Question

A spool (consider it as a double disc system joined by a short tube at their centre) is placed on a horizontal surface as shown in figure. A light string wound several times over the short connecting tube leaves it tangentially and passes over a light pulley. A weight of mass m is attached to the end of the string. The radius of the connecting tube is r and mass of the spool is $M$ and radius is $R$ . Find the acceleration of the falling mass $m$ . Neglect the mass of the connecting tube and slipping of the spool.

Accepted Answer

We have to find out the acceleration of the falling mass $m$ . Note down the details given in the question. Linear velocity $v_{C M}$ is shown in the diagram and $ω$ is the angular velocity in a clockwise direction.

Since there is no slipping, the velocity of the point of contact of the string with the pulley is the vector sum of the velocity of the point of contact due to rotation plus the velocity of the center of mass of the spool. That is
$v_{B} = v_{C M} - ω r, ω$ is in a clockwise direction
But $v_{C M} = ω R$
$v_{B} = v_{C M} - v_{C M} \frac{r}{R} = v_{C M} (\frac{R - r}{R})$
This is also the velocity $(v)$ of the rolling mass.
$v_{B} = v_{C M} (\frac{R - r}{R}) = v (\frac{R - r}{R})$
Here no dissipation of energy so conservation of energy is applicable. Considering the conservation of energy when $m$ lowers by $h$ we get
$m g h = \frac{1}{2} m v^{2} + (\frac{1}{2} M v_{C M}^{2} + \frac{1}{2} I ω^{2})$

(Moment of inertia $I = \frac{1}{2} M R^{2}$ )

$= \frac{1}{2} m v^{2} + (\frac{1}{2} M v_{C M}^{2} + \frac{1}{2} \times \frac{1}{2} M R^{2} \frac{v_{C m}^{2}}{R^{2}})$
$= \frac{1}{2} m v^{2} + \frac{3}{4} M v_{C m}^{2}$
$= \frac{1}{2} m v^{2} + \frac{3}{4} M v^{2} (\frac{R}{R - r})$
$v^{2} = \frac{4 m g h}{2 m + 3 M {[\frac{R}{(R - r)}]}^{2}}$
by comparing with what $v^{2} = 2 a h$ we get,
$a = \frac{v^{2}}{2 h} = \frac{2 m g}{2 m + 3 M {[R / (R - r)]}^{2}}$

Important Questions on Rigid Body Dynamics: Part 2

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