Variable angular acceleration and velocity

[REVIANNA]

Homework Statement

This is a general query not a specific problem.

In several questions in rotational dynamics they ask the trend in angular acceleration and angular velocity.In one of them they released a rod from horizontal position and asked if the angular velocity and angular acceleration are increasing/decreasing In another one they release a square sheet but the trend is different.
Is there a concrete and easy way to figure this out.have a very happy new year!

 

[haruspex]

Homework Statement

This is a general query not a specific problem.

In several questions in rotational dynamics they ask the trend in angular acceleration and angular velocity.In one of them they released a rod from horizontal position and asked if the angular velocity and angular acceleration are increasing/decreasing In another one they release a square sheet but the trend is different.
Is there a concrete and easy way to figure this out.have a very happy new year!

I'd need to see the full questions.

 

[REVIANNA]

I'd need to see the full questions.

The square is pivoted at vertex D and initially held at rest so that sides AB and CD are horizontal. After it is released, the plate swings downward, rotating about the pivot point.

https://courses.edx.org/asset-v1:MI...k/pivoted_falling_square_with_overlay_cac.svg
A rod of mass m is pivoted in the horizontal position as shown (black point). The rod is at rest and then released.

A uniform bar of mass m and length L is pivoted at one end and is held vertically as shown. After the bar is released, it swings downward with no friction in the pivot.

these questions are getting on my nerves.Because they seem the same but are not.

for the fist one α and ω both are increasing and for the second one α decreases while ω increases .(I don't know why)and the third one is the weirdest as the answer says that the α=0
so I thought I could calculate ω with mechanical energy conservation considering the intermediate state and the final state but that's wrong too.
$mg(L/2)=0.5 *((mL^2/12)+(ml^2/4)) *ω^2$
$ω=sqrt(3*g/L)$

but the answer has 6g/L

what am I missing?

edit:
if in the 3rd question I consider the topmost position(1st state) as 0 than the centre of mass falls by L
(l/2 from vertical to horizontal and l/2 again from horizontal to vertical )
and put L instead of L/2 in the above equation the answer is correct but if the angular velocity is const every state must work ,right?

edit 2:
does the velocity vector points radial inward at all points in the third problem?

 

[cnh1995]

If I understand it correctly, in the 3rd problem, centre of mass of the rod undergoes circular motion with variable linear acceleration 'a', hence with variable angular acceleration α(al/2). It's not clear to me what exactly is asked here in the 3rd problem.

 

[haruspex]

The difference between the three arrangements is where the pendulum starts in its motion. In the first, it starts at 45 degrees above horizontal. In all cases, maximum angular acceleration will be in the horizontal position, so in the first one it initially increases. In the second, I would have guessed the rate of change of angular acceleration was initially zero, but soon goes negative, giving a reducing angular acceleration.

I think you have not completely stated the third question. Without some initial nudge, there is no reason for it to fall. Certainly there is no initial angular acceleration. Why would there be any? There is no torque about the pivot when the rod is vertical. But the angular acceleration soon increases, so the angular velocity increases too.

Didn't understand your Edit 2. Do you mean to ask whether the acceleration vector is radial?

 

[REVIANNA]

maximum angular acceleration will be in the horizontal position

why?
and what about ω ?
I don't seem to understand ω and α properly.
when I was learning just linear acceleration it used to be mostly const.

the rate of change of angular acceleration was initially zero

how should I think about it -in terms of forces,energy or intuition or their linear brothers v and a.

 

[REVIANNA]

I think you have not completely stated the third question. Without some initial nudge, there is no reason for it to fall. Certainly there is no initial angular acceleration. Why would there be any?

A uniform bar of mass m and length L is pivoted at one end and is held vertically as shown. After the bar is released, it swings downward with no friction in the pivot.
At the instant the bar is at the bottom of the swing, find the magnitudes of the following quantities:
1-ω

I consider the topmost position(1st state) as 0 than the centre of mass falls by L
(l/2 from vertical to horizontal and l/2 again from horizontal to vertical )

mg(L/2)=0.5∗((mL2/12)+(ml2/4))∗ω2

instead of L/2 I put L and I get the answer

2-The angular acceleration of the bar α and
The horizontal and vertical component of the acceleration of the free end of the bar

what will happen when it becomes vertical again? will it come to rest?

 

[haruspex]

why?
and what about ω ?
I don't seem to understand ω and α properly.
when I was learning just linear acceleration it used to be mostly const.

Just as linear acceleration results from force, angular acceleration results from torque.
Constant torque produces constant angular acceleration. That would occur e.g. with a weight hanging from a rotating drum.

In a pendulum you have a constant force, mg, but the torque about the pivot depends on the angle, so the angular acceleration varies. Maximum torque is when the line of action of mg is as far from the pivot as it can go. That will be when the mass centre is on the same horizontal level as the pivot.

 

[haruspex]

A uniform bar of mass m and length L is pivoted at one end and is held vertically as shown. After the bar is released, it swings downward with no friction in the pivot.
At the instant the bar is at the bottom of the swing, find the magnitudes of the following quantities:
1-ω

instead of L/2 I put L and I get the answer

Does that surprise you? L is clearly the right value to use for the height lost by the mass centre, no?

