Angular velocity of a door as a truck accelerates

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

Let's consider a cylindrical coordinate system whose z-axis coincides with the hinge and origin is the lowest point of the hinge.
Let's say that the truck moves along the x-axis.
W.r.t. this frame,
Torque about the hinge $\vec \tau = \vec r \times \vec F_{n-in} \\= [\frac w 2 \hat s + \frac h 2 \hat s ] \times [ mg (-\hat z )+m A (-\hat x)] \\\hat x = \cos \phi \hat s+ \sin \phi \hat \phi \\\vec \tau = \frac m 2 [ \{ wg - h \cos \phi \} \hat \phi - \hat z~w \sin \phi + h \sin \phi \hat s ]$
The body can rotate only about the hinge, so the torque should be only about the $\hat z$ direction. I can take the component of the torque in the $\hat \phi$ to be 0 as there is both negative and positive terms but how to make the component of the torque in the $\hat s$ zero?

 

[Orodruin]

First of all: You have written $\vec r = (w \hat s + h \hat s)/2$. This does not seem correct, the width and the height are in different directions.

Second: Which point are you computing the torque relative to? Is the inertial force the only force on the door that gives a torque relative to this point? If not: What other forces are there and how will they cancel the torque in the $\hat s$ direction?

Third: You do not need to compute the torque. There is a much more efficient way to solve this problem if you think a bit about it.

 

[J Hann]

Suppose the door were attached to a fixed horizontal axis (instead of the moving vertical axis).
And suppose that the acceleration due to gravity were A (instead of g).
Does that suggest a method of solution?
I think that Einstein postulated that a person in an isolated enclosure had no way of knowing
whether any experienced force would be due to an acceleration of the enclosure or due
to gravitational attraction by a massive object.

 

[alejandromeira]

Suppose the door were attached to a fixed horizontal axis (instead of the moving vertical axis).
And suppose that the acceleration due to gravity were A (instead of g).
Does that suggest a method of solution?
I think that Einstein postulated that a person in an isolated enclosure had no way of knowing
whether any experienced force would be due to an acceleration of the enclosure or due
to gravitational attraction by a massive object.

GREAT!

 

[Pushoam]

Does that suggest a method of solution?

What I understood is:
Let's keep the gate in the x-y plane such that it is free to rotate about x-axis. Its length along x-axis is h and y-axis is w.
Under the effect of gravity, it will rotate towards the earth. I have to calculate its angular velocity when it is in the x -z plane.

Now the center of mass motion is motion of the simple pendulum with length w/2.
So, the angular velocity is equal to the angular frequency : $\omega = \sqrt (\frac {2g}{w/2})$
Is this correct?

You have written ⃗r=(w^s+h^s)/2r→=(ws^+hs^)/2\vec r = (w \hat s + h \hat s)/2. This does not seem correct, the width and the height are in different directions.

It's $\hat z$.

Which point are you computing the torque relative to? Is the inertial force the only force on the door that gives a torque relative to this point? If not: What other forces are there and how will they cancel the torque in the ^ss^\hat s direction?

I was calculating about the origin of the hinge.
The other forces acting are friction forces, normal forces. There must be a component of contact force in the $\hat x$ direction because of which the gate remains attached to the car. The torque due to this component of the force will cancel the torque( earlier calculated) in $\hat s$ direction.
But since I don't have the detail knowledge of these forces, I can't use torque analysis.

There is a much more efficient way to solve this problem if you think a bit about it.

Is the other way already pointed out by J Hann?

 

[Orodruin]

So, the angular velocity is equal to the angular frequency : ω=√(2gw/2)ω=(2gw/2)\omega = \sqrt (\frac {2g}{w/2})
Is this correct?

No. First of all, the moment of inertia for a rod is not the same as the moment of inertia for a mass placed at its centre. Second, the motion is harmonic only for small deviations from the equilibrium point.

But since I don't have the detail knowledge of these forces, I can't use torque analysis.

You can pick a different origin around which the components that are not in the $z$ direction cancel by symmetry.

Is the other way already pointed out by J Hann?

That is part of it. You need to figure out how you can use this to solve the problem.

 

[J Hann]

You wrote:

Now the center of mass motion is motion of the simple pendulum with length w/2.
So, the angular velocity is equal to the angular frequency : ω=(2gw/2)'>ω=√(2gw/2)ω=(2gw/2) \omega = \sqrt (\frac {2g}{w/2})
Is this correct?

