Vertical semicircular guide uniformly accelerated
Problem text
A vertical semicircular guide, with radius \(R\) and without friction, is constrained to a platform that moves with constant acceleration \(a\) along the horizontal direction (see figure). A point body of mass \(m\), initially stationary with respect to the guide and in contact with it at the left extreme horizontal diameter, is allowed to slip. Calculate at the instant when the body reaches the lowest point:
- the maximum speed of the body with respect to the guide;
- the force that the guide exerts on the body.
Compare the results with those obtained when the guide is at rest.
Solution
Force diagram for the point body at rest
With the aim to solve this problem, we prefer to reason in the accelerated reference frame. Moreover, we choose to depict what forces act on the point body when it is considered at rest with the respect to the guide (see next figure).
In such conditions, the figure below shows the force diagram for the body of mass \(m\).
In detail, there are the following vectors:
- \(m\vec{a}\), the pseudo-force due to the fact that the body is at rest in an accelerated reference frame;
- \(\vec{N}\), the force that the semicircular guide exerts on the point body;
- \(m\vec{g}\), the force due to the Earth’s gravity.
\(g\) indicates the gravity acceleration.
It is easy to remind that when the point body is at rest with respect to the guide, the sum of the three forces vanishes \(\vec{N}+m\vec{g}+m\vec{a}=0\). Differently, if the acceleration \(a\) were null, we would have that \(\vec{N}+m\vec{g}=0\).
The introduction of an apparent vertical
As we know, we live on a non-inertial reference frame being the Earth a rotating planet. Consequently, what we feel, it is an apparent gravity. It is given by the vector sum of the gravity acceleration due to the only gravitational attraction and the centrifugal acceleration. Here, in the considered problem, we have a similar case inasmuch as the point body at rest feels the vector sum of the two acceleration \(\vec{g}\) and \(\vec{a}\).
The latter consideration allows us to treat the point body as it were on an inertial reference frame having an effective gravity acceleration. The latter one is given by \(\vec{g^*}=\vec{g}+\vec{a}\), whose intensity is equal to \(g^*=\sqrt{g^2+a^2}\).
As a consequence of that choice, we consider the direction of \(g^*\) as a reference for the gravitational potential energy by setting the latter one to zero in corresponding to the center \(O\) of the semicircular guide (see figure).
To better to identify the direction of \(g^*\), chosen as apparent vertical, we indicate with
\(\displaystyle \alpha=\tan^{-1}\frac{a}{g}\)
the angle formed by the directions of \(\vec{g^*}\) and \(\vec{g}\). In particular, as it will be useful hereafter, we remind the following goniometric identities
\(\displaystyle \cos{\alpha}=\frac{1}{\sqrt{1+\tan^2{\alpha}}}\) and \(\displaystyle \sin{\alpha}=\frac{\tan{\alpha}}{\sqrt{1+\tan^2{\alpha}}}\).
The latter ones can also be rewritten in terms of \(g\) and \(a\)
\(\displaystyle \cos{\alpha}=\frac{1}{\sqrt{1+\left(a/g\right)^2}}\) and \(\displaystyle \sin{\alpha}=\frac{a/g}{\sqrt{1+\left(a/g\right)^2}}\).
The maximum speed of the body with respect to the guide
With this purpose, being the system conservative, we apply the mechanical energy conservation principle, considering, as abovesaid,
… the direction of \(g^*\) as a reference for the gravitational potential energy by setting the latter one to zero in corresponding to the center \(O\) of the semicircular guide (see figure).
In detail, the mechanical energy conservation principle reads
\(\displaystyle -mg^*R\sin{\alpha}=-mg^*R\cos{\alpha}+\frac{1}{2}mv^2\).
The left side term describes the initial gravitational potential energy of the point body when it is placed at the left extreme of the guide. While the right side terms represent the final gravitational potential energy and the kinetic energy when the body reaches the lowest point (along the direction of \(\vec{g}\)).
From the last equation, after replacing the expression of \(\sin{\alpha}\), \(\cos{\alpha}\) and \(g\), it is straightforward to derive that the maximum speed reached by the point body is given by
\(v=\sqrt{2R\left(g-a\right)}\).
The force that the guide exerts on the point body
With this aim, we apply Newton’s law of the motion when the point body reaches the lowest point. Writing directly the scalar version of the equation, we get
\(\displaystyle N-mg=m\frac{v^2}{R}\).
The term \(\displaystyle \frac{v^2}{R}\) on the right side of the equation indicates the centripetal acceleration. At last, by isolating \(N\) from the latter equation and substituting the expression of \(v\) derived on the previous section, we easily obtain
\(N=m\left(3g-2a\right)\).
Comparison with the case \(a=0\) when the guide is at rest
First, when \(a=0\), the effective gravity acceleration felt by the point body is \(g\). Moreover, the angle \(\alpha\) reduces to zero. Lastly, the maximum speed and the force intensity exerted by the guide become
\(v=\sqrt{2Rg}\) and \(N=3mg\).
Remarks
- The last equations correspond exactly with those that one can easier find directly in the case the guide is at rest or moving with constant speed.
- The problem, here solved, has the same solutions as that of a simple pendulum suspended from a pivot moving in a uniformly accelerated way.
©Stefano Spezia. This work is licensed under a Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 International License.