Another electric field question, =/

[Ush]

Homework Statement

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
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).

Homework 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
r = radius
v = velocity

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 = ∑m_ir_i²
I = 0139(1.3)²

ɑ = 0.032rad/s²

ѡf² = wi² + 2ɑΘ
ѡf² = 2ɑΘ
ѡf² = 2(0.032)(58.8π/180)
ѡf = 0.256rad/s

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

the computer says this is the wrong answer =/

 

[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.]

 

[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

this is the correct answer

 

[collinsmark]

=o

E = -dV/dx

dV = -E∫dx

dV = -294(1.3cos58.8)

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 ΔL_x between the cases where θ = 58.8^o and θ = 0^o. The use ΔV = (E)(ΔL_x) which applies for a uniform electric field point along the positive x-axis.]

 

[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 =/

thanks for your help