Physics help

[hotcommodity]

I really appreciate the help, thank you.

 

I was working a problem on forces last night and got stumped, so I used my other physics text as a reference to help me out. It had the same problem I was working, only the exercise used arbitrary masses M and m to find the acceleration of two blocks attached by a massless rope, and the tension between them. This is the exercise:

 

 

There are two things I dont understand about this example. First, on the left hand side it shows two axes, one is upright, but the other one was turned upside down to make "mg" a positive quantity, and I dont see why they did that. Secondly, I used the acceleration equation that they derived for the arbitrary masses, and it worked for the problem I was working on. But when I used the equation for the tension, it gave me the wrong answer. If I leave "g" out of the tension equation, it gives me the answer that the book has. Was the equation for the tension derived correctly, is my other book wrong, or is there some other concept that I'm missing?

[swansont]

The writing is a little small for me, but the most likely reason is this: you have two objects accelerating (with the same magnitude). If you want to combine the equations, you have to choose your coordinate systems so that the acceleration has the proper value. If the acceleration of the top block is to the right, and this is the positive direction, then the direction the other block accelerates (down), has to be the positive direction for that equation. Otherwise you can't combine the equations, because you'll be saying a = -a, and that will only be true if a=0.

[hotcommodity]

Thanks for the reply above

 

I have two questions:

 

The first involves uniform circular motion, and it's somewhat of an extension of what I've posted above. The Earth is a rotating mass, and would therefore contain different points in its mass that display particular tangential velocities, and centripetal accelerations. If I pick a point on the outer edge of the Earth, the radius being 3.38E6m, and the period being 8.64E4s, I get that its tangetial velocity is about 245 m/s. This doesn't seem correct to me. Are objects, or points of mass, on the outer edge of the Earth experiencing this kind of tangential velocity, or am I applying the concepts and equations incorrectly?

 

Secondly, I was reading through the derivation of the work-energy theorem, and it shows that the kinematic equations were used, and that it was assumed that the net force was always in the same direction as the displacement such that the cosine of the angle between the displacment vector and the force vector does not come in to play. So my question is if the work-energy theorem, or more specifically the equation that the work done on an object is equal to the objects change in kinetic energy, can be applied to situations where the force is not in the same direction as the displacement. Any help is appreciated.

[swansont]

Thanks for the reply above

 

I have two questions:

 

The first involves uniform circular motion, and it's somewhat of an extension of what I've posted above. The Earth is a rotating mass, and would therefore contain different points in its mass that display particular tangential velocities, and centripetal accelerations. If I pick a point on the outer edge of the Earth, the radius being 3.38E6m, and the period being 8.64E4s, I get that its tangetial velocity is about 245 m/s. This doesn't seem correct to me. Are objects, or points of mass, on the outer edge of the Earth experiencing this kind of tangential velocity, or am I applying the concepts and equations incorrectly?

 

The centripetal force assumes that it's the net force, i.e. that's what would happen in orbit. An object on or in the earth is experienceing other forces.

 

Secondly, I was reading through the derivation of the work-energy theorem, and it shows that the kinematic equations were used, and that it was assumed that the net force was always in the same direction as the displacement such that the cosine of the angle between the displacment vector and the force vector does not come in to play. So my question is if the work-energy theorem, or more specifically the equation that the work done on an object is equal to the objects change in kinetic energy, can be applied to situations where the force is not in the same direction as the displacement. Any help is appreciated.

 

If you start at rest, the displacement has to be in the direction of the net force as long as it's constant. But in general, you have to take the dot product. A force perpendicular to the displacement can change the direction, but does no work (no change in speed)

[hotcommodity]

 

If you start at rest, the displacement has to be in the direction of the net force as long as it's constant. But in general, you have to take the dot product. A force perpendicular to the displacement can change the direction, but does no work (no change in speed)

 

Thank you for the reply. I'm curious why the displacement must be in the direction of the net force if the mass starts at rest. Aside from that, lets say someone is pulling a sled, and lets say the sled is accelerated from rest, but the net force acting on the object was at angle (less than 90 degrees for instance). Could I still use the work-energy theorem if I only consider the force acting in the direction of the displacement, instead of the net force alone?

[insane_alien]

I'm curious why the displacement must be in the direction of the net force if the mass starts at rest.

 

because thats the direction the acceleration will be in.

 

if you applied a force (lets use basic directions here) forward onto a block you would not expect it to go shooting off to the left.

 

use vectors in the equations and you'll see the displacement will be in the direction of the force.

[swansont]

In the case of the sled, it will still move in the direction of the net force — note that the applied force is not the net force: you have to account for gravity, friction and the normal force.

