Question

A rock is sitting at the edge of a flat merry-go-round at a distance of 1.6 meters from the center. The coefficient of static friction between the rock and the merry-go-round is 0.70. What is the minimum angular velocity of the merry-go-round that would cause the rock to begin sliding off?

Answer 1

Answer:

ω = 2.1 rad/sec

Explanation:

• As the rock is moving along with the merry-go-round, in a circular trajectory, there must be an external force, keeping it on track.
• This force, that changes the direction of the rock but not its speed, is the centripetal force, and aims always towards the center of the circle.
• Now, we need to ask ourselves: what supplies this force?
• In this case, the only force acting on the rock that could do it, is the friction force, more precisely, the static friction force.
• We know that this force can be expressed as follows:

       $f_{frs} = \mu_{s} * F_{n} (1)$

      where μs = coefficient of static friction between the rock and the merry-

      go-round surface = 0.7, and Fn = normal force.

• In this case, as the surface is horizontal, and the rock is not accelerated in the vertical direction, this force in magnitude must be equal to the weight of the rock, as follows:
• Fn = m*g (2)
• This static friction force is just the same as the centripetal force.
• The centripetal force depends on the square of the angular velocity and the radius of the trajectory, as follows:

       $F_{c} = m* \omega^{2}*r (3)$

• Since (1) is equal to (3), replacing (2) in (1) and solving for ω, we get:

       $\omega = \sqrt{\frac{\mu_{s} * g}{r} } = \sqrt{\frac{0.7*9.8m/s2}{1.6m}} = 2.1 rad/sec$

• This is the minimum angular velocity that would cause the rock to begin sliding off, due to that if it is larger than this value , the centripetal force will be larger that the static friction force, which will become a kinetic friction force, causing the rock to slide off.