eco's physics help

[ecoli]

My 1st physics test in coming up, so I'll probably have a few questions... so might as well make one thread for it...

 

 

OK, first problem should be relatively simply, but I keep hitting dead ends.

 

A race car constantly accelerating travels a 140 m distance in 3.6 seconds. If the final velocity is 53 m/s, a) what was the initial velocity of the interval. b) how far has the car traveled from rest.

 

 

Now, it looks simple enough, but without a given value of acceleration I seem to be stuck. I try to solve for velocity in terms of a, and plug it in to a different equation to solve for a. But, it's not going in nicely, and I'm not getting an answer. I don't see how you can get the initial velocity without knowing the acceleration of the body.

[Ducky Havok]

I think the key here is that the car has a constant acceleration. From the info given, you can find the average velocity (140/3.6=38.89) and from that and the final velocity, you can get the initial velocity (53-38.89=14.11). Also, you can divide the average velocity by the time to get the acceleration (since it's a constant). Hope that helped and good luck on your test.

[ecoli]

I don't think that's right... the textbook gives the initial velocity as 24.8 m/s.

 

Because there's acceleration, isn't the average velocity meaningless?

[timo]

For a):

T = 3.6 s

v(T) = v(0) + a*T = 53 m/s

s(T) = 0 m + v(0)*T + a*T²/2 = 140 m

 

How many unknowns, how many equations? => No problem solving that.

 

EDIT: To clearify: t=0 s means the start of the 140 meter distance crossed, not the point where the car was at rest (which is asked in b).

 

I think the key here is that the car has a constant acceleration. From the info given, you can find the average velocity (140/3.6=38.89) ...

Yes

...and from that and the final velocity, you can get the initial velocity ...

Yes

(53-38.89=14.11)

No. I suppose we agree that maximum velocity, minimum (starting at the beginning of the 140 m) and average velocity are related by [math] v_{avg} = \frac {v_{max} + v_{min}}{2} [/math]. The correct calculation follows directly from that.

 

The idea is nice, but it's beaten by one of the most important rules in physics: Never try being smart! If there is a dumb way of solving a problem (brainless writing down of the relevant equations and solving for the unknowns, here), then by all means take that way. Being clever only means opening paths to a lot of more or less artistic ways to shoot yourself in the foot.

[ecoli]

to clarify...

 

does V(0) mean initial velocity or V times t when t=0?

 

edit: never mind, I see how to get it now. Thanks for the help...

 

don't worry, I'll be posting here again ;-)

[timo]

It means v(t) for the case t=0 s, i.e. the velocity at time t=0 s. I have the nasty habit not to add units to zeroes (because the zeroes remain zeroes under scaling) - the correct expression would be v(0 s) or, if you prefer that, [math]v_0[/math]. You should really know the two equations I posted, especially since they are the only two relevant for the typical "constant acceleration"-problems (the only two I can currently think of, at least).

[ecoli]

yes, I know those, just you notation confused me a bit... but I cleared it up, thanks.

 

Ok, I'm having a problem with acceleration vectors...

 

And asteroid is heading towards earth at 15km/s. An international team manages to attach a giant rocket to the asteroid. The rocket fires for 10 minutes, after which the asteroid is moving at 28 degrees to it's original path at 19 km/s... find ave. acceleration.

 

The answer is given as 3.0i +15j m/s^2.

 

I keep trying to find the answer by multiplying velocity by cos and sin of 28, respectively to find the i and j values... but it's not working out. I think I may be approaching this problem incorrectly.

[MrSandman]

Try using algebra. I think it might work. I'm not sure. Try using what you have put it in the formula. Then see if you have anything left. i'll try doing it later, but I have to go.

[ecoli]

you're dealing with components, however, so I'm pretty sure you need trig.

[Klaynos]

you need to firstly draw a diagram, and then split everything into two compantants:

 

x and y

 

You'll do this by using trig as you say, so you'll have an acceleration in the x direction, and one in the y direction...

[ecoli]

This question was designed to test the tought process, rather then math skills.

 

 

A plane flies a distance of 6,000 km round trip from city A to city B (so that's 3,000 km one way). City B lies west of city A. In this trip, the plane flies at constant velocit both ways, and the trip takes 10 hours.

 

On a different trip (same plane, same cities) there is wind blowing west to east, and this time the pilot takes 12 hours to fly the round trip. What is the speed of the wind?

 

MY answer:

 

First measure the average velocity for the first trip, 166.7 m/s. This is the average velocity, but because the velocity it constant, we know that this was the velocity for any given point on this trip.

