# My teacher assigns too much work

1. Nov 8, 2007

### john84

So my teacher handed out a practice test in preperation for our test tomorrow, and I understand a majority of the problems, but need help with a few problems. Any help would be greatly appreciated.

1. A mass m is attached to a light string of length R, and the other end of the string is attached to a fixed point above, making a pendulum. The mass is pulled out to the side so the string makes an angle (theta) with the vertical

a. If the mass is released in this position, what is the tension in the string immediately after the release? (I got that T=mgcos(theta). )
b. If the mass is released in this position, what is the tension in the string at the moment when the mass passes through the bottom of the swing? (No idea)
c. Instead of being released from rest, the mass is given an initial speed tangent to the circle downwards (SE in the diagram). Find the minimum value of v so that the mass travels in a complete vertical circle of radius R (No idea whatsoever)

2.A special spring is constructed in which the restoring force is in the opposite direction to the displacement, but is proportional to the cube of the displacement; ie, F=-kx^3

This spring is placed on a horizontal frictionless surface. One end of the spring is fixed, and the other end is fastened to a mass M. The mass is moved so that the spring is stretched a distance A and then released. Determine each of the following in terms of k, A, and M.
a. The potential energy in the spring at the instant the mass is released. (U=((KA^4)/4)
b.The maximum speed of the mass (square root ((KA^4)/2M))
c. The displacement of the mass at the point where the potential energy of the spring and the kinetic energy of the mass are equal (Once again, no idea)

3. I've got this problem down and don't need any help, although I'll post it if anyone's interested.

Again, thanks for any help at all.

Last edited: Nov 8, 2007
2. Nov 9, 2007

### Zorodius

you're almost there. The potential energy at an arbitrary x is 1/4 kx^4. The potential energy lost between that position and the initial position is equal to the kinetic energy gained, and since it was released from rest, the initial kinetic energy was zero.

So, K = U0 - U kinetic energy is the difference between U initial and U(x)

and you want to solve for:

U = K
U = U0 - U
2U = U0

1/2 kx^4 = 1/4 ka^4
x^4 = 1/2 a^4

x = + or - a * the fourth root of 1/2

I hope that's right

3. Nov 9, 2007

### john84

Wow. Thank you so much, I finally get it. I just reached that point where everything clicks and I understand the problem better now. Thank you very much. Anyone else?

4. Nov 9, 2007

### Staff: Mentor

The pendulum problem is the basis of the Charpy impact test, so it is a rather practical problem.

T = mg cos $\theta$ is correct.

Well there is the weight and there is the centripetal force since the pendulum is moving.

Assuming that the string has negligible mass, we are only interested in the change in elevation of the mass, i.e. the change in gravitational potential energy between the starting elevation and the bottom of the arc. The bottom of the arc is at R from the pivot and the starting distance below the pivot is simply R cos $\theta$.

c. In order for the mass to go half way around, it must increase in elevation by D = 2R from the bottom of the arc. What velocity would be needed to have the kinetic energy equivalent to the change in gravitational potential energy associated with a change in elevation of 2R?