# Work and kinetic energy (dynamics)

1. Feb 5, 2012

### xzibition8612

1. The problem statement, all variables and given/known data

See attachment.

2. Relevant equations

(1) F=ma
(2.13) W=ΔT
(2.20) (weight) x (decrease in altitude of mass center of body)

3. The attempt at a solution

I don't get where |v(t1)| and |v(t2)|^2 comes from. The book says |v(t1)| = 0, which I suppose is because initially there was no movement, so kinetic energy is 0? Does |v(t1)|=0 have anything to do with T=mgsinθ+mlθ'^2?

Then it says |v(t2)|^2 = l^2θ'^2, and this totally confused me. First, why is v(t2) squared and not v(t1)? Second, obviously v(t2) is somehow related to T=mgsinθ+mlθ'^2, but I don't see how you get v(t2) from that. Also originally θ'^2 was already squared, so wouldn't |v(t2)|^2 = l^2θ'^4??
Very confused on this.

Any help would be appreciated thanks.

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2. Feb 5, 2012

### rock.freak667

|v(t1)| = 0 as no movement at t=t1.

|v(t2)|^2 = l^2θ'^2 because v(t2)= radius x angular velocity. So it is just squared. (I am ignoring the modulus signs now).

For the second part, the change in PE = change in KE

so the height went from lsin(30) to lsin(θ) so the change in PE is mg[lsinθ-lsin30]

similarly change in KE is 1/2mv22-1/2mv12 = 1/2mv22 = 1/2m(lω)2 = ½ml2ω2

Do you understand how they the first part for the tension T at least?

3. Feb 5, 2012

### xzibition8612

I understand where T=mgsinθ+mlθ'^2 came from. But for |v(t2)|^2, why is l squared but theta dot not to the 4th power? I totally don't get this. And PE = ΔKE is the same as W=ΔKE right? Thanks.

4. Feb 5, 2012

### rock.freak667

Right, for the velocity, they multiply the radius (l) by the angular velocity (theta dot or ω)

so v(t2)=lω so that v(t2)2=l2ω2

Yep, same idea.

5. Feb 5, 2012

### xzibition8612

so |v(t2)|^2 has absolutely nothing to do with T=mgsinθ+mlθ'^2? Meaning it wasn't derived from there or anything? It's just something you should intuitively know?

6. Feb 6, 2012

### rock.freak667

Well the kind of did it backwards in a way.

to get mg[lsinθ-lsin30]= 1/2mv22

and it would be better to have everything in terms of θ so putting $v_2= l \dot{\theta}$ is easier.

so you'd get

$$mg[lsin\theta - lsin30]=\frac{1}{2}ml^2 \theta ^2 \Rightarrow mg[sin\theta -sin 30]=\frac{1}{2}ml \dot{\theta}^2$$

and if we look at the the $ml \dot{\theta}^2$ you'd see that is mass*radius*angular velocity squared i.e. centripetal force.

The only way that relates to tension T is from the first line

$$T+mgsin\theta = ml \dot{\theta}^2$$