No problem :)
re: comparing it with other vehicles, I'm not sure what you're thinking of comparing. Top speed? Time to get to that speed? Max power output? Total power output? Energy efficiency? (this would have to be experimentally-tested)
Re: velocity, the final velocity right before the...
That's right. And yes, the change in gravitational potential energy is mgh, where h is the vertical distance from the "up" position to the "down" position.
Now, the formula for rotational kinetic energy of an object is 1/2 I omega^2, where I is the rotational inertia. So you can get expressions...
That's right if you're measuring angular velocity in revolutions/second, but for most physics we use radians/second, so the factor of 2 pi is canceled out and we have:
linear velocity = angular velocity x r
v = r * omega
Now, you need to find the linear kinetic energy of your vehicle when...
You forgot the constant in your integration. What you have is an initial condition problem: v(0)=5. So when you integrate both sides, you get
ln|v| = -1/2 t + c
v=e^(-1/2 t+c)
It makes things neater to get that constant out of the exponent, using the laws of exponents:
v=e^c*e^(-1/2 t)...
To address georgeh's question to AKG, the rank of a matrix is the number of linearly independent vectors it contains. So let's say you have an MxN matrix of rank R. Then it contains R linearly independent columns (or rows), and the rest, if any, are dependent on that set. All you have to show is...
When solving a separable differential equation, you treat the derivative (dv/dt) as a fraction. Get all the v's and dv's on one side and all the t's and dt's on the other side:
dv/dt=-1/2 v
1/v dv=-1/2 dt
Then integrate both sides and solve for v.
Alright. Well, if there are no numbers, then you're just supposed to find a symbolic expression for the quantities. You gave an expression for mu in your first post: mu = N (A x I). Plug that into the formula for torque, and you'll have your expression for torque. You can assume that A is a...
Lack of knowledge is okay :) It just looks like you're trying to do something that's a little beyond your physics level. But you can do it.
re: velocity, it will have one when the weight drops off, won't it?
Your first step will have to be finding a general formula for the vehicle's linear...
I would approach this problem using conservation of energy: the change in gravitational potential energy of the block is equal to the change in kinetic energy of the (block+vehicle) system. But it's a bit more complicated than it looks because you need to include both the rotational kinetic...
I'm sure you can get it. Have you figured out v in terms of r and omega? Think about it: the block is moving at the same speed as the string holding it, which is moving at the same speed as the outer edge of the reel. The reel is spinning at omega radians per second, and the outer edge is moving...
Take a close look at the problem. Does it appear to be (e^x)^(5/3) or is it e^(x^(5/3))?
The first is relatively easy (simplify and substitute). If in doubt, I'd assume this interpretation.
The second is really, really ugly. I can't figure out how to do it, my TI-89 chokes on it, and Maple...
Well, I would start by solving for the linear velocity of the block in terms of omega and the vertical distance the block moves in terms of d and theta.
Then I'd set up a conservation of energy equation: (rotational kinetic energy of reel in terms of I and omega) + (linear kinetic energy of...
I'm assuming the units on the three numbers in your first line are kg, m, and s respectively?
You're going to want to find the acceleration a first using one of your kinematics equations. Then identify the mass from your given information and plug m and a into your equation for force.
The angular speed of the reel is related to the linear speed of the block. Conservation of mechanical energy might tell you the total kinetic energy of the system, from which you can get the speeds.