Why is Radial Acceleration Ignored for a Rotating Board?

[momo1111]

Hey! In this setup, https://sciencedemonstrations.fas.harvard.edu/presentations/falling-faster-g ,a board rotating about a fixed axis under gravity, The board's angle θ is measured from the horizontal, Angular acceleration is given by α = MgRcosθ / I Then we're asked to find the acceleration of the free end of the board in the downward direction. The solution provided uses a = αLcosθ for the downward acceleration. However, I'm puzzled about something: Shouldn't we also consider the downward component of this radial acceleration of the end point too? I missing something.
Thanks in advance for any insights!

 

[PeroK]

The acceleration calculated in this case is in the direction tangential to the line of the rod. The magnitude of this acceleration is $\alpha L$. The vertical component, therefore, is $\alpha L \cos \theta$, as given.

 

[hutchphd]

The solution provided uses a = αLcosθ for the downward acceleration.

This is probably wrong (your description is a bit sparse!) What solution? Where?
There is also a force on the board: the table pushes up. The Center of Mass of the board will respond to both this force and gravity.
Rather than pointing to an invisible solution, please provide your best effort for discussion starting with a free body diagram for the board.
REVISION: Oh I just looked at the video. So I don't believe they screwed it up.....but you need to actually solve the problem yourself to tell.

 

[A.T.]

Shouldn't we also consider the downward component of this radial acceleration of the end point too? I missing something.

Is there a hinge joint on the left end of the board? That would provide the force for radial acceleration, not just gravity.

 

[momo1111]

The acceleration calculated in this case is in the direction tangential to the line of the rod. The magnitude of this acceleration is $\alpha L$. The vertical component, therefore, is $\alpha L \cos \theta$, as given.

The acceleration at the rod's end can be divided into radial and tangential components.

However, I’m still unclear on why only the tangential component is considered when discussing vertical acceleration. Shouldn’t the radial component also contribute to the vertical direction? What am I missing?

 

[PeroK]

The acceleration at the rod's end can be divided into radial and tangential components.

However, I’m still unclear on why only the tangential component is considered when discussing vertical acceleration. Shouldn’t the radial component also contribute to the vertical direction? What am I missing?

I think you're right! The vertical acceleration is greater than that calculated.

 

[PeroK]

The point is this. The radial acceleration depends on the speed of the rod ($L\dot \theta^2$). If you release the rod from rest, then the critical angle is as on the page. At that angle, the rod will immediately accelerate vertically faster than gravity.

If you release the rod from a greater angle, then it will initially accelerate vertically less than gravity, but will exceed gravity before it reaches the above critical angle - due, as you say, to the radial component, which itself has a downwards vertical component.

Well spotted!

 

[momo1111]

Thanks for your response!, I think I understood part of what you said, but not everything... I'm still a bit confused.
By ‘critical angle,’ do you mean a specific angle at which the board, when released from rest, will immediately start accelerating vertically faster than gravity? At such an angle, the radial acceleration doesn't contribute to the total acceleration of the mass?

My line of reasoning is (a simple one), let's say we consider the end of the rod as a material point with mass (dm), this point moves along a circular path (which is its only possible trajectory). However, this motion isn’t uniform. why? because it starts from rest and definitely doesn’t end at rest, so the motion cannot be steady. So the acceleration vector of this point can’t be described solely by the tangential acceleration along the circle, no?

So the total downward acceleration (a_d) is the sum of the downward components of both the tangential and radial accelerations:
$$
a_{d}=a_{t,\text{ down }}+a_{r,\text{ down }}=\alpha L\cos\theta+\alpha L\sin^{2}\theta
$$

For reference, this is the setup in the experiment:
A 1-meter long board is held vertically with one end resting on a table and the other end propped up by a vertical stick. A plastic cup is attached 85 cm from the hinged end, and a golf ball is placed on a tee at the free end. When the vertical stick is suddenly removed, the board falls, and the ball falls into the cup.

And the (not complete?) derivation is:
The angular acceleration of the falling board is given by
$$
\alpha=\frac{M g R \cos \theta}{I}
$$

The acceleration of the free end of the board in the downward direction is
$$
a_{d}=\alpha L\cos\theta\:\qquad\boldsymbol{(?)}
$$

Making the substitution, this becomes
$$
a_d=\frac{M g R L \cos ^2 \theta}{I}
$$

Finally, we have
$$
I=\frac{1}{3} M L^2
$$
and
$$
a_d=\frac{3}{2} g\left(\cos ^2 \theta\right)
$$

 

[Lnewqban]

My line of reasoning is (a simple one), let's say we consider the end of the rod as a material point with mass (dm), this point moves along a circular path (which is its only possible trajectory). However, this motion isn’t uniform. why? because it starts from rest and definitely doesn’t end at rest, so the motion cannot be steady. So the acceleration vector of this point can’t be described solely by the tangential acceleration along the circle, no?

We need to consider that material point with mass (dm) located at the end of the rod as only one of many similar material points with mass (dm) located along the rod, all solidly linked to the next and to the grounded pivot.

That is equivalent to having an imaginary mass formed by the summation of all the dm's located at the middle point of the actual rod.

If we could imagine that the board (or all the dm's) had no rotational inertia, that imaginary mass located at half the length of the rod would move with vertical acceleration equal to g.

In this case, the fall is not free but interconnected to the rotation about that fixed pivot.

Because the rod is resisting the rotation induced by gravity because it has rotational inertia, part of the potential energy would need to be invested in motivating rotation, and our imaginary mass located at half the length of the rod would move with vertical acceleration less than g (0.75g).

As tangential velocity increases as points are located farther from the pivot point, there is one of those, located somewhere between the middle point and the free end of the rod, that will move with vertical acceleration equal to g.

The shown equations show that such point is located 2/3 length from the pivot point.
Therefore, the free end point of the rod would move with vertical acceleration greater than g (1.5g).

 

[PeroK]

By ‘critical angle,’ do you mean a specific angle at which the board, when released from rest, will immediately start accelerating vertically faster than gravity? At such an angle, the radial acceleration doesn't contribute to the total acceleration of the mass?

Yes. Initially the velocity is zero, so there is zero radial acceleration.

The page you linked to is wrong, in that it assumes that there continues to be zero radial acceleration, even when the board is moving.

My line of reasoning is (a simple one),

The correct approach is to use the equation for acceleration in polar coordinates, which is:
$$\vec a = (\ddot r - r\dot\theta^2)\hat r + (r\ddot \theta + 2\dot r \dot \theta) \hat \theta$$In this case, we have the constraint that $r = R$ is constant, so the equation reduces to:
$$\vec a = (- R\dot\theta^2) \hat r + (R\ddot \theta) \hat \theta$$And, initial acceleration when the board is released from rest is:$$\vec a_0 = (R\ddot \theta) \hat \theta$$It's this equation that leads to the analysis on the page you linked to, where only the tangential/angular acceleration is considered.

Now, we can transform these accelerations using:
$$\hat r = \cos \theta \hat x + \sin \theta \hat y, \ \hat \theta = -\sin \theta \hat x + \cos \theta \hat y$$To give:
$$\vec a_0 = (R\ddot \theta)(-\sin \theta \hat x + \cos \theta \hat y) = (-R\ddot \theta \sin \theta) \hat x +(R\ddot \theta \cos \theta) \hat y$$Finally, in this case we have $\alpha = -\ddot \theta$, so we have an initial vertial component of acceleration:
$$\ddot y_0 = -R\alpha\cos \theta$$PS I'll leave it as an exercise for you to calculate the expression for the vertical acceleration more generally, after the rod has started moving.