Kinematics of Rigid Bodies question

[DTskkaii]

Homework Statement

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.

Homework 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)

The Attempt at a Solution

So I eventually want to realize 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!

 

[andrien]

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

 

[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

 

[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.

 

[DeeNos]

I would approach this problem by first analyzing the forces acting on the pole. Since the pole is hinged to the truck bed, it is free to rotate about that point. The only force acting on the pole is the normal force from the truck bed, which will provide the centripetal force to keep the pole in circular motion.

Next, I would use Newton's second law, F=ma, to relate the normal force to the acceleration of the pole. Since the acceleration is given as 0.9m/s^2, we can calculate the normal force as F=m(0.9m/s^2). However, since we do not have the mass of the pole, we cannot solve for the normal force at this point.

To find the angular velocity, we can use the equation ω=(ωo^2+2aθ)^(1/2), where ωo is the initial angular velocity (which is 0 since the pole is at rest) and θ is the angle through which the pole has rotated (which is 90 degrees or π/2 radians). Plugging in the values, we get ω=(0^2+2(0.9m/s^2)(π/2))^(1/2) = 0.9(π/2)^(1/2) ≈ 1.2 rad/s.

In conclusion, the angular velocity of the pole as it reaches the horizontal position is approximately 1.2 rad/s. However, without the mass of the pole, we cannot determine the normal force or the exact value of the angular velocity. It is important to always include all relevant information in a problem to accurately solve it.