Advanced 2-D Motion Problem - Bead on Wire

In summary: Ok so this is what I tried so far. For the first part, 1.1, there are only 2 forces acting on the object, the normal force of the wire, and kinetic friction. Now, I made the assumption that the normal force is the centripital force, which I believe is a safe assumption. Therefore, for the radial direction, it becomes:F = m a_RNormal = m a_R mv^2 / r = m a_R --- > v^2/r = a_RFor the second equation, I assumed the force of friction was mu_k*N. Therefore, it became mu_k * mv^
  • #1
Runelord
6
0

Homework Statement



"Consider a bead of mass m that is free to move around a horizontal, circular ring of wire (the wire passes through a hole in the bead). You may neglect gravity in this problem (assume the experiment is being done in space, far away from anything else). The radius of the ring of wire is r. The bead is given an initial speed v_0 and it slides with a coefficient of friction mu_k. In the subsequent steps we will investigate the motion at later times. You should begin by drawing a free-body diagram at some instant of time. Note that there will be a radial acceleration, a_R, and a tangential acceleration, a_T, in this problem.

1.1 Write Newton's 2nd law for the radial and tangential directions.

1.2 Combine the above equations to write a differential equation for dv/dt, where v is the speed at time t.

1.3 Solve the above differential equation to determine v(t). The solution has the form v = c1/(1+c2*t) - find c1 and c2. Hint: if v_0 = 3 m/s, mu_k = 0.1, r = 10 cm, and t = 3 s, v(3) = 0.3 m/s. "

Thats just the entire problem for reference.


Homework Equations



F_c=mv^2/r, F_f=u_k N, F = ma, a = a_R^2 + a_T^2


The Attempt at a Solution



Ok so this is what I tried so far. For the first part, 1.1, there are only 2 forces acting on the object, the normal force of the wire, and kinetic friction. Now, I made the assumption that the normal force is the centripital force, which I believe is a safe assumption. Therefore, for the radial direction, it becomes:

F = m a_R
Normal = m a_R
mv^2 / r = m a_R --- > v^2/r = a_R

For the second equation, I assumed the force of friction was mu_k*N. Therefore, it became mu_k * mv^2 / r. Using this information, the net force in the tangential direction became:

F = m a_T
mu_k *mv^2/r = m a_t ---> mu_k *v^2/r = a_T

Now things get dicey for part 2. I am not sure how to combine these equations. I'm fairly certain a= root(a_R^2 + a_T^2) (pythagorean theorem), and from there you use a = dv/dt and solve that, but it gets messy. If I try that out, it becomes something like this:

a = root(v^4/r^2 + mu_k^2 * v^4/r^2) <factor our v^4/r^2 out of square root)
dv/dt = v^2/r * root (1 + mu_k^2)

For some reason that doesn't sit well with me. Because for part 1.3, when I'm thus solving for the constants c1 and c2, it simply doesn't work. What am I doing wrong? If I integrate that, the answer I got above, this is what I get:

dv/dt = v^2/r * root (1 + mu_k^2) <--- r = .1 m, mu_k = .1
..
dv/v^2 = 10 root (1.01) dt
..
-1/v = 10 root (1.01) t + C


This doesn't work with the initial conditions. What am I doing here?
 
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  • #2
Don't banish me to page 2 :(
 
  • #3
Have respect for vectors!

Runelord said:
I'm fairly certain a= root(a_R^2 + a_T^2) (pythagorean theorem), and from there you use a = dv/dt and solve that, but it gets messy.

No, you've treated acceleration (a) and velocity (v) as scalars (numbers without direction). They're vectors (numbers with direction), and you must respect that!

You must work out the acceleration and velocity in one direction at a time.

You did it for the radial direction, and correctly used it to calculate the "effective weight" and therefore the friction force.

Now use that value of the friction force to calculate the acceleration in the tangential direction (as a function of the speed in that direction)! :smile:
 

Related to Advanced 2-D Motion Problem - Bead on Wire

1. What is an "Advanced 2-D Motion Problem - Bead on Wire"?

An "Advanced 2-D Motion Problem - Bead on Wire" is a physics problem that involves calculating the movement of a bead on a wire in two dimensions. This type of problem typically requires knowledge of kinematics, forces, and energy to solve.

2. How do I approach solving an "Advanced 2-D Motion Problem - Bead on Wire"?

To solve an "Advanced 2-D Motion Problem - Bead on Wire", you should first draw a free body diagram of the bead and identify all the forces acting on it. Then, use kinematic equations and/or energy equations to solve for the unknown variables.

3. What are some common mistakes when solving an "Advanced 2-D Motion Problem - Bead on Wire"?

Some common mistakes when solving an "Advanced 2-D Motion Problem - Bead on Wire" include forgetting to account for all the forces acting on the bead, using the wrong kinematic equations, and not properly setting up the equations to solve for the unknown variables.

4. Can I use calculus to solve an "Advanced 2-D Motion Problem - Bead on Wire"?

Yes, calculus can be used to solve an "Advanced 2-D Motion Problem - Bead on Wire". In fact, calculus is often necessary when the problem involves non-constant acceleration or the bead's position or velocity is given as a function of time.

5. What are some real-world applications of "Advanced 2-D Motion Problem - Bead on Wire"?

Problems involving a bead on a wire can be found in various fields such as engineering, physics, and biology. For example, the movement of a bead on a wire can model the motion of a roller coaster car on a track, the trajectory of a projectile, or the behavior of a cell membrane in biophysics.

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