Which cylinder reaches the ground first?

[Pushoam]

Homework Statement

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Homework Equations

The Attempt at a Solution

The correct option is d because
which body will fall first depends on the moment of inertia, the less MOI, the fast the body reaches the ground.
COM motion will not be applicable here as the friction force will be different for different material.
But since the motion is pure rolling, work done by the friction is 0. So, conservation of energy for pure rolling gives the speed of both the cylinders in terms of MOI. And it turns out that lesser the MOI, faster the speed. And faster the speed, less time would be taken for reaching the ground. Hence, the correct option is d.

Is this correct?[/B]

 

[mjc123]

In option d, why would A have less MOI than B? It would have less MOI than a hollow cylinder of the same mass, but not than a hollow cylinder of the same radius.

 

[Pushoam]

In option d, why would A have less MOI than B? It would have less MOI than a hollow cylinder of the same mass, but not than a hollow cylinder of the same radius.

Sorry, the answer would depend not upon the MOI, but $\frac{ MOI} {mass}$ of the corresponding cylinder.
The edited answer is
The correct option is d because
which body will fall first depends on the $\frac{ MOI} {mass}$, the less $\frac{ MOI} {mass}$, the fast the body reaches the ground.
COM motion will not be applicable here as the friction force will be different for different material.
But since the motion is pure rolling, work done by the friction is 0. So, conservation of energy for pure rolling gives the speed of both the cylinders in terms of $\frac{ MOI} {mass}$. And it turns out that lesser the$\frac{ MOI} {mass}$, faster the speed. And faster the speed, less time would be taken for reaching the ground. Hence, the correct option is d.

$\frac{ MOI} {mass}$ is known as radius of gyration, right? So, this means that in pure rolling for a given force; lesser the radius of the gyration, faster is the speed. So, this is the significance of the concept of "radius of gyration", right?

Is this correct?

 

[ubergewehr273]

Strictly speaking, time taken to reach bottom must depend on acceleration of the body and not simply velocity. Don't you think you're argueing on the fact that velocity being maintained constant and hence your logic ?

 

[Chestermiller]

Strictly speaking, time taken to reach bottom must depend on acceleration of the body and not simply velocity. Don't you think you're argueing on the fact that velocity being maintained constant and hence your logic ?

I see no indication from the OPs answer that he is not taking into account acceleration.

 

[Pushoam]

I see no indication from the OPs answer that he is not taking into account acceleration.

Is my answer correct?

 

[Chestermiller]

Is my answer correct?

I would like to see your detailed analysis of this problem (i.e., the equations) before I can say.

 

[Pushoam]

Applyng conservation of energy for pure rolling,
$\frac 1 2 ( mv^2 + I {\omega}^2) = mgH \\ v^2(1+\frac I {mR^2}) = 2gH$

 

[haruspex]

Applyng conservation of energy for pure rolling,
$\frac 1 2 ( mv^2 + I {\omega}^2) = mgH \\ v^2(1+\frac I {mR^2}) = 2gH$

Right, and to complete it, fill in the value of I for the two cylinders.
Note that, surprisingly perhaps, both mass and radius are irrelevant.

 

[J Hann]

A simpler form would be a = g sin theta / K
where K is the constant relating M R^2 to the moment of inertia
about the point of contact with the plane.
As an example for a solid cylinder, K = 3/2 and for a hollow cylinder K = 2.

 

[Pushoam]

Note that, surprisingly perhaps, both mass and radius are irrelevant.

It depends on the radius of gyration.
Will you please tell me the physical meaning of this concept "radius of gyration"?
Here, knowledge of the radius of gyration gives us the knowledge that which body will reach the ground first. In general, how does one apply this concept?

 

[Pushoam]

A simpler form would be a = g sin theta / K
where K is the constant relating M R^2 to the moment of inertia
about the point of contact with the plane.
As an example for a solid cylinder, K = 3/2 and for a hollow cylinder K = 2.

But I have to derive the 1st eqn as I don't know the validity of the 1st eqn.
If K = $\frac {mR^2} I \\ \text{ for solid cylinder, I about z- axis is} \frac {mR^2} 2 .$ And so K = 2
Similarly for hollow cylinder without ends, it is $mR^2, K = 1$

 

[haruspex]

It depends on the radius of gyration.

Not exactly. It depends on the ratio of radius, in the normal sense, to radius of gyration. The "shape" (uniform solid cylinder, uniform hollow cylinder, uniform sphere...) determines that ratio.

The radius of gyration of an object about an axis through its mass centre is the radius of the ring (or hollow cylinder) of the same mass that would have the same moment of inertia.

But I have to derive the 1st eqn as I don't know the validity of the 1st eqn.
If K = $\frac {mR^2} I \\ \text{ for solid cylinder, I about z- axis is} \frac {mR^2} 2 .$ And so K = 2
Similarly for hollow cylinder without ends, it is $mR^2, K = 1$

In J Hann's approach using force and acceleration, J is taking moments about the instantaneous centre of rotation, i.e. the point of contact with the plane. Using the parallel axis theorem, the MoI about that is 2mr² for the hollow cylinder and (3/2)m² for the solid one. Note that this corresponds to the "1+" in your formula.

To get the equation, taking moments gives I_{point of contact}α=mgr sin(θ). Writing a=rα and I_{point of contact}=Kmr² yields a=g sin(θ)/K.