks to figure out its Velocity
 
  • #34
For question 26:

To solve for the speed of the iceberg, set the initial momentum of the system (originally, just the ship is moving) equal to the final momentum of the system (both ship and iceberg). The only unknown will be the speed of the iceberg.
 
  • #35
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp05

Question 14 , i know i am doing it right but i can't get the answer

so all it is is

Force of Friction - F Cos 64 = 0

(.1)(9.80(7) - F (cos 64)

F = 15.648
 
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  • #36
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp05 Question 17 , um you so 45/270 = Omega = .16667

How do i go about knowing the lap time? DO i find the Area? and would that be The change in X
 
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  • #37
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp05

Question 14 , i know i am doing it right but i can't get the answer

so all it is is

Force of Friction - F Cos 64 = 0
This is correct so far.

(.1)(9.80(7) - F (cos 64)
This is not. Hint: Did you answer Question 13?
 
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  • #38
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp05 Question 17 , um you so 45/270 = Omega = .16667

How do i go about knowing the lap time? DO i find the Area? and would that be The change in X
Snap out of it! :smile: Try using Distance = speed * time.
 
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  • #39
Doc Al said:
Snap out of it! :smile: Try using Distance = speed * time.
what is Distance? O man i don't even know anymore
 
  • #40
last 2 of the day, then I am hitting the bed since I am losing it

http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp05 Question 22 and 23


So i know that at Point H it takes .75 Sec to get there but how do i figure the time it takes to land back at 4m. and i guess if i get that then 23 will help itself
 
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  • #41
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp06

QUestion 17,

So

-Fpush Cos 30 - Force of Friction = 0

Fpull Cos 30 - Force Of Friction = 0

Fpull should be larger but it is not, is my math wrong?
 
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  • #42
The sandbag has vertical speed zero at the top (H) .

how far above the thrower is this?

method a) v_o t + 1/2 a t^2 ?

method b) average vertical speed , for how long upward?
 
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  • #43
F_push is not equal to F_pull .

You'll need to look at the CAUSE of the friction Force to determine that
(not just the observable effect of the TOTAL Force).
 
  • #44
Doc Al said:
This is correct so far.


This is not. Hint: Did you answer Question 13?
yes i answerd question 13 since it is F sin Theta - mg + normal = 0

Normal = mg - F sin theta which becomes Normal is smaller than mg

Edit: i answerd my own question
 
  • #45
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp05 Question 17 , um you so 45/270 = Omega = .16667

How do i go about knowing the lap time? DO i find the Area? and would that be The change in X
So what would be the circular Distance for this bad boy? It is a circle so 360 degrees am i thinking too much?
 
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  • #46
degrees are usually used to measure ANGLES, not DISTANCES.
hint : see what units the velocity (or acceleration) is measured in

Are you asking how far it is around a circle?
The distance around a hexagon (equilateral triangles of length R) is 6R ...
 
  • #47
Alt+F4 said:
So what would be the circular Distance for this bad boy? It is a circle so 360 degrees am i thinking too much?
okay well this is what i did, i don't even know what it means but

A= pi R^2 = 229022

229022/ (45 * 270/2))
 
  • #48
lightgrav said:
degrees are usually used to measure ANGLES, not DISTANCES.
hint : see what units the velocity (or acceleration) is measured in

Are you asking how far it is around a circle?
The distance around a hexagon (equilateral triangles of length R) is 6R ...
well omega is 45/270 = .1666

I need to find the time, the answer is 37.7
 
  • #49
i got it, i forgot there was such a thing as circumference
 
  • #50
Did you get the box push vs pull when you realized the answer to prob.13?
 
  • #51
lightgrav said:
Did you get the box push vs pull when you realized the answer to prob.13?
well i understood it from the beginning, i just wanted to know if there was an equation in case it is asked on the final to find one of them
 
  • #52
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp05 I still don't get how to do 21 and 23, so i got .7536 seconds to reach at its highest

Actually i found the answer to 22 which was 4.3 4.0 - 4.3 = .3 meters left to fall

. 3 = .5 9.8 T^2

T = .2474 + .7536 = 1.001

Now 23
 
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  • #53
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp05 I still don't get how to do 21 and 23, so i got .7536 seconds to reach at its highest

Actually i found the answer to 22 which was 4.3 4.0 - 4.3 = .3 meters left to fall

. 3 = .5 9.8 T^2

T = .2474 + .7536 = 1.001

Now 23
For 23, you have two choices:

you can do it with the equations of kinematics (v_x stays constant. For v_y you may use v_y(t) = v_{y,initial} - g t. Then the final speed is given by Pythagora's theorem.

OR you could use conservation of energy
 
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  • #54
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp05 I still don't get how to do 21 and 23, so i got .7536 seconds to reach at its highest

Actually i found the answer to 22 which was 4.3 4.0 - 4.3 = .3 meters left to fall

. 3 = .5 9.8 T^2

T = .2474 + .7536 = 1.001

Now 23
What's the problem with 23? To find speed at some point, use kinematics or energy conservation. (Horizontal component of velocity remains constant.)
 
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  • #55
nrqed said:
For 23, you have two choices:

you can do it with the equations of kinematics (v_x stays constant. For v_y you may use v_y(t) = v_{y,initial} - g t. Then the final speed is given by Pythagora's theorem.

OR you could use conservation of energy
so V y intitial = 7.5 Cos 80 = 1.30236
so v (t) = 1.30236 - (9.8)(.2474)
V X intial = 7.5 sin 80 = 7.386

Then what? i am not getting how to use the forumla
 
  • #56
Doc Al said:
What's the problem with 23? To find speed at some point, use kinematics or energy conservation. (Horizontal component of velocity remains constant.)
so horizontal componet is 7.5 cos 80 = 1.3
 
  • #57
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp05

Question 25

So what i did was 4 cos 22 = 3.7087 M/s

Time * Velocity = Distance

(120)(3.7087) = 445.04

Ans: 450, did they just round?
 
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  • #58
Alt+F4 said:
so V y intitial = 7.5 Cos 80 = 1.30236
so v (t) = 1.30236 - (9.8)(.2474)
V X intial = 7.5 sin 80 = 7.386

Then what? i am not getting how to use the forumla
You must use sine for the y component and cos for the x component (because of the way the chose their angle).

In the end you are getting the x and y components of the final velocity vector. To find the final speed, just calculate the magnitude of the final velocity vector (so use v_f = {\sqrt{v_{x,f}^2+v_{y,f}^2}})
 
  • #59
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp04

Question 9

Well i found the weight of the other mass which is also 2.5, so all i did was add 15 N + 15N = 30 which is the answer is this the correct way or did i luck out
 
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  • #60
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp04 Question 11, why are they equal there is a verticle component for the second box which is mgsin 45 while for the first it is 0
 
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