z

Young Writers Society


Physics (Mechanics)



User avatar
377 Reviews



Gender: Female
Points: 22732
Reviews: 377
Fri Aug 27, 2010 10:18 am
seeminglymeaningless says...



*sad*

I have to complete these questions by my next physics tutorial, due on the date when my physics prac report is due + I'm presenting a reading in an English unit + I have an actual physics exam the following week.

I don't know why I have such problems in solving mechanic questions, but I guess I'm just really lame...

Could anyone help me with these questions? I've even changed the values so I couldn't just copy down the help you guys provide, as I really do want to learn - I just would like to see how people would solve these equations. Maybe there is an easier way than the way I am trying to learn.

Sorry to dump this on you... if you have the time, please help :) I've included the formula I believe is supposed to be with each equation and what I've done, but I could be/am most likely wrong :/

- Jai

Questions

1) A wheel has a constant angular acceleration of 2.5 rad s-2. During a 3 second interval it turns through an angle of 140 radians. Assuming the wheel started from rest, how long had it been in motion at the start of the given time interval?
http://en.wikipedia.org/wiki/Angular_acceleration

w = 2.5
t = 3
theta = 140
u = 0

a = w/t = theta/t^2
= 2.5/3
= 0.833

I'm so confused...

2) A rear-wheel-drive car supports 40% of its weight on its two drive wheels and has a coefficient of static friction of 0.60 with a horizontal straight road. Find the vehicle's maximum acceleration.
http://en.wikipedia.org/wiki/Coefficien ... c_friction

u = 0.6
--> horizontal
a = ?

f = u Fn

Where does the 40% come in??? aaaarsh. Am I retarded?

3) An object attached to a spring executes simple harmonic motion with an amplitude of 20 cm. What percentage of its total energy is potential energy when it is 14 cm from the mean position?
http://en.wikipedia.org/wiki/Simple_harmonic_motion

Amplitude = 0.2m
x = .14m

F = -kx

.... :(

4) A disk with a moment of inertia of 5 kg m2 rotates like a merry-go-round while undergoing a constant torque of 3 N m. At time t = 1 s, it's angular momentum is 2 kg m2 s-1. What is its angular momentum at t = 3.6 s?
http://en.wikipedia.org/wiki/Angular_momentum

I = 5
T = 3
t = 1, w = 2
t = 3.6, w = ?

L = I w

... ?

5) A 3 kg particle starts from rest at x = 0 and moves under the influence of a single force Fx = 5.8 + 4.3x - 3x^2, where Fx is in Newtons and x is in metres. Find the work done by the force as the particle moves from x = 1.7 to x = 3.6.
???

I'm sorry if I appear really idiotic. I really really just don't know what to do, and am too embarrassed to ask when everyone else seems to know what to do already. I know that's stupid, but yeah...

Thanks again, guys :D
I have an approximate knowledge of many things.
  





User avatar
108 Reviews



Gender: Female
Points: 15484
Reviews: 108
Fri Aug 27, 2010 2:34 pm
View Likes
Shepherd says...



I'm going to solve these conceptually if you don't mind. For most of these, there is more than one way to go about solving them, so if you'd rather see something explained in terms of energy/momentum conservation, kinematics, or calculus, let me know and I can switch things around a little. Mostly I'm going to do things the easy way =)

1) Here I think you are neglecting your basic kinematic equations. Don't do that! Kinematics are (often) the most straight-forward way to go about solving a problem that would otherwise involve far too much complex calculus to be imminently solve-able.

First you've got to recognize that this problem has two real sections to it. The initial angular velocity (we'll call it wi), cannot be used in the kinematic equation that we are going to use, because it thus far unconnected to the other information that you've got (which is occurring some time later).

Really, we can use the equation: theta = wi(t) + (1/2)at(^2) where all quantities are presumed to be angular. Now if you solve for wi, you will have the initial speed with which your wheel was rotating at the start of the three-second interval. Although you already had a value for wi, it's important to recognize that THAT wi refers to how quickly the wheel was moving before the interval in question began, and is of little use here.

Now that you have wi, you can input that value into a second kinematic equation:

t = (change in velocity)/a
Paramedic
Writer
Crazy
Nije vas zahvatila druga kušnja osim ljudske. Ta vjeran je Bog: neæe pustiti da budete kušani preko svojih sila, nego æe s kušnjom dati i ishod da možete izdržati.
  





User avatar
108 Reviews



Gender: Female
Points: 15484
Reviews: 108
Fri Aug 27, 2010 2:51 pm
Shepherd says...



2) Hm.

