Find the Maximum Angular Velocity of the Quarter Circle with Energy

[Northbysouth]

Homework Statement

The uniform quarter-circular sector is released from rest with one edge vertical as shown. Determine its subsequent maximum angular velocity. The distance b is 560 mm.

Homework Equations

The Attempt at a Solution

I know that I need to use:

T₁ + V₁ + U'_1-2 =T₂ + V₂

This reduces to:

V₁ = T₂

mgb = 0.5*I_Ow²

w = [(2gb)/I_O]^1/2

But I'm not not sure about the I_O. I found the mass moment of inertia (http://engineering-references.sbainvent.com/dynamics/mass-moment-of-inertia.php#.UX3U9MokSf4) for a thin disk to be:

1/2*mr² and I had thought that maybe if I divided it by 4 that would give me the correct mass moment of inertia, but this did not work.

I'm fairly certain that it's my I_O that is wrong.

Any help would be appreciated. Thank you

 

[Sunil Simha]

I know that I need to use:

T₁ + V₁ + U'_1-2 =T₂ + V₂

This reduces to:

V₁ = T₂

mgb = 0.5*I_Ow²

w = [(2gb)/I_O]^1/2

Firstly, is the change in potential energy mgb? Check that out.

And in the second equation that I have reddened, you have missed out m ( but I guess that's a typo from you and you have originally included while solving the problem). The moment of inertia is mb²/8. There's nothing wrong with that.

 

[Northbysouth]

So, if the change in potential energy is not mgb, then how would I go about finding out what it is?

Would it be reasonable to assume that the quarter circle will begin to slow down once it's center crosses the vertical, and as such I need to calculate where the center of the quarter circle is?

 

[Sunil Simha]

So, if the change in potential energy is not mgb, then how would I go about finding out what it is?

Would it be reasonable to assume that the quarter circle will begin to slow down once it's center crosses the vertical, and as such I need to calculate where the center of the quarter circle is?

Yes the center of mass lies on the bisector of the quarter circle and it would slow down once it has crossed the vertical. ( this can be deduced by considering the symmetry of the mass distribution).

To find the center of mass, I would advise you to first try to derive the position of the center of mass of a thin quarter ring of uniform mass distribution. Then the quarter disk can be assumed to be made of such concentric quarter rings and thus the center of mass can be be found by a bit of integration.

Once you've found the center of mass, it is simple to calculate the change in PE.

 

[Nickie]

In order to find the maximum angular velocity of the quarter circle, we can use the conservation of energy principle. This means that the initial total energy (kinetic + potential) will be equal to the final total energy (kinetic + potential).

The initial kinetic energy is zero since the quarter circle is released from rest, so we can ignore that term. The initial potential energy is given by mgh, where m is the mass of the quarter circle, g is the acceleration due to gravity, and h is the height of the center of mass of the quarter circle above the ground. In this case, h = b/2 = 280 mm.

The final kinetic energy is given by 1/2*I*w^2, where I is the moment of inertia and w is the angular velocity. The moment of inertia for a quarter circle can be found using the parallel axis theorem, which states that I = Icm + md^2, where Icm is the moment of inertia about the center of mass and d is the distance from the center of mass to the axis of rotation. In this case, the moment of inertia about the center of mass can be found using the formula for a thin disk (1/2*mr^2) and d is equal to the radius of the quarter circle (b/2). Therefore, the final kinetic energy can be written as 1/2*(1/2*mb^2 + mb^2)*(dw)^2, where dw is the angular velocity in radians per second.

The final potential energy is equal to zero since the center of mass of the quarter circle is at ground level.

Setting the initial and final energies equal to each other, we can solve for dw to find the maximum angular velocity:

mgh = 1/2*(1/2*mb^2 + mb^2)*(dw)^2

Solving for dw:

dw = sqrt(2gh/(1/2*mb^2 + mb^2))

Substituting in the values given in the problem (g = 9.8 m/s^2, b = 0.56 m, m = 1 kg), we get:

dw = sqrt(2*9.8*0.28/(1/2*1*0.56^2 + 1*0.56^2))

dw = 3.32 rad/s

Therefore, the maximum angular