# Uneven Seesaw -> angular acceleration of motion

## Homework Statement

http://img713.imageshack.us/img713/9151/screenshot20110501at113.png [Broken]

## Homework Equations

Angular acceleration = Torque/Inertia
Torque = Fd
Mass_moment_Inertia=mr^2

## The Attempt at a Solution

->assume masses on the end of seesaw are point particles for calculating inertia. is this the correct thing to do?
->is this formula for angular acceleration fine for the instant at release?

I=mr^2
I=m[d-a]^2-ma^2
I=m[(d-a)^2 - a^2]

Torque = Fd
Torque = mg[(d-a)-a]

Angular acceleration= Torque/Inertia
[(d-a)-a]mg/[(d-a)^2 - a^2]m ---->cancel m from top and bottom
=g/(d-2a) ---->canceled (d-a)-a from the top and bottom (because equivalent to rg/r^2)

I think my maths at the end might be dodgy, maybe my assumptions are incorrect too.

Edit: Or should I somehow be using the equations of motion for this one? I can't see how though :'(

Last edited by a moderator:

PhanthomJay
Homework Helper
Gold Member

## Homework Statement

http://img713.imageshack.us/img713/9151/screenshot20110501at113.png [Broken]

## Homework Equations

Angular acceleration = Torque/Inertia
Torque = Fd
Mass_moment_Inertia=mr^2

## The Attempt at a Solution

->assume masses on the end of seesaw are point particles for calculating inertia. is this the correct thing to do?
In the absence of other data, this assumption is valid.
->is this formula for angular acceleration fine for the instant at release?
the problem asks for the angular acceleration of motion, not just the angular acceleration at the point of release
I=mr^2
I=m[d-a]^2-ma^2
I=m[(d-a)^2 - a^2]
Why are you subtracting here?
Torque = Fd
Torque = mg[(d-a)-a]

Angular acceleration= Torque/Inertia
[(d-a)-a]mg/[(d-a)^2 - a^2]m ---->cancel m from top and bottom
=g/(d-2a) ---->canceled (d-a)-a from the top and bottom (because equivalent to rg/r^2)

I think my maths at the end might be dodgy, maybe my assumptions are incorrect too.

Edit: Or should I somehow be using the equations of motion for this one? I can't see how though :'(
You should write the expression for torque as a function of theta.

Last edited by a moderator:
So...
I=m(d-a)^2+m(a)^2?
Torque=d*F*sin(theta)
angular accel=Torque/I
m*g*sin(theta)*((d-a)-a)/m(d-a)^2+m(a)^2
angular accel=g*sin(theta)*((d-a)-a)/(d-a)^2+a^2 ------>factorised denominator, got rid of m

Is this any better? Is that the expression for Torque required?

gneill
Mentor
Torque=d*F*sin(theta) when theta is the angle between the force vector and the radius vector (rotation arm). In this case the angle θ as shown in the diagram is not this angle.

It might be better to think in terms of finding the component of the force that is perpendicular to the lever arm, multiplying that by the radius vector magnitude. In this case I think you'll find that the perpendicular components of the force are given by m*g*cos(θ).

#### Attachments

• Fig1.jpg
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Ahh I see thanks gneill. Yeah, the diagram illustrates your point nicely.*blushes* silly mistake.