 

The average velocity in the second scenario is 138.9 m/s. The trip takes longer. Obviously, since the wind is travelling from west to east, the trip back, from B to A is going to take longer. But since the wind speed does not change, this sort of "cancels out" from the trip there, which is exactly the opposite.

 

Therefore, we only have to deal with the average velocities, and not worry about individual variations due to wind speed.

 

I simply subtracted the trip with no wind, 166.7 m/s, from the second velocity, 138.9 m/s, for a wind velocity of 27.8 m/s.

 

Is this line of reasoning correct? It's confusing me because I think you don't have to know the separate velocites in the second trip, I think they'll cancel out... but I wouldn't bet my life on that.

 

What do you guys think?

[D H]

For the first trip, the constant speed (not velocity) is 6000km/10hr = 600 km/hr, or 166.7 m/s. This is presumably without a wind. The key point here is that the plane flies 600km/hr with respect to the wind, not with respect to the land. During the outbound trip, the plane flies against the wind, reducing it's ground speed to 600km/hr - w. On the return trip the ground speed increases to 600km/hr + w. Solve for the wind speed w that yields a total trip time of 12 hours.

[ecoli]

Ok, but this is what I'm not getting.

 

If you do what you suggested, then you get the expression

 

[math] (600km/hr - w) + (600 km/hr + w) = \frac{6000km}{12 hrs} [/math]

 

 

The wind speed would cancel you, and the equation wouldn't make any sense. And, you can't really split the equation in half, because you don't know how long each half of the trip took, just the total.

[D H]

What you just did is not correct. You need to do it like this:

 

[math]

\frac{3000\;\text{km}}{600\;\text{km/hr}-w} +

\frac{3000\;\text{km}}{600\;\text{km/hr}+w} = 12\;\text{hr}[/math]

[timo]

EDIT: I somehow didn't notice the posts after #11 so maybe part of the following might be redundant:

 

 

Obviously, since the wind is travelling from west to east, the trip back, from B to A is going to take longer. But since the wind speed does not change, this sort of "cancels out" from the trip there, which is exactly the opposite.

. At least you're not proposing this to be the newest theory of everything. Perhaps I should mention why I had to grin here: You use the very definite attribute "exactly" in a completely indefinite context - what is exactly opposite?

 

Therefore, we only have to deal with the average velocities, and not worry about individual variations due to wind speed.

Just try not being smart and just look at the equations. If you rearrange s=v*t into t = s/v, the application on this scenario would read

 

[math] 10 h = \frac{3000 km}{v} + \frac{3000 km}{v} [/math] for the first case and

 

[math] 12 h = \frac{3000 km}{v + w} + \frac{3000 km}{v - w} [/math] (with w being the wind speed) for the 2nd case.

 

I would say you better look at the individual variations; I don't see how they'd easily cancel.

 

I simply subtracted the trip with no wind, 166.7 m/s, from the second velocity, 138.9 m/s, for a wind velocity of 27.8 m/s.

Why? Substracting the velocities would make sense if the wind was always blowing against the movement of the plane. I don't think that's the case here.

 

Is this line of reasoning correct?

I don't think so, but I don't understand it - perhaps try being less smart...

[D H]

The problem with you approach is that you are doing something that is inherently invalid. Suppose the wind is blowing at 400 km/hr. The outbound trip takes 15 hours since the plane's ground speed is a paltry 200 km/hr. The return flight takes a mere 3 hours since the plane's ground speed is 1000 km/hr. The average of the two speeds is 600 km/hr. The average speed, on the other hand, is 266.7 km/hr.

[ecoli]

EDIT: I somehow didn't notice the posts after #11 so maybe part of the following might be redundant:

 

 

 

. At least you're not proposing this to be the newest theory of everything. Perhaps I should mention why I had to grin here: You use the very definite attribute "exactly" in a completely indefinite context - what is exactly opposite?

 

Sorry... sometimes I get wordy, and that leads to inaccurate semantics. I meant that the plane is going exactly opposite to the wind.

 

 

Just try not being smart and just look at the equations.If you rearrange s=v*t into t = s/v, the application on this scenario would read

 

[math] 10 h = \frac{3000 km}{v} + \frac{3000 km}{v} [/math] for the first case and

 

[math] 12 h = \frac{3000 km}{v + w} + \frac{3000 km}{v - w} [/math] (with w being the wind speed) for the 2nd case.

