Question

2
A small block of mass 1 kg is connected by a light string passing through
a hole in a smooth table with a block of mass 3kg which hangs vertically.
Find the angular velocity of the block on the table moving in a circle of
radius 0.1 m in order to keep the other block at rest?
A) 10rad's B) 105 radls C) 102 radis D radis​

Answer 1

Answer:

Angular speed of the block on the table is given as 17.1 rad/s

Explanation:

As we know that 1 kg block is placed on table which is revolving in circle

So here Tension force in the string will provide the required centripetal force on it

So we have

$T = m\omega^2 R$

Also we know that the block which is suspended by string is also in equilibrium

so we have

$Mg = T$

so we have

$Mg = m\omega^2 R$

$3(9.81) = 1 \omega^2 (0.1)$

$\omega = 17.1 rad/s$

#Learn

Topic : Circular motion

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