[hotcommodity]

because thats the direction the acceleration will be in.

 

 

I appreciate the replies above. Insane, I understand the object in question will move in the same direction as the net force, but I'm curious why it must "accelerate from rest." For instance, I would assume that the object could move at a constant velocity, and then accelerate over some distance, and the work-energy theorem would still be valid because 1) a net force occured, and 2) a constant acceleration occured over some distance. I don't see why the object would have to accelerate from rest only, if that what swansont was getting at, but maybe he was just giving it as an example.

 

I do have another question however. I ran across a problem where the work-energy theorem was applied to an object in uniform circular motion, and they substituted the tangential velocity in for the velocity in the equation Work= 1/2*m*(vf^2-v0^2). I'm curious how this is possible since the work-energy theorem was derived using the kinematics equations, which apply to straight line motion with respect to x, y, and z components.

[swansont]

I appreciate the replies above. Insane, I understand the object in question will move in the same direction as the net force, but I'm curious why it must "accelerate from rest." For instance, I would assume that the object could move at a constant velocity, and then accelerate over some distance, and the work-energy theorem would still be valid because 1) a net force occured, and 2) a constant acceleration occured over some distance. I don't see why the object would have to accelerate from rest only, if that what swansont was getting at, but maybe he was just giving it as an example.

 

I do have another question however. I ran across a problem where the work-energy theorem was applied to an object in uniform circular motion, and they substituted the tangential velocity in for the velocity in the equation Work= 1/2*m*(vf^2-v0^2). I'm curious how this is possible since the work-energy theorem was derived using the kinematics equations, which apply to straight line motion with respect to x, y, and z components.

 

Accelerate from rest is used precisely because of things like circular motion, where a force can result in a change in direction but not speed. Start simple, then move to more complex.

 

You ignore the radial force because it does no work. The tangential force is then the force that is always in direction of motion. At any given instant, you can break it down into a kinematics problem, so that infinitesimal motion can be seen as straight-line.

[hotcommodity]

Thank you for the reply. I have a question about the impulse that acts upon an object. The impulse is defined as an object's change in momentum, and so initial and final velocities must be given values, in addition to the mass, to solve for the impulse. I know how initial and final velocities in the kinematic equations are dependent upon the displacement of the object in question, but what are they dependent upon when it comes to impulse? I would assume it's the objects initial velocity when the force begins to act, and its final velocity the moment the force stops acting, but I'm not quite sure.

 

Additionally, I was looking at the bullet and pendelum example, where you're given the masses of the two objects and the height the pendelum swings, and asked to find the bullets initial velocity. When looking at the part where the two masses are combined and swing to some height, the book assumes that total mechanical energy is conserved, in other words, the work done on the object by non-conservative forces is zero. My question is why the bullets velocity is not considered a propulsive non-conservative force.

 

And finally, I have a question that somewhat relates to the two questions above. I'm curious what determines the "initial velocity" of the bullet. I would assume it's the speed the bullet assumes after the force of the exploding gun powder stops acting on it, but I'm not certain. It wouldn't have any forces acting on it horizontally, so I wonder if the kinematic equations play a role since it has a constant velocity in both the x and y directions. Any help is appreciated.

[swansont]

Thank you for the reply. I have a question about the impulse that acts upon an object. The impulse is defined as an object's change in momentum, and so initial and final velocities must be given values, in addition to the mass, to solve for the impulse. I know how initial and final velocities in the kinematic equations are dependent upon the displacement of the object in question, but what are they dependent upon when it comes to impulse? I would assume it's the objects initial velocity when the force begins to act, and its final velocity the moment the force stops acting, but I'm not quite sure.

 

Impulse depends on the length of time a force acts. [math]F\Delta t = \Delta p[/math]

 

Additionally, I was looking at the bullet and pendelum example, where you're given the masses of the two objects and the height the pendelum swings, and asked to find the bullets initial velocity. When looking at the part where the two masses are combined and swing to some height, the book assumes that total mechanical energy is conserved, in other words, the work done on the object by non-conservative forces is zero. My question is why the bullets velocity is not considered a propulsive non-conservative force.

 

ME is conserved after the collision, but not during. The collision is completely inelastic, so KE is not conserved. You need to use conservation of momentum to solve for the KE right after the pendulum captures the bullet.

 

And finally, I have a question that somewhat relates to the two questions above. I'm curious what determines the "initial velocity" of the bullet. I would assume it's the speed the bullet assumes after the force of the exploding gun powder stops acting on it, but I'm not certain. It wouldn't have any forces acting on it horizontally, so I wonder if the kinematic equations play a role since it has a constant velocity in both the x and y directions. Any help is appreciated.

