Rotation 5 Question 15
25. The minimum value of $\omega _0$ below which the ring will drop down is
(a) $\sqrt{\frac{g}{2 \mu(R-r)}}$
(b) $\sqrt{\frac{3 g}{2 \mu(R-r)}}$
(c) $\sqrt{\frac{g}{\mu(R-r)}}$
(d) $\sqrt{\frac{2 g}{\mu(R-r)}}$
(2017 Adv.)
The general motion of a rigid body can be considered to be a combination of
Passage 2
(i) a motion of its
centre of mass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass. These axes need not be stationary. Consider, for example, a thin uniform disc welded (rigidly fixed) horizontally at its rim to a massless stick, as shown in the figure. When the disc-stick system is rotated about the origin on a horizontal frictionless plane with angular speed $\omega$, the motion at any instant can be taken as a combination of (i) a rotation of the centre of mass of the disc about the $Z$-axis, and (ii) a rotation of the disc through an instantaneous vertical axis passing through its centre of mass (as is seen from the changed orientation of points $P$ and $Q)$. Both these motions have the same angular speed $\omega$ in this case.
Now, consider two similar systems as shown in the figure : Case (a) the disc with its face vertical and parallel to $x-z$ plane; Case (b) the disc with its face making an angle of $45^{\circ}$ with $x$ - $y$ plane and its horizontal diameter parallel to $X$-axis. In both the cases, the disc is welded at point $P$, and the systems are rotated with constant angular speed $\omega$ about the $Z$-axis.
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Answer:
Correct Answer: 25. (c)
Solution:
- If height of the cone $h \gg r$
$$ \begin{aligned} \text { Then, } \mu N & =m g \\ \mu m(R-r) \omega _0^{2} & =m g \\ \omega _0 & =\sqrt{\frac{g}{\mu(R-r)}} \end{aligned} $$
26-27. (i) Every particle of the disc is rotating in a horizontal circle.
(ii) Actual velocity of any particle is horizontal.
(iii) Magnitude of velocity of any particle is
$$ v=r \omega $$
where, $r$ is the perpendicular distance of that particle from actual axis of rotation (Z-axis).
(iv) When it is broken into two parts then actual velocity of any particle is resultant of two velocities
Here,
$$ v _1=r _1 \omega _1 \text { and } v _2=r _2 \omega _2 $$
$r _1=$ perpendicular distance of centre of mass from $Z$-axis.
$\omega _1=$ angular speed of rotation of centre of mass from $Z$-axis.
$r _2=$ distance of particle from centre of mass and $\omega _2=$ angular speed of rotation of the disc about the axis passing through centre of mass.
(v) Net $v$ will be horizontal, if $v _1$ and $v _2$ both are horizontal. Further, $v _1$ is already horizontal, because centre of mass is rotating about a vertical $Z$-axis. To make $v _2$ also horizontal, second axis should also be vertical.
