Physics help

[iNow]

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.

 

Thank you, swansont.

[hotcommodity]

I just have a few conceptual questions, and the first is regarding the direction of the torque vector of an object accelerating angularly. I was told in class that the torque vector points perpendicular to the vector of the angular momentum, and I have no idea why that is. There's lots of directions the torque vector can point and still be perpendicular to the vector of angular momentum, how can you tell? My professor mentioned something about the weight force of the object having something to with which way the rotating object will precess. I was late to class and only caught the tail end of the discussion. You always miss the good stuff when you're late.

 

My second question is about vector notation. I'm a bit lost on how to write forces in vector form, as we tend to use magnitudes when we're adding up the forces that act on a body. When is it proper to write vector quantities in vector notation? For instance, if theres a frictional force decelerating a sliding box, and it's the net force acting on the body, how can I write that in vector form? I'm sorry if I appear a bit lost, but I'm trying to get what's going on here. So, we have the frictional force opposing the motion of the sliding box, and I know by looking at a free body diagram that it points in the negative x-direction, and I want to write the frictional force in vector notation. So, can I say f = ma, and a = -ai, and therefore the frictional force is -f ? I'm wondering if it would make sense to just place the negative in front of the force to begin with such as -f = ma, where a = -ai, but that would make f point in the positive direction. I know I've done it assbackwards a few times so I just want to get it right, any help is appreciated.

[swansont]

I just have a few conceptual questions, and the first is regarding the direction of the torque vector of an object accelerating angularly. I was told in class that the torque vector points perpendicular to the vector of the angular momentum, and I have no idea why that is. There's lots of directions the torque vector can point and still be perpendicular to the vector of angular momentum, how can you tell? My professor mentioned something about the weight force of the object having something to with which way the rotating object will precess. I was late to class and only caught the tail end of the discussion. You always miss the good stuff when you're late.

 

Actually, the torque vector doesn't have to relate to the angular momentum vector at all, and certainly isn't perpendicular to it. If you start an object from uniform rotation and speed it up, but maintain the same rotation axis, the torque and angular momentum will be in the same direction. So you must have missed some specific scenario where this is true.

 

(You have the cross products, r x F and r x p which by the right-hand-rule, have the resultant perpendicular to whatever plane is defined by those two vectors. If p and F are in the same direction, the torque and angular momentum will be, too)

 

My second question is about vector notation. I'm a bit lost on how to write forces in vector form, as we tend to use magnitudes when we're adding up the forces that act on a body. When is it proper to write vector quantities in vector notation? For instance, if theres a frictional force decelerating a sliding box, and it's the net force acting on the body, how can I write that in vector form? I'm sorry if I appear a bit lost, but I'm trying to get what's going on here. So, we have the frictional force opposing the motion of the sliding box, and I know by looking at a free body diagram that it points in the negative x-direction, and I want to write the frictional force in vector notation. So, can I say f = ma, and a = -ai, and therefore the frictional force is -f ? I'm wondering if it would make sense to just place the negative in front of the force to begin with such as -f = ma, where a = -ai, but that would make f point in the positive direction. I know I've done it assbackwards a few times so I just want to get it right, any help is appreciated.

 

 

Something like a is already a vector, and is a variable, so it will have a magnitude and direction. a = -ai now has a specific direction, and a is a scalar. But you shouldn't have both the directional "- sign" and the vector notation.

 

F = ma so F = -mai (m and a are scalars, -i is the vector)

[hotcommodity]

I see, I thought the torque vector would be along the axis of rotation, just as the angular momentum vector is, as you mentioned. Maybe I read the notes wrong, I'll have to go back and see. I remember a picture in my text book with the angular momentum pointing along the axis of rotation, the force pointing down, and the torque vector at a right angle to the force. I'll double check. I think I get what you're saying about the direction of vectors regarding f = ma. If I understand correctly, "a" is always a positive value, but as a vector a, it's a positive value "a" pointing in the negative i direction. That makes sense. Thank you for the help.

[swansont]

The torque vector doesn't have to be along the axis of rotation, though. That will cause a change in the angular momentum direction, much like a force not along the direction of your velocity will change your direction of motion.

 

If your prof was discussing precession (gyroscopic precession, from the mention of gravity), then he was talking about a specific situation that does not generally hold.

 

http://en.wikipedia.org/wiki/Precession

http://hyperphysics.phy-astr.gsu.edu/hbase/gyr.html

[hotcommodity]

That's exactly what he was talking about. I watched the part of the lecture I missed online and my professor did a demonstration in class with the rope and the bicycle wheel, where it spins vertically without falling sideways due to gravity. I had to watch it 3 times to get a decent grasp on it, and it boggled my mind while I was trying to get to sleep. I took the test today tho', and all went well, I really do appreciate your help. I'll definitely check out the links.

[hotcommodity]

I have a question on a homework problem having to do with a block accelerating toward a spring down an inclined plane. The problem is this:

 

A 1.0 kg block starts at rest and slides a distance of 0.75 m down a frictionless 30 degree incline (with respect to the horizontal) where it hits a spring whose spring constant is 9.8 N/m.

 

(No need to comment on my windows paint skills).

 

Part "a" asks me to find the distance the spring is compressed, and I figured that out by using the the fact that the total mechanical energy of the system is conserved. Part "b" is what I'm having trouble with.

 

It asks "where does the block achieve maximum speed during this slide." One of the hints my professor gave was to find a function of velocity and find its maximum by taking the derivative of that function. So this is what I did:

 

I used the fact that mechanical energy is conserved from the time the block begins accelerating to right before it touches the spring. The energy terms below only deal with the block.