 

You'd have to look at the expansion of the gases in the gun, and look at the pressure as the volume changes, which involves a lot of thermodynamics.

[hotcommodity]

Thanks for the reply. About the bullet and pendelum problem, I understand that the problem must be broken into two parts: analyzing the system before the collision, and analyzing the new system (the bullet and pendelum combined) after the collision. After the collision, the book says that total ME is conserved, and I know that to be true if and only if the work done by non-conservative forces on the bullet and pendelum is zero. I'm curious why the accelerating bullet isn't considered to be a propulsive force acting on both the bullet inside the pendelum and the pendelum itself. I know I'm overlooking a key concept, but I can't quite put my finger on it.

 

And in case I don't get to it later, I'd like to thank you swansont for the help and insight you've given me throughout the summer, I know I'll put it to good use in the coming semester.

[swansont]

Thanks for the reply. About the bullet and pendelum problem, I understand that the problem must be broken into two parts: analyzing the system before the collision, and analyzing the new system (the bullet and pendelum combined) after the collision. After the collision, the book says that total ME is conserved, and I know that to be true if and only if the work done by non-conservative forces on the bullet and pendelum is zero. I'm curious why the accelerating bullet isn't considered to be a propulsive force acting on both the bullet inside the pendelum and the pendelum itself. I know I'm overlooking a key concept, but I can't quite put my finger on it.

 

At what point in the problem are you worried about the bullet?

 

During the collision, you are basically assuming that the time it takes to come to rest inside the pendulum is small, and the pendulum doesn't move much. So while the bullet and pendulum are exerting non-conservative force on each other, you can ignore that. The fact that the collision is completely inelastic actually means you can figure out how much KE is lost to other forms; what you end up doing is actually calculating the remaining KE, which then gets transformed into PE.

 

And in case I don't get to it later, I'd like to thank you swansont for the help and insight you've given me throughout the summer, I know I'll put it to good use in the coming semester.

 

You're quite welcome.

[hotcommodity]

Hello again, I have another question but this time concerning my physics lab. In lab we took measurements of a pendulum including its period in seconds, the length of the string the ball was attached to, the diameter of the ball, and the mass and weight of the ball. We considered the uncertainty in these measurements, and it was said that we could only measure the value of mass, period, etc. to some degree. I didn't understand this part very well. How do I know what degree of certainty I can measure an object to with tools like the meter stick, stopwatch, caliper, and so forth? Any help is appreciated.

[swansont]

It depends on the resolution of your measuring device. Your last digit will be uncertain, and you have to estimate just how uncertain it is, which will depend on the instrument.

 

Let's say you have a ruler marked down to the mm level (though this could be the markings on a scale or balance, or whatever)

 

You might be able to estimate the length of something to less than a mm, but that last digit is uncertain. So if something is between 23 and 24 mm long, and say you can tell that it's closer to the 23 than the 24. That means your resolution is about a half of a mm. So you might be able to say that the value is 23.3 ± 0.3 mm.

 

If you have a digital readout, the last digit is uncertain (you don't know if it was rounded up or down by the electronics) so if it were e.g. a stopwatch that read out down to tenths of a second you'd go with ± 0.1

 

 

Another way of coming up with an uncertainty is to make multiple measurements, which will have a range of values if the errors are random. You take the average and determine the standard deviation, which would be a measure of your uncertainty.

[hotcommodity]

I see, I think I have a better grasp on it now, thank you

 

I was reading through my lab book a little further, and I'm still having a bit of trouble with uncertainties. Sometimes a value is given with ± 0.1, but sometimes they give a value with ± 0.02 or ± 0.04 and I'm just not getting where those numbers come from. Oddly enough, I'm getting the part about error propagation, but the basic determination of the uncertainty of variables using lab tools is throwing me. I know it's a bit hard to explain over the internet, so I'm wondering if there are any tutorial sites that lay all of this stuff out. Any help is appreciated.

[swansont]

Check these out and see if they help

 

http://phys.columbia.edu/~tutorial/

http://instructor.physics.lsa.umich.edu/ip-labs/tutorials/errors/vocab.html

http://www.lhup.edu/~dsimanek/scenario/errorman/measures.htm

[hotcommodity]

Thank you so much, these look like really good sites

[hotcommodity]

I'm stuck on a homework problem. There's a bicycle wheel, and we're given the mass of the wheel and the rim as well as their distance from the axis of rotation, and then we're asked to calculate it's moment of inertia. I understand that part. Then we're asked why the contributions due to the hub and the spokes can be ignored.

