Why Is Deriving the Motion Formula for Hoops More Complex?

In summary, deriving the motion formula for hoops is complex due to the interplay of multiple factors, including the angular momentum, the friction between the hoop and the surface, and the gravitational forces acting on the object. The dynamics of the system require a comprehensive understanding of rotational motion and energy conservation principles, making the derivation more intricate than standard linear motion equations.
  • #1
lsie
8
1
Homework Statement
Conservation of mechanical energy proof
Relevant Equations
a = 1/2 g sin theta
a = 3/5 g sin theta
I've worked out how to derive the formulas for a solid cylinder and a solid sphere rolling down a hill.

E.g., for a cylinder:
Emech = KE + PE
mgh = 1/2 mv^2 + 1/2 Iw^2
gh = 1/2 v^2 + 1/2 (1/2r^2) v^2/r^2
gh = 3/4 v^2
v^2 = 4/3 gh
I then performed a derivative with respect to time and found a = 2/3 g sin theta

But I'm having trouble proving the formula for spherical shells and hoops. The rub is that extra radius nested in the inertial moment (1/2m (r^2 + R^2). I get to gh = 3/4 v^2 + 1/2 R^2 v^2 and I don't know how to deal with the R^2.

I've tried using various equations (v = wr, a = αr, and α = w/t) but can never get rid of that radius.

Any pointers would be appreciated.

Cheers!
 
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  • #2
Neglect the difference between r and R.
 
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  • #3
That worked! Thanks.

I was certain the solution would be more elegant. Out of curiosity, are there good reasons why we can neglect the difference between r and R? Surely there are hoops where r and R are (how's that for a tongue twister) significantly different?
 
  • #4
lsie said:
That worked! Thanks.

I was certain the solution would be more elegant. Out of curiosity, are there good reasons why we can neglect the difference between r and R? Surely there are hoops where r and R are (how's that for a tongue twister) significantly different?
(It sounds like) You're looking for a "thin hoop approximation". Thats why it can be neglected. If moment of inertia never mattered, then they wouldn't bother with it. Perhaps I'm not seeing what you are asking.

i.e. I can't think of a reason to say ##r \approx R## unless it's true for the model you are analyzing?
 
Last edited:
  • #5
Got it. I saw some google-images of a hoop with some thick-ish edges and it threw me off. Cheers!
 

FAQ: Why Is Deriving the Motion Formula for Hoops More Complex?

Why is deriving the motion formula for hoops more complex than for other shapes?

Deriving the motion formula for hoops is more complex because hoops have a unique distribution of mass. Unlike solid objects, the mass of a hoop is concentrated at a fixed radius from the center. This affects the moment of inertia and requires a different approach when applying Newton's laws of motion and rotational dynamics.

What role does the moment of inertia play in the complexity of the motion formula for hoops?

The moment of inertia is crucial because it quantifies how mass is distributed relative to the axis of rotation. For a hoop, the moment of inertia is \(I = MR^2\), where \(M\) is the mass and \(R\) is the radius. This differs from other shapes like disks or spheres, leading to different rotational characteristics and complicating the derivation of the motion formula.

How does the rolling motion of a hoop differ from that of other objects?

When a hoop rolls, it involves both translational and rotational motion. The relationship between these motions is governed by the condition of rolling without slipping, which states that the linear velocity \(v\) is related to the angular velocity \(\omega\) by \(v = \omega R\). This dual nature of motion adds to the complexity of deriving the formula.

Why do energy considerations make the derivation more complex for hoops?

Energy considerations add complexity because the total kinetic energy of a rolling hoop includes both translational kinetic energy \(\frac{1}{2}Mv^2\) and rotational kinetic energy \(\frac{1}{2}I\omega^2\). For a hoop, this becomes \(\frac{1}{2}Mv^2 + \frac{1}{2}MR^2\omega^2\), which simplifies to \(Mv^2\) due to the relationship \(v = \omega R\). Balancing these energy forms accurately requires careful analysis.

Are there specific examples that illustrate the complexity of deriving the motion formula for hoops?

Yes, one example is analyzing a hoop rolling down an inclined plane. Unlike a solid sphere, which has a different moment of inertia, the hoop's motion must account for its unique mass distribution. This affects the acceleration and requires solving differential equations that incorporate both translational and rotational dynamics, illustrating the added complexity.

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