How Do Rotating Reference Frames Affect Forces on a Merry-Go-Round?

[jimmy.dude]

Looking down from a stationary tree branch, a merry-go-round spins in a counterclockwise
direction with an angular velocity of 1 radian per second. a squirrel of mass 0.2 kg sits on the outer rim of the merry-go-round, at a radius of 2.0 meters.

a) what is the magnitude and direction of the vector omega

b) What is the magnitude and direction of the Coriolis pseudo-force as felt by the squirrel?
(possible equation: -2m(omega x v_r)

c)What is the magnitude and direction of the centrifugal pseudo-force as felt by the squirrel?
(possible equation: -m(omegax(omegaxr_r)

I have no idea where to start!

 

[jimmy.dude]

wrong

 

[jimmy.dude]

wrong

 

[PhanthomJay]

your answer to part a is correct, but for part b,it is not.

For part b, your equation is OK, but what is v_r of the squirrel?

For part c, your magnitude is correct, but you need to explain what is meant by the minus sign regarding the direction of the pseudo centrifugal force.

 

[rwisz]

I would first clarify the concept of rotating reference frames. A rotating reference frame is a coordinate system that rotates with a certain angular velocity relative to an inertial frame. In this case, the merry-go-round is the rotating reference frame, while the stationary tree branch is the inertial frame.

Now, let's address the questions:

a) The magnitude of the angular velocity, denoted as omega (ω), is given in the problem as 1 radian per second. The direction of the vector is counterclockwise, as stated in the problem.

b) The Coriolis pseudo-force is a fictitious force that appears in a rotating reference frame. It is given by the equation -2m(ω x v), where m is the mass of the object, ω is the angular velocity, and v is the velocity of the object relative to the rotating reference frame. In this case, the squirrel is sitting on the outer rim of the merry-go-round, so its velocity relative to the rotating reference frame is tangential to the circle with a magnitude of 2mω. Therefore, the magnitude of the Coriolis pseudo-force is -2m(ω x 2mω) = -4mω^2, and its direction is perpendicular to both the angular velocity and the velocity of the squirrel, pointing towards the center of the merry-go-round.

c) The centrifugal pseudo-force is another fictitious force that appears in a rotating reference frame. It is given by the equation -m(ω x (ω x r)), where m is the mass of the object, ω is the angular velocity, and r is the position vector of the object relative to the rotating reference frame. In this case, the position vector of the squirrel is 2m, and the direction of the angular velocity is counterclockwise. Therefore, the magnitude of the centrifugal pseudo-force is -m(ω x (ω x 2m)) = -2mω^2, and its direction is opposite to the angular velocity, pointing away from the center of the merry-go-round.

I hope this explanation helps you understand rotating reference frames and how to approach these types of problems. As a scientist, it is important to have a strong understanding of fundamental concepts and equations in order to solve complex problems in various fields of study.