Do you draw force diagrams when you work these? I think you might find them to be extremely helpful when solving this type of problem, because holding all this information in your head can be tricky.

f = qN

We have q and we can figure out N using variables. N, on a horizontal surface with no vertical acceleration, is going to be equal to mg. In this case, because only 40% of the car's weight is resting on its rear wheels, you could even go as far to say that N = (m)(.4)(9.8 ). Make sense? Now we can put all this into the original equation.

f = .6(.4)(9.8 )m

Now you have the frictional force in terms of m.

The maximum acceleration of the car is going to be the point at which the tires are exerting the same amount of force as the frictional force, only in the opposite direction. This is the only way that the car can accelerate without slippage. As such, you can set the equation F = ma equal to the frictional force equation that you just figured out.

ma = .6(.4)(9.8 )m. Mass should cancel out, and you will be left with a maximum acceleration.
Paramedic
Writer
Crazy
Nije vas zahvatila druga kušnja osim ljudske. Ta vjeran je Bog: neæe pustiti da budete kušani preko svojih sila, nego æe s kušnjom dati i ishod da možete izdržati.
  





User avatar
108 Reviews



Gender: Female
Points: 15484
Reviews: 108
Fri Aug 27, 2010 3:00 pm
Shepherd says...



3) Let's start here by writing down what we have.

A = .2
x = .14

It doesn't seem like enough, right? So let's think about what we know about simple harmonic motion. Because energy is conserved during SHM, we know that when x is .2, all of the energy in the spring is going to be potential energy. So maybe we should approach this as an energy conservation problem.

Total energy is going to be kinetic energy added to potential energy. In order to simplify this, we can make one of those values zero by imagining that the spring is at one apex or the other. So let's say x is .2. The potential energy of a spring is determined by the equation that reads (1/2)kx^2. Don't panic yet. =) We don't need the spring constant to solve this problem.

(1/2)k(.2)^2 is going to be the total energy of the system. Now, to find the amount of potential energy when x is .14, we can rewrite the equation like this:

(1/2)k(.14)^2. Divide this by the total value and you should get the correct answer! The k's should cancel out, of course, as they are divided by each other.

(Edited: I'll be back to answer the other questions in a bit if someone doesn't beat me to it! Let me know if you have any questions!)
Paramedic
Writer
Crazy
Nije vas zahvatila druga kušnja osim ljudske. Ta vjeran je Bog: neæe pustiti da budete kušani preko svojih sila, nego æe s kušnjom dati i ishod da možete izdržati.
  





User avatar
377 Reviews



Gender: Female
Points: 22732
Reviews: 377
Sat Aug 28, 2010 12:12 am
seeminglymeaningless says...



Thank you so much for your help, but I don't know if I'm right with my first answer (haven't attempted the other questions yet, will post them up as I do).

theta = wi(t) + .5(a)(t^2)
140 = 0(3) + .5(a)(3^2)
a = 140 x 2 /3^2
a = 31.11 <---------- which I really doubt :P

t = change in velocity / a
t = 2.5 - 0 / 31.11
t = 0.08

---------------

Then I did it another way, if wi didn't start from rest, the way the question said it did

---------------

theta = wi(t) + .5(a)t^2
140 = wi(3) + .5(a)3^2

a = theta/t^2
a = 140/9
a = 15.55

140 = wi(3) + .5(15.55)9
140 - .5(15.55)9 / 3 = wi
wi = 116.675 <------------- which seems like a ridiculously high value

t = change in velocity/a
t = 116.675 - 2.5/15.55
t = 7.366 <----------------- which seems like a reasonable time.

-----------

Are either of those right? Or did I screw it up entirely? :3

- Jai (will do the others when I get back from the markets)
I have an approximate knowledge of many things.
  





User avatar
377 Reviews



Gender: Female
Points: 22732
Reviews: 377
Sat Aug 28, 2010 1:25 am
View Likes
seeminglymeaningless says...



So I just got back from the markets.

Q2)

f = uFn
u = 0.6

Fn = mg
F = ma = umg

ma =.6(.4)9.8m
a = 2.352 m/s^s

Wow, you made that look so easy... Is that about right?

-----------

Q3)

A = 0.2 <------- max
x = 0.14 <------ what it has been stretched to

Ef = KE + PE

PE = 1/2kx^2
Find PE for max extension, so x = .2
= 1/2k(.2^2)
= 0.02k

Find PE for x = 0.14
= 1/2k(.14^2)
= 0.0098

To find how much of PE is used total as a percentage
PE(x=0.14)/Energy Total (PE, x=0.2) x 100
0.0098/0.02 = 0.49 x 100
= 49%

Riiiiiight?

-------------------------

Whee, I hope I've done the right things... Can anyone else help me with the last two?

Thank you so much, Shepherd!
I have an approximate knowledge of many things.
  








The person who has no opinion will seldom be wrong.
— Anonymous