2-The angular acceleration of the bar α and
The horizontal and vertical component of the acceleration of the free end of the bar

what will happen when it becomes vertical again? will it come to rest?

Yes and no. The question statement offers no reason for the bar to have left the vertical in the first place. Usually it would say it was not quite vertical, or that is was vertical and given a slight nudge. If there is no work lost, what will happen after one revolution depends critically on why it started moving. If it was stationary but not quite vertical in the first place then it will not have enough energy to reach the vertical, so will fall back down, and continue to oscillate. If, on the other hand, it did start vertical but was given a nudge then it will have more than enough energy to reach the vertical and continue to rotate in the same direction forever.
The third possibility is that it did not start vertical, or stationary, but with exactly as much KE as if it had fallen from vertical. In this case it will just make it back up to vertical and, theoretically, stop.

 

[SammyS]

A uniform bar of mass m and length L is pivoted at one end and is held vertically as shown. After the bar is released, it swings downward with no friction in the pivot.
At the instant the bar is at the bottom of the swing, find the magnitudes of the following quantities:
1-ω

instead of L/2 I put L and I get the answer

Of course, L is correct. Why?

The center of mass drops by a total distance, L .

2-The angular acceleration of the bar α and
The horizontal and vertical component of the acceleration of the free end of the bar

what will happen when it becomes vertical again? will it come to rest?

 

[REVIANNA]

I think this is the case with this problem

If it was stationary but not quite vertical in the first place then it will not have enough energy to reach the vertical, so will fall back down, and continue to oscillate.

it asks for
The vertical component of the acceleration of the free end of the bar
$v^2/R=(6gL)/L=6g$
and similarly I found the The vertical component of the acceleration of the midpoint of the bar (3g)
the horizontal components of acceleration are zero.

now the problem:
what are the forces acting on the rod "at the bottom of the swing" .more specifically "The vertical component of the force exerted on the bar by the hinge"

I know that linear acceleration occurs due to net force experienced by the centre of mass (up is +ve)
therefore,
$F_n=F_h-F_e$
from linear acceleration of the mid point (centre of mass) calculated above $F_n=3mg$
so $F_h=4mg$

does this mean:
the ω is same everywhere(depending on the height ), but linear velocity varies because of varying moment arm , therefore the radial acceleration varies ,that's why the net force varies.

also the problem gives this hint(which I didn't read initially)

"Energy is constant (frictionless pivot), which can be used to find the angular velocity at any angle(okay)
, and therefore the centripetal acceleration of the center of mass at that angle.(okay) The forces are found from dynamics: F=m(a_cm) and $∑τ_x=I_xα$. Note that x has to be a fixed point (e.g. the pivot) except that it can be the c of m even if the c of m accelerates."(what does that mean:is centre of mass special?)"

 

[REVIANNA]

@haruspex

for the first two problems ω is increasing because of the direction of rotation- clockwise .
are the trends in α (which I understand now ) related to the trends in ω?

 

[SammyS]

@haruspex

for the first two problems ω is increasing because of the direction of rotation- clockwise .
are the trends in α (which I understand now ) related to the trends in ω?

@REVIANNA ,
haruspex resides in Australia, so he might not get back to you until tomorrow, therefore, I will attempt an answer. (I have been watching this thread.)

Actually, clockwise rotations are considered to be negative.

I suppose the references to ω and α increasing or decreasing refers to their magnitudes.

 

[REVIANNA]

haruspex resides in Australia, so he might not get back to you until tomorrow

oh I thought that the difference is only about two hrs.

clockwise rotations are considered to be negative.

I know that.I thought that like some others they have taken rightward and clockwise as positive but you're right we can't assume it unless its mentioned .

I understand that for the square sheet α increases and for the blue rod α decreases.but in these questions we are either using Newton's second law (a and α) or energy conservation and transformation.
How do I know that ω increases for both of them ?
back in linear motion if a and v had the same sigh it meant "speeding up" .

 

[SammyS]

oh I thought that the difference is only about two hrs.I know that.I thought that like some others they have taken rightward and clockwise as positive but you're right we can't assume it unless its mentioned .

I understand that for the square sheet α increases and for the blue rod α decreases.but in these questions we are either using Newton's second law (a and α) or energy conservation and transformation.
How do I know that ω increases for both of them ?
back in linear motion if a and v had the same sigh it meant "speeding up" .

Yes, the sense of the problem is if α and ω have the same sign, then the rotational speed is increasing.

 

[haruspex]

I understand that for the square sheet α increases and for the blue rod α decreases.but in these questions we are either using Newton's second law (a and α) or energy conservation and transformation.
How do I know that ω increases for both of them ?
back in linear motion if a and v had the same sigh it meant "speeding up" .

I get the impression you feel there's an inconsistency there. It's completely consistent. As Sammy writes, in these problems $\alpha$ and $\omega$ have the same signs, so the rotation is speeding up. Whether $\alpha$ is increasing or decreasing in magnitude is another matter; that would correspond to a increasing or decreasing in magnitude.