A coin placed on a rotating table just slips when it is placed at a distance of 1 cm from the center. If the angular velocity of the table in halved, it will just slip when placed at a distance of _________ from the centre :
Question
A coin placed on a rotating table just slips when it is placed at a distance of 1 cm from the center. If the angular velocity of the table in halved, it will just slip when placed at a distance of _________ from the centre :
Solution
The slipping of the coin is due to the lack of enough frictional force to keep it in circular motion. The frictional force is given by the equation:
F = m * r * w^2
where: F is the frictional force, m is the mass of the coin, r is the radius (distance from the center of the table), w is the angular velocity.
In this case, the frictional force is just enough to keep the coin from slipping, so we can say that the frictional force equals the gravitational force:
m * r * w^2 = m * g
where g is the acceleration due to gravity. We can cancel out the mass of the coin from both sides:
r * w^2 = g
We know that the coin just slips when it is placed at a distance of 1 cm from the center and the angular velocity is w. So we can write:
1 * w^2 = g
Now, if the angular velocity is halved, the new angular velocity is w/2. We want to find the new radius r' where the coin just slips. We can write:
r' * (w/2)^2 = g
Solving for r', we get:
r' = 4 * (1 * w^2) / (w/2)^2 = 4 cm
So, if the angular velocity of the table is halved, the coin will just slip when placed at a distance of 4 cm from the center.
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