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.
  • #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
 
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  • #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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  • #61
Alt+F4 said:
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

Well, how do you find the vertical component?

~H
 
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  • #62
Stupid question, but i get soo confused when do i know the say acceleration will be negative in this equation like Friction - Tension = ma

Is there key words in the problem or what cause sometimes it is postive some times it is negative
 
  • #63
Hootenanny said:
Well, how do you find the vertical component?

~H
well verticle component is just if there is an angle then you have one
 
  • #64
Alt+F4 said:
well verticle component is just if there is an angle then you have one

What would happen if there was an unbalanced force acting upwards?

HINT: What other force is acting upwards in BOTH cases.

~H
 
  • #65
Hootenanny said:
What would happen if there was an unbalanced force acting upwards?

HINT: What other force is acting upwards in BOTH cases.

~H
Normal Force is acting upward
 
  • #66
Alt+F4 said:
Normal Force is acting upward

Yep, but what would happen if there was a net force acting upwards on a block?

~H
 
  • #67
Hootenanny said:
Yep, but what would happen if there was a net force acting upwards on a block?

~H
then it would equal
 
  • #68
Alt+F4 said:
then it would equal

What? Using Newton's second law, what would happen if there was a net upwards force?

~H
 
  • #69
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/fa03

Question 8, why couldn't it not have been done like this 1.4/9.8 = .14
 
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  • #70
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/fa03

Question 8, why couldn't it not have been done like this 1.4/9.8 = .14

The co-efficent of friction is given by;

\mu = \frac{F}{mg}

Where F is the maximum frictional force. You have to take into account the force that the tractor is pulling.

~H
 
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  • #71
Hootenanny said:
The co-efficent of friction is given by;

\mu = \frac{F}{mg}

Where F is the maximum frictional force. You have to take into account the force that the tractor is pulling.

~H
um so what i did was add all the weights multiply by 9.8 is that even right and then i divided by the tractor weight or am i missing something
 
  • #72
Alt+F4 said:
um so what i did was add all the weights multiply by 9.8 is that even right and then i divided by the tractor weight or am i missing something

No, I'm afraid it isn't right. The reaction force (mg) is due to the tractor's weight only. The F is the force required to accelerate the total mass at 1.4 m.s-2. Do you follow?

~H
 
  • #73
Hootenanny said:
No, I'm afraid it isn't right. The reaction force (mg) is due to the tractor's weight only. The F is the force required to accelerate the total mass at 1.4 m.s-2. Do you follow?

~H
o yes, thanks
 
  • #74
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp03

Question 23

I did Centripital Acceleration * weight

50 * .5 = 25 is this the right way?
 
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  • #75
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp03

Question 23

I did Centripital Acceleration * weight

50 * .5 = 25 is this the right way?
That's fine. (I assume you meant to write mass, not weight.)
 
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  • #76
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/fa05 Last question, i still don't get how to set it up incase he asks the same question and wants the resultant velocity

so

(1.5*10^5)(8) = (-4*1.5*10^5) + (1.5*10^6*X)

Is this the setup
 
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  • #77
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/fa05 Last question, i still don't get how to set it up incase he asks the same question and wants the resultant velocity

so

(1.5*10^5)(8) = (-4*1.5*10^5) + (1.5*10^6*X)

Is this the setup
That's how I'd do it.
 
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  • #78
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp05

Question 6

So what i was thinkin is to do it this way


so work done by friction = W = *.43*9.8*12 = 50.568

Total Kinetic Energy = MGH + .5 MV^2 = (9.8)(9) + .5 (20^2) = 288.2

288.2-50.568 = 237.632

Edit: well it works if i don't take in effect the PE of the hill but why not?
 
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  • #79
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp05

Question 6

So what i was thinkin is to do it this way


so work done by friction = W = *.43*9.8*12 = 50.568

Total Kinetic Energy = MGH + .5 MV^2 = (9.8)(9) + .5 (20^2) = 288.2

288.2-50.568 = 237.632

Edit: well it works if i don't take in effect the PE of the hill but why not?
Why should the 9 m hill matter? Since there's no friction, the KE is the same before and after climbing that hill.
 
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  • #80
http://online.physics.uiuc.edu/cgi/c...ice/exam1/sp99 Question 25

here was my plan divide the thing into 2 so 55 M on left, 55 M on right

do Tan 35 = X / 55

The Height = 38.5114

Okay then i decided to just do

38.5114 = .5 * 9.8 * T^2
T = 2.80

Okay another way

(2)(9.8)* 38.5114 = V^2

V = 27.47405

so what i did was divide that be Cos 35 and got 33.53 why am i off?
 
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  • #81
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/fa04

Question 16

So i know that Change in kinetic = Uk g * D

So what i did was (22)(9.8) + (9)(9.8) = X * 9.8 *53

wat am i doing wrong
 
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  • #82
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/fa04

Question 16

So i know that Change in kinetic = Uk g * D

So what i did was (22)(9.8) + (9)(9.8) = X * 9.8 *53

Your problem is this equation. Why have you added additional GPE [+ (9)(9.8)]?

