Final Exam Q6: Finding Problem with Torque Equation

[Alt+F4]

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

 

[Doc Al]

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?

 

[Doc Al]

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! Try using Distance = speed * time.

 

[Alt+F4]

Snap out of it! Try using Distance = speed * time.

what is Distance? O man i don't even know anymore

 

[Alt+F4]

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

 

[Alt+F4]

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?

 

[lightgrav]

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?

 

[lightgrav]

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).

 

[Alt+F4]

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

 

[Alt+F4]

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?

 

[lightgrav]

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 ...

 

[Alt+F4]

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))

 

[Alt+F4]

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

 

[Alt+F4]

i got it, i forgot there was such a thing as circumference

 

[lightgrav]

Did you get the box push vs pull when you realized the answer to prob.13?

 

[Alt+F4]

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

 

[Alt+F4]

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

 

[nrqed]

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

 

[Doc Al]

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.)

 

[Alt+F4]

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

 

[Alt+F4]

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

 

[Alt+F4]

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?

 

[nrqed]

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 $v_f = {\sqrt{v_{x,f}^2+v_{y,f}^2}}$)

 

[Alt+F4]

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

 

[Alt+F4]

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

 

[Hootenanny]

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

 

[Alt+F4]

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

 

[Alt+F4]

Well, how do you find the vertical component?

~H

well verticle component is just if there is an angle then you have one

 

[Hootenanny]

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

 

[Alt+F4]

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

 

[Hootenanny]

Normal Force is acting upward

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

~H

 

[Alt+F4]

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

~H

then it would equal

 

[Hootenanny]

then it would equal

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

~H

 

[Alt+F4]

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

 

[Hootenanny]

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