Angular acceleration of the bar

[Saitama]

Homework Statement

(see attachment)

Homework Equations

The Attempt at a Solution

I haven't yet attempted the b) and c) parts.
Part a) for the case i): When the spring 2 breaks, the rod should rotate about the point A and the only torque acting is due to weight of bar. (α is the angular acceleration of bar)
Therefore Iα=mg\frac{L}{2}
Solving, i get α=\frac{3g}{2L}
But the answer says, its \frac{3g}{L}.

Where am i wrong?

 

[ehild]

Point A is neither a fixed axis nor the CM. You have to write the torque with respect either to the hinge (and then use the parallel axis theorem to get the moment of inertia) or the CM of the bar.

ehild

 

[Saitama]

Point A is neither a fixed axis nor the CM. You have to write the torque with respect either to the hinge (and then use the parallel axis theorem to get the moment of inertia) or the CM of the bar.

ehild

Thanks ehild, i get the answer if i calculate torque about the CM.
If i try to calculate the torque about the hinge, i get the wrong answer. Here's my attempt:
Moment of inertia of bar about the hinged point is \frac{mL^2}{6}
From the initial condition, before the spring breaks,
mg=2kxsin(30^o)
The only torque acting is due to the force of the spring i.e
kxsin(30^o)\frac{L}{2}=Iα
\frac{mgL}{4}=\frac{mL^2}{6}α
Solving this, i doesn't end up with the right answer.

 

[ehild]

Thanks ehild, i get the answer if i calculate torque about the CM.
If i try to calculate the torque about the hinge, i get the wrong answer. Here's my attempt:
Moment of inertia of bar about the hinged point is \frac{mL^2}{6}
From the initial condition, before the spring breaks,
mg=2kxsin(30^o)
The only torque acting is due to the force of the spring i.e
kxsin(30^o)\frac{L}{2}=Iα
\frac{mgL}{4}=\frac{mL^2}{6}α
Solving this, i doesn't end up with the right answer.

Well, it was not a good idea to suggest the hinge. The lines of both forces go through the hinge, so the torque is zero with respect to it. I think the whole bar has some angular momentum with respect to the hinge (it is the angular momentum associated with the motion of the CM) and it has some angular momentum with respect to its CM.


ehild

 

[Saitama]

Well, it was not a good idea to suggest the hinge. The lines of both forces go through the hinge, so the torque is zero with respect to it. I think the whole bar has some angular momentum with respect to the hinge (it is the angular momentum associated with the motion of the CM) and it has some angular momentum with respect to its CM.


ehild

Woops, i should have taken care of that, the torque is zero about the hinge.

I am facing some difficulty for part b). I need to calculate the acceleration of point A, how will i calculate the radial acceleration? Radial acceleration is v^2/r but how will i find the tangential velocity?

 

[ehild]

The speed of A with respect to the CM is ωL/2, but ω=0 at the beginning. Decompose the acceleration to vertical and horizontal components. The acceleration of A is the sum of acceleration of the CM and the acceleration of A with respect to the CM.

Why do you need to solve such terrible problems? They make me totally confused .

ehild

 

[Saitama]

The speed of A with respect to the CM is ωL/2, but ω=0 at the beginning. Decompose the acceleration to vertical and horizontal components. The acceleration of A is the sum of acceleration of the CM and the acceleration of A with respect to the CM.

There are two forces acting on the bar(in the vertical direction) when the spring 2 breaks.
mg-kxsin(30°)=ma_{CM}
mg-\frac{mg}{2}=ma_{CM}
a_{CM}=\frac{g}{2}
Tangential acceleration of point A is in the vertical direction and its direction is upward.
\frac{αL}{2}=\frac{3g}{2}
Therefore net acceleration of point A in vertical direction is 3g/2-g/2=g

Since, there is a horizontal component too of the force due to spring, therefore
kxcos(30°)=ma_{CM}
a_{CM}=\frac{\sqrt{3}g}{2}
(This time a_CM denotes the acceleration of bar in horizontal direction)
So net acceleration of point A is:
\frac{\sqrt{3}g}{2}i+gj
(i and j denotes the unit vector in horizontal and vertical direction respectively, i would like to know how can we write unit vectors in latex)
This is the answer mentioned in the answer key.
Thanks ehild!

Why do you need to solve such terrible problems? They make me totally confused .

Just to get a decent score in my engineering entrance exam, i need to do these terrible problems, i myself don't like to do them but i can do nothing about them. There are still more questions to come.

 

[ehild]

This last problem is really annoying. I have to check my book on Theoretical Physics. But all right, just post them. I need some training for my little grey brain cells...

ehild

 

[Saitama]

This last problem is really annoying. I have to check my book on Theoretical Physics. But all right, just post them. I need some training for my little grey brain cells...

ehild

Just curious, are you talking about this book:
http://mirtitles.org/2012/10/22/theoretical-physics-kompaneyets/

 

[ehild]

No, I meant Hungarian books of Budó, both on Experimental Physics and theoretical Mechanics.

But I also have Kompaneyets' Theoretical Physics on my bookshelves. It is a good book, but only a small part is Mechanics.

ehild