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Centripetal Acceleration

Two blocks of...

Question

Two blocks of masses $m_{1} = 10 kg$ 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 a turn table. The coefficient of static friction between the table and block $m_{1}$ is $μ = 0.5$ while there is no friction between block $m_{2}$ and the table. The table is rotating with an angular speed of $10 rad/s$ about a vertical axis passing through its centre $O$ . The masses are placed along the diameter of turn 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 rotating with the turn table without slipping. Calculate the frictional force acting on block $m_{1}$ .

48 N

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36 N

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24 N

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Frictional force is not required as tension force is sufficient to provide centripetal force

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Solution

The correct option is B $36 N$
Given, $m_{1} = 10 kg, m_{2} = 5 kg, ω = 10 rad/s$
Length of string connecting the blocks is $r = 0.3 m$
Distance of mass $m_{1}$ from centre $O$ of the turn table $r_{1} = 0.124 m$
$∴$ Distance of mass $m_{2}$ from centre $O$
$r_{2} = r - r_{1} = 0.176 m$

Since masses $m_{1}$ and $m_{2}$ are rotating with the turn table without slipping, static friction $(f)$ will be acting to support the tension acting on mass $m_{1}$ to avoid the slipping.
i.e Tension $T$ and friction $f$ will be directed towards centre of circular path to provide necessary centripetal force.

From FBD of $m_{1}$ and $m_{2}$ :

All the vertical forces are balanced due to equilibrium in vertical direction.
Along radial direction:
For block $m_{2}$ :
$T = m_{2} ω^{2} r_{2}$ ... (i)

For block $m_{1}$ :
$f + T = m_{1} ω^{2} r_{1}$ ... (ii)

Subtracting Eq. (i) from Eq. (ii), we get:
$f = m_{1} r_{1} ω^{2} - m_{2} r_{2} ω^{2}$
$\Rightarrow f = (m_{1} r_{1} - m_{2} r_{2}) ω^{2}$ ... (iii)

$∴ f = (10 \times 0.124 - 5 \times 0.176) (10)^{2} N = 36 N$

Therefore, frictional force on $m_{1}$ is $36 N$ towards centre of turn table.

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