 

I'm pretty sure that the contribution to the moment of inertia from the hub can be ignored because the hub is at the center of rotation, and would effectively have a distance of zero from the center of rotation. I'm stuck on the part about the spokes however. Unless they were of negligible mass, I don't see how their contribution to the rotational inertia can be ignored. I don't want the answer, I'm just looking for a push in the right direction. Any help is appreciated.

[D H]

Unless they were of negligible mass, I don't see how their contribution to the rotational inertia can be ignored.

 

There's the answer. The spokes are of nearly negligible mass. In reality, they add about 5-10% to the rotational inertia. Not truly neglible, but close enough for government and introductory physics.

[swansont]

Also, the geometry of the spokes reduces their contribution (per unit mass) by a factor of 3 (since I= 1/3 MR² for a rod rotating in that fashion). So, even if the mass of the spokes were equal to the mass of the rim and wheel, they would contribute only 25% to the total moment of inertia. So you could actually estimate what spoke mass, relative to the other mass, would make a 10% contribution (assuming that's a the estimation threshold where we ignore it), which IMO would be a good way to answer the question. What do you get?

[iNow]

Also, the geometry of the spokes reduces their contribution (per unit mass) by a factor of 3 (since I= 1/3 MR² for a rod rotating in that fashion). So, even if the mass of the spokes were equal to the mass of the rim and wheel, they would contribute only 25% to the total moment of inertia.

Sorry for stepping in with a question instead of answer, but can you clarify for me...

 

Is the contribution of the spoke less (per your quote above) due to the fact that all of it's mass is not in one equal distance away from the center, but instead some of it's mass is closer to the hub while some of it's mass is closer to the rim, and therefore each different area on the spoke contributes a different portion of it's mass? So... what I'm thinking is that as you traverse the spoke farther away from the hub (toward the rim) then the contribution of spoke mass continually increases. Said another way, as you go down the spoke from the rim to the hub, the contribution of mass decreases.

 

Sorry if this is completely off base. That's just the visual I had while reading the quoted text above. I'm trying to determine if it's at all valid or if it's time for another cup of coffee this morning (although likely, it's some combination of both). Thanks.

[swansont]

Sorry for stepping in with a question instead of answer, but can you clarify for me...

 

Is the contribution of the spoke less (per your quote above) due to the fact that all of it's mass is not in one equal distance away from the center, but instead some of it's mass is closer to the hub while some of it's mass is closer to the rim, and therefore each different area on the spoke contributes a different portion of it's mass? So... what I'm thinking is that as you traverse the spoke farther away from the hub (toward the rim) then the contribution of spoke mass continually increases. Said another way, as you go down the spoke from the rim to the hub, the contribution of mass decreases.

 

Sorry if this is completely off base. That's just the visual I had while reading the quoted text above. I'm trying to determine if it's at all valid or if it's time for another cup of coffee this morning (although likely, it's some combination of both). Thanks.

 

Yes, that's it. The 1/3 MR² accounts for the varying distance from the point of rotation; the mass closer to the rim contributes more than the mass at the hub. If you integrate over the whole length, you get the 1/3 factor. The moment of the rim and tire is given by MR².

[hotcommodity]

Thanks for the replies.

 

Also, the geometry of the spokes reduces their contribution (per unit mass) by a factor of 3 (since I= 1/3 MR² for a rod rotating in that fashion). So, even if the mass of the spokes were equal to the mass of the rim and wheel, they would contribute only 25% to the total moment of inertia. So you could actually estimate what spoke mass, relative to the other mass, would make a 10% contribution (assuming that's a the estimation threshold where we ignore it), which IMO would be a good way to answer the question. What do you get?

 

Well, the mass given for the rim and tire is 1.3 kg, and it's distance from the axis of rotation is .34 meters. Using I = MR², the moment of inertia is .15028 kg*m². I assume that you're saying the spokes would contribute 10% of the total rotational inertia, so that's 10% of .15028 kg*m² which gives .015028 kg*m². Then using the equation I = 1/3 MR² with .015028 kg*m² for the moment of inertia due to the spokes, and with a distance .34 meters from the axis of rotation, I get that the total mass of the spokes is .39 kg, which is about a third of the mass of the tire and rim. Am I following this correctly?

[swansont]

Yes. (You could also have solved it with just the equations and found that 1/3 of the mass gives a 10% contribution to the moment.) So as long as the spokes have less than that amount of mass (or whatever your standard happens to be), their contribution is small enough that we ignore it.