Sliding off from the circular plate

[sheeproars]

Homework Statement

Toby is on a merry-go-round. The coefficient of static friction between his feet and
the merry-go-round is 0.5. Toby is located 2 m from the center.
a. Once Toby starts to slide he is going to slide outward until he falls off the merry-go-round at radius = 3m. How many revolutions will the merry-go-round undergo during this part of his journey?

Homework Equations

f(friction) = u(coefficient of Friction) * N(Normal Force) = u*mg
Net F = ma_c = mv^2/r

The Attempt at a Solution

I thought I had to use kinetic friction but I wasn't sure because it wasn't given and there's no way to find it...
so I just used static friction.. f = u * mg
and friction is the only force working in this one.. so F net = f
u*mg = mv^2/r
m's cancel out and I got v = 3.1305m/s as the speed that he will start slipping off the merry-go-round. Then I thought may be I can use Kinematic equations to find the time and find the total revolution.. but I have no idea how to approach it.. help!

 

[anathema]

Thank you for your post. It seems like you are on the right track with your solution so far. To find the number of revolutions, we can use the fact that the circumference of the circle is equal to 2πr, where r is the radius. Since Toby starts at a radius of 2m and ends up at a radius of 3m, we can calculate the distance he travels as 2π(3m) - 2π(2m), or 4πm. This is equal to the distance traveled in one revolution (2πr) multiplied by the number of revolutions, so we can set up the equation 4πm = 2πr * n, where n is the number of revolutions. Solving for n, we get n = 2. This means that Toby will undergo 2 revolutions before sliding off the merry-go-round.

I hope this helps clarify the solution for you. Keep up the good work! If you have any further questions or need any additional assistance, please don't hesitate to ask. Science is all about asking questions and finding answers. Best of luck with your studies!