The door is not a simple pendulum but can be considered a compound pendulum.
Also, the basic pendulum equations only pertain to small displacements ( sin theta = theta).

 

[kuruman]

To elaborate on post #3 by @J Hann: Imagine being the truck driver. You are at rest with respect to the truck and so is the door hinge. As the truck accelerates, you are pinned to the back of your seat. You interpret this as an additional horizontal "gravitational" acceleration A. This acceleration A also acts at the center of mass of the door (see figure below). Thus, the situation is equivalent to one in which the truck is hanging vertically at rest, the door is open horizontally and the acceleration of gravity is A instead of g.

 

[Pushoam]

Let's keep the gate in the x-y plane such that it is free to rotate about x-axis. Its length along x-axis is h and y-axis is w.
Under the effect of gravity, it will rotate towards the earth. I have to calculate its angular velocity when it is in the x -z plane.

The door is not a simple pendulum but can be considered a compound pendulum.

The motion of the gate can be taken as pure rotation about x - axis.
Applying work kinetic energy theorem,
Work done by the torque = change in its rotational kinetic energy
$Mgw/2 = I \omega^2 /2 \\ \omega = \sqrt { \frac { Mgw}{I} }$
I about x-axis = $\frac {Mw^2 } 3$
$\\ \omega = \sqrt { \frac { 3g}{w} }$
Is this correct?
Here g = A.

 

[Orodruin]

That is correct.

 

[Pushoam]

In part b, is the horizontal force (being asked ) that component of contact force which keeps the door together with the truck? If this is so, then the horizontal force is in the direction of acceleration.
Or is it the force exerted by the truck on the door when the door is shut? Then it will be perpendicular to the acc. of the truck.

 

[Orodruin]

They are asking for the force from the truck on the door when it has swung 90 degrees. This would mean the door is pointing in the "back" direction. I suggest you again analyse the problem in the accelerated frame.

 

[Pushoam]

I suggest you again analyse the problem in the accelerated frame.

I solved it.
Thanks for the guidance.

 

[J Hann]

The motion of the gate can be taken as pure rotation about x - axis.
Applying work kinetic energy theorem,
Work done by the torque = change in its rotational kinetic energy
$Mgw/2 = I \omega^2 /2 \\ \omega = \sqrt { \frac { Mgw}{I} }$
I about x-axis = $\frac {Mw^2 } 3$
$\\ \omega = \sqrt { \frac { 3g}{w} }$
Is this correct?
Here g = A.

Are you sure you want to use torque?
The force producing the torque is only perpendicular to the door when the door is wide open.
Also, is there not a change in the potential energy since it would take work to open the door
while the truck is accelerating?
Although, I agree with the answer you obtained.

 

[Orodruin]

Also, is there not a change in the potential energy since it would take work to open the door
while the truck is accelerating?

He has taken this into account. The potential energy is on the left-hand side.

Are you sure you want to use torque?

He never used the torque itself, although that is possible, but only its potential energy.

 

[Pushoam]

Another attempt to solve part (a)
wrt accelerated frame,
Considering COM motion,
Applying work - kinetic energy theorem,
$MAw/2 = ½ Mv^2 = ½M {\omega}^2 (w/2)^2 \\ \omega = \sqrt {(2A/w) }$

What is wrong here?

 

[TSny]

The kinetic energy $\frac{1}{2}Mv^2$ due to motion of the COM is only part of the total kinetic energy. There is additional kinetic energy due to rotation about the COM. The total of these two contributions should equal the kinetic energy you calculated in post #9 (edited) as rotational kinetic energy about the hinges (x axis).

 

[kuruman]

The kinetic energy 12Mv2\frac{1}{2}Mv^2 due to motion of the COM is only part of the total kinetic energy. There is additional kinetic energy due to rotation about the COM. The total of these two contributions should equal the kinetic energy you calculated in post #14 as rotational kinetic energy about the hinges (x axis).

Or you can use $\frac{1}{2}I_h\omega^2$ for the kinetic energy, where $I_h$ is the moment of inertia about the hinge. Either way should give the same answer.

 

[Orodruin]

Or you can use $\frac{1}{2}I_h\omega^2$ for the kinetic energy, where $I_h$ is the moment of inertia about the hinge. Either way should give the same answer.

He already did that in post #9.