 

Ok... so I understand why your equation is right. But what I don't understand is how the equation I wrote is wrong. Logically, what does the equation I wrote mean? If I can understand why I wrote the wrong thing, hopefully the next time this sort of thing comes up, I'll know what sort of fallacious logic to avoid.

 

I would say you better look at the individual variations; I don't see how they'd easily cancel.

Well, in the equation I wrote, +w and -w are on the same side of the equation, so that would equal zero.

 

 

Why? Substracting the velocities would make sense if the wind was always blowing against the movement of the plane. I don't think that's the case here.

My logic went like this: if the ave. velocity of the plane without wind is X, and the ave velocity of plane with wind is Y, then X - Y equals the velocity of the wind.

 

 

I don't think so, but I don't understand it - perhaps try being less smart...

 

Trust me, that won't be a problem

[D H]

My logic went like this: if the ave. velocity of the plane without wind is X, and the ave velocity of plane with wind is Y, then X - Y equals the velocity of the wind.

You are comitting two faults here: Confusing speed and velocity, and computing the average speed by averaging the speeds. I'm sure you know the difference between speed and velocity. Its any easy trap to get caught in because colloquially these two terms mean the same thing.

 

You can average speeds to get the average speed, but you have to use a time-weighted average. You were right in that the average speed for the second trip was 6000km / 12 hr = 500 km/hr. Thus another way to solve this problem is to compute the average speed as a time-weighted average and equate this result to the known 500 km/hr average speed.

 

Doing this, the time-weighted average speed is

 

[math]\bar s = \frac{s_1t_1 + s_2t_2}{t_1+t_2}[/math]

 

where the suffixes 1 and 2 refer to the outbound and return flights respectively. The outbound speed is v-w while the return speed is v+w. Note that the products [math]s_1t_1[/math] and [math]s_2t_2[/math] are the distance flown during each leg: 3000 km in both cases. Denoting this as d, the time-weighted average speed becomes

 

[math]\bar s = \frac{2d}{d/(v-w)+d/(v+w)}[/math]

 

Simplifying and solving for w,

 

[math]w = \sqrt{v(v-\bar s)}[/math]

 

which is the same result that obtains from post #14.

[ecoli]

thanks a lot for the help, DH and Atheist. My test is tomorrow, so hopefully I'll be able to reason these things out in high pressure situation. I'll probably post some more problems tomorrow, as I study.

 

The test goes down like this. There are three multipart problems, with plenty of partial credit. What I like about the system, is that the professor said if we think we're doing a problem incorrectly, and we can't figure out what the issue is, we write down why we think our answer is wrong. And you can get more partial credit that way. I think it's a good system. Anyways, wish me luck

[ecoli]

well I did shit on the first exam anyway... hopefully tomorrow's test will be easier.

 

Couple of questions from this practice test...

 

http://skipper.physics.sunysb.edu/%7Eabhay/phy131/exams/2006/midterm2.pdf

 

for question 3, is the tension in the string just dependent upon the y and x components, and is that component related to the mv^2/R ?

I can draw the free body diagram, but I'm not sure what it means, because Should I have to subtract that Tension from the force due to gravity?

 

Question 4b - Is the acceleration of the block just due to gravity? Or is it related to the angular velocity of the turning disk? I have a feeling that the moment of inertia comes into play in the conservation of angular momentum, but I'm not really sure how. I don't really know how to approach this problem.

 

Thanks for the help.

[swansont]

well I did shit on the first exam anyway... hopefully tomorrow's test will be easier.

 

Couple of questions from this practice test...

 

http://skipper.physics.sunysb.edu/%7Eabhay/phy131/exams/2006/midterm2.pdf

 

for question 3, is the tension in the string just dependent upon the y and x components, and is that component related to the mv^2/R ?

I can draw the free body diagram, but I'm not sure what it means, because Should I have to subtract that Tension from the force due to gravity?

 

Question 4b - Is the acceleration of the block just due to gravity? Or is it related to the angular velocity of the turning disk? I have a feeling that the moment of inertia comes into play in the conservation of angular momentum, but I'm not really sure how. I don't really know how to approach this problem.

 

Thanks for the help.

 

 

For 3, at 45º the pendulum is momentarily at rest, so there's no centripetal term. Just sum the forces, and you know there's no net force along the radius. At the vertical, the net force must be the centripetal force, so yes, you have to add the weight and the tension (vectorally) to give the net force.