 

[math]E_0 = E_f[/math]

 

[math]U_f + K_f = U_0 + K_0[/math]

 

The final potential energy, and the initial kinetic energy are zero if I take the point of the spring to be height = 0.

 

[math]K_f = U_0[/math]

 

[math]0.5*m*v^{2}_f = mgh[/math]

 

[math]v_f = \sqrt{2*g*y}[/math]

 

[math]\frac{dv_f}{dy} = \frac{g}{\sqrt{2*g*y}}[/math]

 

 

 

 

It appears that it would achieve its maximum speed when the acceleration of the block is zero. If that's the case, this would be the point where the block is stopped by the spring. But if it's stopped, how can it have a maximum speed. Am I doing this wrong or am I missing a concept? Any help is appreciated.

[ydoaPs]

It appears that it would achieve its maximum speed when the acceleration of the block is zero. If that's the case, this would be the point where the block is stopped by the spring. But if it's stopped, how can it have a maximum speed. Am I doing this wrong or am I missing a concept? Any help is appreciated.

 

When it hits the spring, it still has a positive acceleration. As the spring compresses, it adds a force countering that causing the positive acceleration which slowly makes the net acceleration smaller until the acceleration is zero. This is where the block will be moving the fastest. At this point, the block is still compressing the spring adding a negative acceleration. This will slow the block down and eventually reverse it's direction of motion. There will be some harmonic motion until the block comes to a rest.

[hotcommodity]

Thank you for the reply. I'm still having a bit of trouble with this problem. If I understand what you're saying, the block will achieve its maximum speed right before it begins decelerating (and graphically, this makes sense), and it begins decelerating as it hits the spring. For part "a" of the problem, I found that the block does not compress the spring, x is equal to zero. So I think I only have to consider the movement of the block from the time it begins accelerating to the part right before it hits the spring. I'm not sure how to find this maximum speed, because in the dv/dy equation above, I have "g" in the numerator, and that's just a constant, it can never equal zero....

[swansont]

[math]E_0 = E_f[/math]

 

[math]U_f + K_f = U_0 + K_0[/math]

 

The final potential energy, and the initial kinetic energy are zero if I take the point of the spring to be height = 0.

 

[math]K_f = U_0[/math]

 

[math]0.5*m*v^{2}_f = mgh[/math]

 

[math]v_f = \sqrt{2*g*y}[/math]

 

[math]\frac{dv_f}{dy} = \frac{g}{\sqrt{2*g*y}}[/math]

 

 

 

 

It appears that it would achieve its maximum speed when the acceleration of the block is zero. If that's the case, this would be the point where the block is stopped by the spring. But if it's stopped, how can it have a maximum speed. Am I doing this wrong or am I missing a concept? Any help is appreciated.

 

[math]0.5*m*v^{2}_f = mgh[/math]

 

You have very significant oversight here. Does this actually describe the situation?

 

[math]v_f = \sqrt{2gh}[/math]

 

v must be positive, so what direction will make this be true? This is a case where your choice of h=0 defines the coordinate directions for you.

 

 

[math]\frac{dv_f}{dy} = \frac{g}{\sqrt{2gy}}[/math]

 

The only variable here is y. This tells you that when y is infinite, the derivative is zero, but your earlier mistake makes this equation incorrect.

[hotcommodity]

[math]0.5*m*v^{2}_f = mgh[/math]

 

You have very significant oversight here. Does this actually describe the situation?

 

[math]v_f = \sqrt{2gh}[/math]

 

v must be positive, so what direction will make this be true? This is a case where your choice of h=0 defines the coordinate directions for you.

 

 

[math]\frac{dv_f}{dy} = \frac{g}{\sqrt{2gy}}[/math]

 

The only variable here is y. This tells you that when y is infinite, the derivative is zero, but your earlier mistake makes this equation incorrect.

 

 

Thank you for the reply. If I define the downward slope of the incline to be the positive x-axis, then the block will have a positive velocity. I defined h=0 to be the point at which the block starts accelerating, and [math]h_f[/math] to be the point at which the block begins decelerating. So I'd have:

 

[math]0.5*m*v^{2}_f + mgh_f= 0[/math]

 

[math]0.5*m*v^{2}_f = -mgh_f[/math]

 

Now v final is a function of [math]h_f[/math]

 

[math]v^{2}_f = \sqrt{-2gh_f}[/math]

 

 

[math]\frac{dv_f}{dy} = \frac{-g}{\sqrt{-2gh_f}}[/math]

 

Is this correct?

[swansont]

It's usually best to leave the variables in and put the actual values in at the end.

 

[math]0.5*m*v^{2}_f = -mgy[/math]

 

 

[math]v= \sqrt{-2gy}[/math]

 

[math] \frac{dv}{dy} = \sqrt{\frac{-g}{2y}}[/math]

 

(and, of course, y is a negative number)

 

So the slope never reaches zero; what you have to do is put in the spring as well, but you won't get a nice differentiable function. You could see it graphically, though.

[hotcommodity]

Thank you for the reply. Ok, I get that it won't reach an absolute maximum, but if I consider "x" to be the distance that the spring compresses due to the block, can I think of that point as an end point graphically? In other words, can I plug in y = -[(.75m + x) sin(30°)] for the velocity function to obtain the maximum speed?

[swansont]

Max speed occurs at the point of contact (no compression), so I'm not sure why you'd want to do that.