# Homework Help: Kinematics of Rigid Bodies question

1. May 13, 2012

### DTskkaii

1. The problem statement, all variables and given/known data
The uniform 3.6m pole is hinged to the truck bed and released from the vertical position as the truck starts from rest with than acceleration of 0.9m/s^2. If the acceleration remains constant during the motion of the pole, calculate the angular velocity of the pole as it reaches the horizontal position.
Diagram attached.

2. Relevant equations
I believe these equations are relevant, however, I am not given a mass for the pole, so i'm not entirely sure.
a(tangential)=mrθ''
a(normal) = mrω^2
ƩMo=Iθ''+Ʃma(vector)d
I=k^2m
ω=2Vx
ω=(ωo^2+2aθ)^(1/2)

3. The attempt at a solution
So I eventually want to realise the angular velocity ω.
I have done a similar question that utilised energy and momentum methods, however, with a negligible mass, i'm not sure whether that will affect the equations, since the example question used mass.

So essentially, I believe it may be using the last formula I provided, since it is not determined by time or mass.
So, all I need to find is the acceleration, since I already know that ωo is at rest, and θ=90. Finding the acceleration, from the given positive direction of acceleration of the truck body, is something that I can't figure out.

Thankyou for any feedback!

#### Attached Files:

• ###### 000002.jpg
File size:
66.5 KB
Views:
300
2. May 13, 2012

### andrien

your last formula is valid for constant angular acceleration,so from vertical to horizontal position does it remain the same.

3. May 13, 2012

### DTskkaii

@ Andrien
Well, the pole accelerates to the left (opposite the direction of the car) at 0.9m/s^2, and accelerates downward at 9.8m/s^2.
These are both constants, but I don't know which to use in the formula.
It can't possibly as simple as angular vel=(0^2+2*0.9*90)^1/2

4. May 14, 2012

### andrien

consider taking moment about hinge point, and write ma as a pseudo force along with gravity which will act towards left i.e. opposite to acceleration of truck.integrate from 0 to pi/2,you will get
w=[3(g+a)/l]^.5, where l= length of rod.assuming k^2=l^2/12,
w is around 3.