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Standard XIII

Physics

Centripetal Acceleration

The pulley sy...

Question

The pulley system (shown in the figure) is placed on a smooth horizontal surface which is rotating at an angular velocity of $1 rad/s$ in a circle of radius $10 m$ (in a horizontal plane). The pulley is fastened to the table. Two blocks of masses $m_{1}$ = $3 kg$ and $m_{3} = 10 kg$ are placed on the table. A third block (rough) of mass $m_{2} = 2 kg$ is placed over $m_{1}$ Find the minimum frictional force between the blocks $m_{1}$ and $m_{2}$ such that they move together. Assume the lengths of the strings to be very small when compared to the radius of the circle of rotation.

2 N

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3 N

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4 N

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6.66 N

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Solution

The correct option is D $6.66 N$
Assume that $m_{1} + m_{2}$ is moving radially inward with acceleration $a$ and $m_{3}$ is moving radially outward with acceleration $a$ (since the string is the same, common acceleration of the system is $a$ )

From the frame of reference of the pulley, forces acting on the masses are

$T - (m_{1} + m_{2}) r ω^{2} = (m_{1} + m_{2}) a - (1)$
$m_{3} r ω^{2} - T = m_{3} a - (2)$

Adding both equations:
$a = \frac{r ω^{2} (m_{3} - m_{1} - m_{2})}{m_{1} + m_{2} + m_{3}}$

Since the only force on $m_{2}$ is friction,
For $m_{1}$ and $m_{2}$ to move together, minimum value of friction force $f = m_{2} a = \frac{r m_{2} ω^{2} (m_{3} - m_{1} - m_{2})}{m_{1} + m_{2} + m_{3}}$

Putting $m_{3} = 10 kg$ , $m_{1} = 3 kg$ , $m_{2} = 2 kg$ , $r = 10 m$ , $ω = 1 rad/s$
$f = \frac{10 \times 2 \times 1^{2} \times 5}{15} = 6.66 N$

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Centripetal Acceleration

Standard XIII Physics

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