1. Jul 19, 2010

### Ush

1. The problem statement, all variables and given/known data
A particle having charge q=+2.30 µC and mass m=0.0139 kg is connected to a string that is L=1.30 m long and is tied to the pivot point P in the figure below.
[PLAIN]http://capa.physics.mcmaster.ca/figures/sb/Graph25/sb-pic2515.png [Broken]
The particle, string, and pivot point all lie on a horizontal table. The particle is released from rest when the string makes an angle θ=58.8° with a uniform electric field of magnitude E=294 V/m. Determine the speed of the particle when the string is parallel to the electric field (point a in the figure).

2. Relevant equations

Note: These are equations I think might apply, they are no equations given specifically for this question

E = -dV/dx
F = qE
τ = rFsinθ = Iɑ
U = qV
ѡf2 = wi2 + 2ɑΘ
v = ѡr

F = force
V = voltage
E = electric field
I = moment of inertia
ɑ = angular acceleration
q = charge
τ = torque
ѡ = angular velocity
d = angular displacement
v = velocity

3. The attempt at a solution

The question states that the system rests on top of a horizontal table. Therefore, I ignored the gravity / normal forces.

It looks like a torque question, so.

τ = rFsinθ
τ = r(qE)sinθ
τ = 7.519 x 10-4

T = Iɑ
ɑ = T/I

I = ∑miri2
I = 0139(1.3)2

ѡf2 = wi2 + 2ɑΘ
ѡf2 = 2ɑΘ
ѡf2 = 2(0.032)(58.8π/180)

v = wr
v = 1.3(0.256)
v = 0.333m/s

the computer says this is the wrong answer =/

Last edited by a moderator: May 4, 2017
2. Jul 19, 2010

### collinsmark

I haven't gone through your calculations in detail (I approached the problem a different way). But this problem should be a lot easier by invoking conservation of kinetic and potential energy.

What is the charge's electrical potential energy difference between the cases where θ=58.8° and θ=0°?

[Edit: I think I see where you went wrong in your calculations. It looks like you were assuming a constant angular acceleration, -- i.e. a constant torque. But that is not the case with this problem. If you want to use kinematics to solve this, you've have to use calculus and perform an integral. Alternately, you can use conservation of energy like I mentioned above and avoid the calculus.]

Last edited: Jul 19, 2010
3. Jul 19, 2010

### Ush

=o

E = -dV/dx

dV = -E∫dx

dV = -294(1.3cos58.8)

U = qV
U = -4.237E-4

-4.237E-4 = K
-4.237E-4 = 1/2mv2
how do I cancel out the negative sign??
If i pretend that negative isn't there..
v = 2.47m/s

4. Jul 19, 2010

### collinsmark

That should be ΔV = -[294(1.3cos58.8) - 294(1.3cos0)]

Remember, you're calculating the potential difference. You need to take both points into account and find the difference in voltage between them.

[Edit: or another way to think about it is to find the difference ΔLx between the cases where θ = 58.8o and θ = 0o. The use ΔV = (E)(ΔLx) which applies for a uniform electric field point along the positive x-axis.]

Last edited: Jul 19, 2010
5. Jul 19, 2010

### Ush

ah, okay perfect!
thanks!
ΔV = -E[(1.3cos58.8) - 1.3(294)]
ΔV = 294(1.3) - 294cos58.8)

I wish I had gotten a hint or something saying "apply conservation of energy"... but that never happens =p

I was hoping the torque method also worked =/