Question

A small block is supported by a turn-table. The friction coefficient between block and surface is $μ_{s}$ .
(a) If turn-table rotates at constant angular speed $ω,$ what can be the maximum angular speed for which the block does not slip?
(b) If the angular speed is increased uniformly from rest with an angular acceleration $α,$ at what speed will the block slip?

Accepted Answer

(a) The weight of the block should be balanced by the normal reaction by the turn table on the block. The block moves in the circular path, the friction force $f$ acting on the block provides required centripetal force for its circular motion to keep the block in the circle.

Since angular speed is constant, the block has centripetal acceleration only. From Newton's second law in radial direction,
$F_{net} = f = m a_{r} = m ω^{2} r$ $. . . (i)$
If the block does not slip the friction should be static in nature, then,

$f \leq μ_{s} m g$
$m ω^{2} r \leq μ_{s} m g$

$\Rightarrow ω \leq \sqrt{\frac{μ_{s} g}{r}}$
Hence, maximum angular speed, $ω_{\max} = \sqrt{\frac{μ_{s} g}{r}}$

(b) When the turn-table rotates with angular acceleration, the block has centripetal as well as tangential acceleration.
Therefore, the net acceleration of the box will be,

${\vec{a}}_{net} = {\vec{a}}_{c} + {\vec{a}}_{t}$
$|{\vec{a}}_{R}| = \sqrt{a_{c}^{2} + a_{t}^{2}} = \sqrt{{(ω^{2} r)}^{2} + (r α)^{2}}$
Resultant acceleration of block is parallel to surface of turntable. The only force that is parallel to surface is force of friction.

Net force acting on the block,
$F_{n} = m a_{net} = m \sqrt{{(ω^{2} r)}^{2} + (r α)^{2}}$
If the block does not slip the friction should be static in nature $f \leq μ_{s} m g$ $m \sqrt{{(ω^{2} r)}^{2} + (r α)^{2}} \leq μ_{s} m g x$

$ω \leq {[{(\frac{μ_{s} g}{r})}^{2} - α^{2}]}^{1 / 4}$
Hence, maximum angular speed, $ω_{\max} = {[{(\frac{μ_{s} g}{r})}^{2} - α^{2}]}^{1 / 4}$ .

B M Sharma Solutions for Chapter: Circular Motion, Exercise 35: ILLUSTRATION 9.34

B M Sharma Physics Solutions for Exercise - B M Sharma Solutions for Chapter: Circular Motion, Exercise 35: ILLUSTRATION 9.34

Questions from B M Sharma Solutions for Chapter: Circular Motion, Exercise 35: ILLUSTRATION 9.34 with Hints & Solutions

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