 

 

For 4b you have to calculate the torque the falling mass will exert and find the angular acceleration of the disks. Then you can convert that to a linear acceleration of the rim of the smaller disk, which is the same as the acceleration of the falling mass.

[martianxx]

going back to the original question i like to look at the data that you have. you have:

 

s (displacement): 140m

t (time): 3.6s

v (final velocity): 53 ms

 

you can then use s = [(u + v)/2]t

 

140 = [(u + 53)/2]3.6

140 = [(u/2) + 26.5]3.6

140 = 1.6u + 95.4

44.6 = 1.6u

27.875 = u

 

 

Then you can use this to work out the acceleration:

 

v = u + at

53 = 27.875 +3.6a

25.125 = 3.6a

6.9792 = a

 

Hope this helps - ps may have made a slip up on the maths in the first part it is late and i am about to go to bed

[ecoli]

For 4b you have to calculate the torque the falling mass will exert and find the angular acceleration of the disks. Then you can convert that to a linear acceleration of the rim of the smaller disk, which is the same as the acceleration of the falling mass.

 

So let me see if this is right...

 

[math]\tau = R X F[/math]

This part I'm not sure about... it's the cross product, but one component doesn't matter because it's parallel, right?

So we can just treat it like a 'regular' multiplication.

[math] \tau = I*\alpha [/math]

 

[math] \tau = \frac{1}{2} mr^3 * \alpha [/math]

 

[math] \alpha = \frac{\tau}{\frac{1}{2}}{mr^2} [/math]

[swansont]

So let me see if this is right...

 

[math]\tau = R X F[/math]

This part I'm not sure about... it's the cross product, but one component doesn't matter because it's parallel, right?

So we can just treat it like a 'regular' multiplication.

[math] \tau = I*\alpha [/math]

 

[math] \tau = \frac{1}{2} mr^3 * \alpha [/math]

 

[math] \alpha = \frac{\tau}{\frac{1}{2}}{mr^2} [/math]

 

The torque from the mass will be mgr (the contact is tangential, so the angle is 90º, r is the radius of the small disk)

 

and ^{[math] \tau = I\alpha [/math]} as you note, but the I of the system is the sum of the individual moments of the two disks, which will be ^{[math]\frac{1}{2}mr^2 + \frac{1}{2}MR^2[/math]}

[mooeypoo]

There is certainly a way to calculate this using tension, but I would do it using energies (but since I'm taking physics for the first time since highschool, please double check it.. I'm almost sure this is right, but not 100% sure)

 

 

When the pendulum is 45 degrees up, it has potential energy:

[math]mgh[/math]

When the pendulum is at the bottom (90 degrees from the horizontal) it has no potential energy, but just kinetic energy:

[math]1/2*mv^2[/math]

 

The variables are:

m = 0.12 kg

g = 9.8 m/s^2

 

I calculated the h using trigonometry. The rope length is 0.8, so when the pendulum is 45 degrees, the mass is 0.566m distance from the horizontal 'ceiling' (0.8cos(45)) and the difference between the two is 0.8-0.566=0.234

so:

 

h = 0.234

 

Conservation of energy says E1=E2

So:

[math]mgh=(1/2)*mv^2[/math]

[math]v=sqrt(2*g*h)[/math]

[math]v=sqrt(2*(9.8 m/s^2) * (0.234 m))[/math]

so:

[math]v=2.14 m/s[/math]

 

I must say, I am not entirely sure you can use conservation of energies with a pendulum, but the question specified the string has no mass and there is no air resistence, so there shouldn't be any external forces to 'ruin' the conservation of eneregy.

 

If anyone thinks this is wrong, though, please correct me, and I would appreciate knowing why would conservation not work here, if it doesn't. It should.. I would think.

 

In any case, drawing a free body diagram is always a good idea. It's the first thing I do with any question, even for the sole purpose that it sorts things out in my head and shows me what I do have and don't have. Like here, for instance, just drawing the diagram showed me that I have all variables for mgh and 1/2mv^2 that are needed to solve using energies..

 

Free body diagram also helps with the Tension and other forces that aren't supplied, because you can see right away which forces affect the Tension..

 

Anyways, I hope that helps..?

 

~moo

 

 

---- EDIT -----

 

Okay I missed the fact that lots of other people answered already (I didn't notice there's a second page so firsts off, this is question 3 part a, and second, I see no one used conservation of energy, so I would love to know if I was right or not.

 

G'luck anyways, ecoli.. I'm in it too.. only my college calls it "Physics 207".. egh

 

~moo