How Do You Calculate Centripetal Forces on a Merry-Go-Round?

[imnotsmart]

A 50.0 kg child stands at the rim of a merry-go-round of radius 1.90 m rotating with an angular speed of 3.00 rad/s.
(a) What is the child's centripetal acceleration?
(b)What is the minimum force between her feet and the floor of the merry-go-round that is required to keep her in the circular path?
(c) What minimum coefficient of static friction is required?

Need some help getting started. Don't really know where to go here?

 

[whozum]

a)
v_{linear} = \omega r

That should give you a velocity, along with radius you can find acceleration.

b) You know that the only force keeping her in circular motion is the centripetal force, and the only forces acting are friction and gravity. Does gravity alone help her stay in circular motion? If not, how does friction do the job?

c) This is just number crunching after you figure out the equation in (b)

 

[thepatientmental]

To solve this rotational motion problem, we can use the formula for centripetal acceleration, which is given by a = ω^2r, where ω is the angular speed and r is the radius of the circular path.

(a) Plugging in the given values, we get a = (3.00 rad/s)^2 * 1.90 m = 17.1 m/s^2. So, the child's centripetal acceleration is 17.1 m/s^2.

(b) To find the minimum force between her feet and the floor, we can use Newton's second law, which states that the net force on an object is equal to its mass multiplied by its acceleration. In this case, the net force is the force between the child's feet and the floor, and the acceleration is the centripetal acceleration we calculated in part (a). So, we have F = ma = (50.0 kg)(17.1 m/s^2) = 855 N. Therefore, the minimum force between her feet and the floor is 855 N.

(c) To find the minimum coefficient of static friction, we can use the formula F = μN, where F is the force of friction, μ is the coefficient of static friction, and N is the normal force (in this case, equal to the force between the child's feet and the floor). Rearranging the formula, we get μ = F/N = 855 N / (50.0 kg * 9.8 m/s^2) = 1.75. So, the minimum coefficient of static friction required is 1.75.

In summary, the child's centripetal acceleration is 17.1 m/s^2, the minimum force between her feet and the floor is 855 N, and the minimum coefficient of static friction required is 1.75.