# Can someone check my answer for this work-energy problem?

1. May 17, 2015

### toesockshoe

1. The problem statement, all variables and given/known data
A mass attached to a string of length L is released from rest from an initial horizontal position (although the diagram implies it is not that high, please ignore that part of the diargram - sorry). From veritcally below where the string is attached at its top, there is a peg a distance x below the top attachment of the string. As the mass swings down to its lowest position, the string then touches the peg and the mass swings up and around from this new point (i.e. the peg's position). The mass continues to swing upward. Find the tension in the string when the mass is at its highest vertical point.

The diagram is here: http://nebula.deanza.edu/~newton/4A/4AHWSet6.html (it is number 5).

2. Relevant equations

$$W_net=\Delta E$$
$$\vec{F} = m\vec{a}$$

3. The attempt at a solution

I think the highest point is when it goes around the peg and is completely vertical. (I think I am supposed to disregard the position when the peg is just released from rest, in which that would be the highest point).

so I first found the velocity at the point:

System: Earth and Mass

$$W_{net}=\Delta E$$
tension force does no work becuase it is normal to the displacement of the mass
so:
$$0 = \Delta GPE + \Delta KE$$
$$0 = -mg(2x-L) + \frac{1}{2}mv_f^2$$
2x-L is the change of vertical displacement from the position at rest and when it vertically at the highest point after going passed the peg. so:

$$v_f=\sqrt{2g(2x-L)}$$

Now, I am trying to find the tension by doing an $$\vec{F}=m \vec{a}$$ problem

Let the system be the mass with the y direction going up and the x direction to the left:

$$F_t - F_g = ma_c$$
$$F_t = \frac{mv^2}{L-x}+mg$$
^the above step is where I am a bit unsure of myself... do I need to include gravity into my centripetal acceleration? if so? can I just say that the centripetal acceleration is as follows:

$$a_c=\frac{mv^2g}{L-x}$$ ???

ok so i continue and I plug in the velocity I found through the energy work above and plug it into the velocity variable in the F=ma solution to get:

$$F_t = \frac{2mg(2x-L)}{L-x} +mg$$

can someone check my work? I'm not sure if my answer makes sense because if L=x, then my Force value would go berserk...

2. May 17, 2015

### haruspex

I gather you are taking g as a positive quantity. Consider the mass at speed v at the top of the loop around the peg. Would a greater g increase or decrease the tension?

3. May 17, 2015

### toesockshoe

my apogies: i would like to change my F=ma coordinate system as y going downard and I would have:

Fg -Ft = ma

then my final answers would be:

$$mg - \frac{2mg(2x-L)}{L-x}$$

is that correct?

4. May 17, 2015

### haruspex

You still have a sign wrong somewhere.
What are the forces on the mass at the point of interest? Which way does each act? Which way does their resultant act?

5. May 18, 2015

### yawaraf

I got
$$F_T = -mg - \frac{2mg*(-L+2*x)}{L-x}$$

As for what happens when L=x.. well... I'm not sure there will be any tension force using the equation you got above with ac, since there will be no centripetal acceleration and the whole thing doesn't make sense (the string isn't long enough to go around the peg: no centripetal acceleration, so your F=ma equation will need to be modified).

See you in class tomorrow! Only a few more weeks then Exam 2, and Final. I will give Newton a nice shiny red apple on exam day.

Last edited: May 18, 2015
6. May 18, 2015

### toesockshoe

haha cool. see ya!

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