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Work-energy questions
By Anonymous on Monday, April 23, 2001 - 05:11 pm:

Hi,

A particle of mass m is attached to end A of a light inelastic string AB of length l which is fixed at B. The particle is held with AB horizontal and is prohjected with speed u. Show that the least value of u for the particle to describe complete circles is independent of the direction of projection. Calculate this least value.

I am a little stuck on this question.

When AB is horizontal and projected with speed u, and final speed at the top of the circle needs to be v>0. So we can use the work-energy principle,

0.5mu2 - 0.5mv2 = mgl
hence u2 - v2 = 2mgl.

What do I do next. I am not sure. It would be nice also if someone could explain how to show that the direction does not matter (vertically down or up - is that what it means?) And how to calculate the value of u.
By Anonymous on Monday, April 23, 2001 - 05:20 pm:

I just did this bit more on the question

Using f=ma inward radius, when it is at the top of the circle,

T + mg = (m.v2)/r
But T=0 when at max height, so
putting this equation and the one in the last post together you get u2=3gl. What have I managed to show? And have I fully answered the question yet, I suspect not. Please help.

By Geoff Milward (Gcm24) on Monday, April 23, 2001 - 06:19 pm:

I think you have most of the ingredients. Just run the problem through logically from there.
We need the particle to be projected with sufficient energy such that it reaches a height l above the point B with sufficient velocity, v, so that the centripetal acceleration (v2/l) matches the acceleration due to gravity (g).

Take your datum at A and thus total energy as particle projected at u
E = KE + PE = ½mu2 + 0 = ½mu2
we want the particle to reach B with velocity v such that we have circular motion, that is so that
mg = mv2/l as the radius of circle is length of string
solve for v...
v2 = lg
put into energy expression at top of circle
E = KE+PE = ½mv2 + 2lmg
but we know v2 = lg for circular motion
and thus
E= 2.5×lmg at top
use conservation of energy to equate E's thus
2.5×lmg = ½mu2
so u2= 5lg, independent of mass

To show direction is independent of direction just use the rotational symmetry of the problem (and I think that they mean that the particle is projected horizontally).

Hope this helps (& DO CHECK MY ALGEBRA; as a theoretical physicist it is notoriously bad)

Geoff

By Anonymous on Monday, April 23, 2001 - 06:47 pm:

It helped very much.
Thank you Geoff.

By Anonymous on Tuesday, April 24, 2001 - 11:08 am:

Hi,

I am getting confused how to set up the equations to answer this question. I want to answer this question by considering Elastic Potential Eneregy, Kinetic Energy, and Gravational Potential Energy.

A particle of mass 0.5kg is attached to one end A of a light elastic string of natural length 2m and modulus 20N. The other end B of the string is fixed to a point on the ceiling. The particle is held at a distance of 1.5m vertically below B and then released.
Calculate (a) the length of the string when the particle reaches its lowest point, (b) the speed of the particle when it passes through its equilibrium position.

I would appreciate it if someone could take me step by step through this question. Thanks.
By Kerwin Hui (Kwkh2) on Tuesday, April 24, 2001 - 04:02 pm:

Initially the particle has GPE -0.5×9.8×1.5 relative to B, and no KE nor EPE.

At the lowest point, the string is of length l>2, the EPE is (1/2)×(20/2)×(l-2)2. There are no KE, and the GPE relative to B is -0.5×9.8×l.

Now equate the energies

-0.5×9.8×1.5=5(l-2)2-0.5×9.8×l

So we solve the quadratic to find l (only the root l>2 is valid).

Similarly, we work out the equilibrium position for (b), and the corresponding EPE and GPE. Equate energies again to obtain KE, and hence the speed of particle at that point.

Kerwin

By Anonymous on Tuesday, April 24, 2001 - 05:26 pm:

Thank you Kerwin.
I shall give that a go.

By Anonymous on Tuesday, April 24, 2001 - 12:00 pm:

Hi,

I am not sure how to set this situation up and answer. I know how to do bits of it but not fully.

A particle P of mass m is attached to one end of a light elastic string of natural length l whose other end is attached to a point A on a ceiling. When P hangs in equilibrium AP has length (5l/3).
Show that if P is projected vertically downwards from A with speed (3gl/2)0.5, P will first come to instantaneous rest after moving a distance (10×l/3).

Hope someone can help me do this question using EPE, EP, KE. Can you show me also your method and reasoning. These work-energy problems are causing me some problems.
By Peter Conlon (P2714) on Tuesday, April 24, 2001 - 03:18 pm:

First, we find the modulus of elasticity of the string:
Tension in the String = Weight of Particle
at equilibrium.

Using Hooke's law
lx/l = mg, but the extension is 2l/3 from the question, so:
l = 3mg/2

Initially, measuring potential from the ceiling:

G.P.E of P = 0
K.E of P = 0.5mv2 = 3mgl/4
E.P.E = 0

At instantaneous rest:

G.P.E of P = mgh = -mg(l+x) (-ve because the particle is lower)
K.E of P = 0 (the particle is at rest)
E.P.E = lx2/2l =3mgx2/4l

where x is the extension of the string.

Now, using the principle of the conservation of energy, the total energy at the beginning must equal the total energy at the end:

G.P.E + K.E + E.P.E = G.P.E + K.E + E.P.E

0 + 3mgl/4 + 0 = -mg(l+x) + 0 + 3mgx2/4l

rearranging and factorising etc gives:

(3x-7l)(x+l) = 0

So the extension at instananeous rest is 7l/3. (we can ignore the negative value)

As the particle started at the ceiling, it travelled l to get to the end of the string, and then, 7l/3 before coming to instantaneous rest. Total distance travelled, therefore, = 10l/3 as required.

Hope this helps,

Peter

By Anonymous on Tuesday, April 24, 2001 - 03:57 pm:

Thank you Peter.

I understand your method of approach for this question.

I know it does not matter, but how could we have also used the fact the we are told that the equilibrium level was (5l/3)?

Also when it asked us the distance it moved till coming to rest, did it mean from the equlibrium position or the natural length? Am a little confused on this part. If it wanted us to get 10l/3, then it must be from the natural length as you said. But I found this to be a little ambiguos. What do you think? Or am I reading the question in the wrong context?

By Peter Conlon (P2714) on Thursday, April 26, 2001 - 10:15 am:

Hi Anonymous,

Sorry this is a bit late.

The reason the question told us that the equilibrium length is 5l/3, is so that we can work out the modulus of elasticity, l.

We needed this to do the rest of the question involving energy. As far as I know, this is the only use for that information.

The question wants to know the distance travelled from where the particle started, to where it was at instantaneous rest.

>Show that if P is projected vertically downwards from A. A is on the ceiling

This tells us that the particle starts at A, on the ceiling. In our calculations, we worked out the extension at instantaneous rest. So the total distance travelled was from the ceiling to the natural length, when the particle was in free fall, plus the distance from the natural length to the extension, when it was being slowed by the elastic string, hence l + 7l/3 = 10l/3

Hope this is clearer,

Peter