Determine angular velocity - Dynamics

[dvep]

Homework Statement

Torsional spring constant k=33 Nm/rad
It is undeflected when it is in the vertical position, it weighs 6.4 kg
When the bar is released from the horizontal position from rest, determine the angular velocity $$\omega$$ as it passess the vertical position.
We can assume that it is frictionless.

The Attempt at a Solution

Am I right to say that I have to use the conservation of energy and find

T1+V1 = T2+V2 which would be my angualr velocity?

I am quite stuck on this question, I would be grateful for some guidance.

 

[Doc Al]

Yes, you'll need to apply conservation of energy. Hint: There are three 'types' of energy involved here. What are they? What are expressions for each?

 

[dvep]

Yes, you'll need to apply conservation of energy. Hint: There are three 'types' of energy involved here. What are they? What are expressions for each?

Thank you for your reply

Elastic potential energy = 1/2kx^2
Rotational kinetic energy = 1/2Iw^2 = 1/2I(v/r)^2

Though I am unsure what the third energy is.

So,

EPE=1/2*33*(0.8/sin45)^2 = 20 J

Am I on the right track?

Sorry to be a pain, I haven't been taught this yet in class.

 

[Doc Al]

Thank you for your reply

Elastic potential energy = 1/2kx^2

OK, but for a torsion spring, the displacement from equilibrium will be in radians.

Rotational kinetic energy = 1/2Iw^2 = 1/2I(v/r)^2

OK.

Though I am unsure what the third energy is.

Hint: The bar has weight.

So,

EPE=1/2*33*(0.8/sin45)^2 = 20 J

Not sure what you did here. What angle (in radians) does the initial position make with the unstretched position?

 

[dvep]

ok, but for a torsion spring, the displacement from equilibrium will be in radians.

Oh yeah, so

EPE=1/2k$$\theta$$^2

hint: The bar has weight.

I wasn't sure but I was thinking gravitational potential energy?

not sure what you did here. What angle (in radians) does the initial position make with the unstretched position?

EPE=1/2*33*1.571^2 J

 

[Doc Al]

Oh yeah, so

EPE=1/2k$$\theta$$^2

Right.

I wasn't sure but I was thinking gravitational potential energy?

Exactly.

EPE=1/2*33*1.571^2 J

Good.

 

[dvep]

Right.


Exactly.


Good.

So,

EPE=40.723 J
GPE=50.176 J

Am I meant to add these together to get the total energy, then at the vertical position they both equal 0 and rotational kinetic energy equals the total energy, then use the rotational kinetic energy formula and rearrange to find the angualr velocity w?

 

[Doc Al]

EPE=40.723 J

OK.

GPE=50.176 J

How did you calculate this? For what position? Initial or final?

Am I meant to add these together to get the total energy, then at the vertical position they both equal 0 and rotational kinetic energy equals the total energy, then use the rotational kinetic energy formula and rearrange to find the angualr velocity w?

To avoid confusion, set up your conservation equation like so:

EPE₁ + GPE₁ + RotKE₁ = EPE₂ + GPE₂ + RotKE₂

(Some terms will end up being zero.)

 

[dvep]

How did you calculate this? For what position? Initial or final?

I thought that might be wrong

GPE1=mglcos(theta)= 6.4*9.8*0.8*cos(1.571) = 50.157 J
GPE2=mglcos(theta)= 6.4*9.8*0.8*cos(0) = 50.157 J

Is that correct, they are both the same.

To avoid confusion, set up your conservation equation like so:

EPE₁ + GPE₁ + RotKE₁ = EPE₂ + GPE₂ + RotKE₂

(Some terms will end up being zero.)

So what I get,

EPE1 + GPE1 + RotKE1 = EPE2 + GPE2 + RotKE2
40.723 + 50.157 + 0 = 0 + 50.157 + 40.723

RotKE2= 1/2Iw^2 = 1/2I(v/r)^2
Angular velocity = w = sqrt(2*RotKE2/mr^2) = sqrt(2*40.723/6.4*0.8^2) = 4.459 rads/s

Am I at least a little bit close?

 

[Doc Al]

I thought that might be wrong

GPE1=mglcos(theta)= 6.4*9.8*0.8*cos(1.571) = 50.157 J
GPE2=mglcos(theta)= 6.4*9.8*0.8*cos(0) = 50.157 J

Is that correct, they are both the same.

No, that makes no sense. The bar rises from horizontal to vertical, so its GPE increases. Hint: Measure height from the surface. Hint 2: Consider what happens to the center of mass.

 

[dvep]

No, that makes no sense. The bar rises from horizontal to vertical, so its GPE increases. Hint: Measure height from the surface. Hint 2: Consider what happens to the center of mass.

I'm still lost. What do you mean measure the height from the surface? Will the center of mass move towards the end of the bar?

 

[Doc Al]

I'm still lost. What do you mean measure the height from the surface? Will the center of mass move towards the end of the bar?

When the bar goes from horizontal to vertical, what happens to its center of mass? That determines the change in gravitational PE.

 

[dvep]

When the bar goes from horizontal to vertical, what happens to its center of mass? That determines the change in gravitational PE.

The GPE is 0 at the initial position and 50.157 J when vertical?
The center of mass gains translation kinetic energy?

 

[Doc Al]

The GPE is 0 at the initial position

OK.

and 50.157 J when vertical?

No. How high does the center of mass rise?

The center of mass gains translation kinetic energy?

Sure, but you need not worry about that. Treat the bar as being in pure rotation about the axis, so all its kinetic energy is rotational KE about that axis.

 

[dvep]

OK.

No. How high does the center of mass rise?

0.4 m?

so,

EPE1 + GPE1 + RotKE1 = EPE2 + GPE2 + RotKE2
40.723 + 0 + 0 = 0 + 25.088 + 15.635

 

[Doc Al]

0.4 m?

Right.

so,

EPE1 + GPE1 + RotKE1 = EPE2 + GPE2 + RotKE2
40.723 + 0 + 0 = 0 + 25.088 + 15.635

OK, but since you are solving for RotKE2, I'd write it as:
EPE1 + GPE1 + RotKE1 = EPE2 + GPE2 + RotKE2
40.723 + 0 + 0 = 0 + 25.088 + RotKE2

Then you'd deduce that RotKE2 = 15.635

And then use RotKE = 1/2Iw^2 to find w.

 

[dvep]

Right.


OK, but since you are solving for RotKE2, I'd write it as:
EPE1 + GPE1 + RotKE1 = EPE2 + GPE2 + RotKE2
40.723 + 0 + 0 = 0 + 25.088 + RotKE2

Then you'd deduce that RotKE2 = 15.635

And then use RotKE = 1/2Iw^2 to find w.

Thank you for your help.

 

[Skittles999]

Right.

Then you'd deduce that RotKE2 = 15.635

And then use RotKE = 1/2Iw^2 to find w.

Just looked through this problem... to calculate the moment of inertia, I, do you use I = 1/3mL^2?

 

[Doc Al]

Just looked through this problem... to calculate the moment of inertia, I, do you use I = 1/3mL^2?

Yes.