Laws of Motion 4 Question 9
9. Two blocks of mass $m _1=10 kg$ and $m _2=5 kg$ connected to each other by a massless inextensible string of length $0.3 m$ are placed along a diameter of turn table. The coefficient of friction between the table and $m _1$ is 0.5 while there is no friction between $m _2$ and the table. The table is rotating with an angular velocity of $10 rad / s$ about a vertical axis passing through its centre $O$. The masses are placed along the diameter of table on either side of the centre $O$ such that the mass $m _1$ is at a distance of $0.124 m$ from $O$. The masses are observed to be at rest with respect to an observer on the turn table. $(1997,5 M)$
(a) Calculate the frictional force on $m _1$.
(b) What should be the minimum angular speed of the turn table, so that the masses will slip from this position?
(c) How should the masses be placed with the string remaining taut so that there is no frictional force acting on the mass $m _1$ ?
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Answer:
Correct Answer: 9. (a) $36 N$ (towards centre) (b) $11.67 rad / s$
(c) $m _1$ should be placed at $0.1 m$ from the centre $O$
Solution:
- Given, $m _1=10 kg$,
$$ \begin{aligned} m _2 & =5 kg, \omega=10 rad / s \\ r & =0.3 m, r _1=0.124 m \\ \therefore \quad r _2 & =r-r _1=0.176 m \end{aligned} $$
(a) Masses $m _1$ and $m _2$ are at rest with respect to rotating table. Let $f$ be the friction between mass $m _1$ and table.
Free body diagram of $m _1$ and $m _2$ with respect to ground
$$ \begin{gathered} m _1 \bullet T+f \quad T \longleftrightarrow m _2 \\ T=m _2 r _2 \omega^{2} \end{gathered} $$
Since,
$$ m _2 r _2 \omega^{2}<m _1 r _1 \omega^{2} $$
Therefore, $\quad m _1 r _1 \omega^{2}>T$
and friction on $m _1$ will be inward (toward centre)
$$ f+T=m _1 r _1 \omega^{2} $$
From Eqs. (i) and (ii), we get
$$ \begin{aligned} f & =m _1 r _1 \omega^{2}-m _2 r _2 \omega^{2} \\ & =\left(m _1 r _1-m _2 r _2\right) \omega^{2} \\ & =(10 \times 0.124-5 \times 0.176)(10)^{2} N=36 \end{aligned} $$
Therefore, frictional force on $m _1$ is $36 N$ (inwards)
(b) From Eq. (iii)
$$ f=\left(m _1 r _1-m _2 r _2\right) \omega^{2} $$
Masses will start slipping when this force is greater than $f _{\max }$ or
$$ \left(m _1 r _1-m _2 r _2\right) \omega^{2}>f _{\max }>\mu m _1 g $$
$\therefore$ Minimum values of $\omega$ is
$$ \begin{aligned} & \omega _{\text {min }}=\sqrt{\frac{\mu m _1 g}{m _1 r _1-m _2 r _2}}=\sqrt{\frac{0.5 \times 10 \times 9.8}{10 \times 0.124-5 \times 0.176}} \\ & \omega _{\min }=11.67 rad / s \end{aligned} $$
(c) From Eq. (iii), frictional force $f=0$
where, $\quad m _1 r _1=m _2 r _2$
or $\frac{r _1}{r _2}=\frac{m _2}{m _1}=\frac{5}{10}=\frac{1}{2}$ and $r=r _1+r _2=0.3 m$
$\therefore \quad r _1=0.1 m$ and $r _2=0.2 m$
i.e. mass $m _2$ should placed at $0.2 m$ and $m _1$ at
$0.1 m$ from the centre $O$.