~H
 
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  • #83
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/fa04

Question 19

So Wnet = 8.6483
Wnet - Force of friction - Force of gravity
I did get the above questions right so i know the numbers are right


8.6483 = .5 mV^2

V = 2.9773


The 3 is the ending velocity
(-2)(2.9773) - (2)(3) = -11.9546 ( Total Momemtum)

Fave T = P

X (.06) = (11.9546)

X = 200

Is this how it is done
 
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  • #84
WHen it says you Double The Frequency does it Actually mean Multiply it by 2 or does it mean Half it?
 
  • #85
Alt+F4 said:
WHen it says you Double The Frequency does it Actually mean Multiply it by 2 or does it mean Half it?

In what context? What is the full question?

~H
 
  • #86
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/fa04

Question 19

So Wnet = 8.6483
Wnet - Force of friction - Force of gravity
I did get the above questions right so i know the numbers are right


8.6483 = .5 mV^2

V = 2.9773


The 3 is the ending velocity
(-2)(2.9773) - (2)(3) = -11.9546 ( Total Momemtum)

Fave T = P

X (.06) = (11.9546)

X = 200

Is this how it is done

Why have you gone through the trouble of calculating the velocity just before impact when it is given in the question? :wink: But yes you working is correct. The change in momentum will be -6m and you were correct to use the impulse relationship.

~H
 
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  • #87
Suppose the mass of the string in the above problem is M0. What is the mass of a new string (in terms of M0 ), if the fundamental frequency of the string is doubled compared to the previous value, but the length and tension in the string are the same as before?

Now does this Mean Frequncy * 2 or frequency / 2
 
  • #88
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/c...ice/exam1/sp99 Question 25

here was my plan divide the thing into 2 so 55 M on left, 55 M on right

do Tan 35 = X / 55

The Height = 38.5114

Okay then i decided to just do

38.5114 = .5 * 9.8 * T^2
T = 2.80

Okay another way

(2)(9.8)* 38.5114 = V^2

V = 27.47405

so what i did was divide that be Cos 35 and got 33.53 why am i off?
let me just quote this so it doesn't get lost
 
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  • #89
Alt+F4 said:
Suppose the mass of the string in the above problem is M0. What is the mass of a new string (in terms of M0 ), if the fundamental frequency of the string is doubled compared to the previous value, but the length and tension in the string are the same as before?

Now does this Mean Frequncy * 2 or frequency / 2

It means the fundemetal fequency is twice what is was before. I.e. If intially the fundamental frequency was 50Hz, the new fundamental frequency would be 100Hz.

~H
 
  • #90
Alt+F4 said:
let me just quote this so it doesn't get lost

I answered this above in post #86 :smile:

~H
 
  • #91
Hootenanny said:
I answered this above in post #86 :smile:

~H
that is a different problem
 
  • #92
Alt+F4 said:
that is a different problem

Ahh, so it is. Perhaps it's time for me to take a break and get my fix of caffine :zzz:

~H
 
  • #93
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/c...ice/exam1/sp99 Question 25

here was my plan divide the thing into 2 so 55 M on left, 55 M on right

do Tan 35 = X / 55

The Height = 38.5114

Okay then i decided to just do

38.5114 = .5 * 9.8 * T^2
T = 2.80

Okay another way

(2)(9.8)* 38.5114 = V^2

V = 27.47405

so what i did was divide that be Cos 35 and got 33.53 why am i off?
That link isn't working.
 
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  • #94
sorry. http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp99
 
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  • #95
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/c...ice/exam1/sp99 Question 25

here was my plan divide the thing into 2 so 55 M on left, 55 M on right

do Tan 35 = X / 55

The Height = 38.5114
Huh? The motion is parabolic, not straight line.

Okay then i decided to just do

38.5114 = .5 * 9.8 * T^2
T = 2.80

Okay another way

(2)(9.8)* 38.5114 = V^2

V = 27.47405

so what i did was divide that be Cos 35 and got 33.53 why am i off?
(http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp99; #25)

Treat x and y components separately and set up two equations for position as a function of time. Solve them together to find the initial speed.
 
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  • #96
Doc Al said:
Huh? The motion is parabolic, not straight line.


(http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp99; #25)

Treat x and y components separately and set up two equations for position as a function of time. Solve them together to find the initial speed.
so what are you pointing too,

I know X componet is

V Cos 35 t


Y componet

V sin T - .5*9.8 * t^2
 
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  • #97
Alt+F4 said:
so what are you pointing too,

I know X componet is

V Cos 35 t


Y componet

V sin T - .5*9.8 * t^2
That's right. You know the coordinates of the landing point at time t.
 
  • #98
so?

V Cos 35 t =110

110/ V cos 35 = T


and then substitute T in that V sin T equation?

and then the distance should equal 0 right/>
 
  • #99
Alt+F4 said:
so?

V Cos 35 t =110

110/ V cos 35 = T


and then substitute T in that V sin T equation?
That will work.

and then the distance should equal 0 right?
When the ball lands, its coordinates are x = 110 m; y = 0.
 
  • #100
A satellite is in circular orbit at a fixed radius from the center of the Earth and with a constant speed. Which one of the following statements is correct about the satellite?

(a) The acceleration is constant but the velocity is not.
(b) Both the acceleration and the velocity are constant.
(c) Neither the acceleration nor the velocity are constant.


Ans:C


WHy so? Constant Speed = Constant Velocity
 

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