Question

You are sitting on the edge of a horizontal disk (for example, a playground merry-go-round) that has radius 3.00 m and is rotating at a constant rate about a vertical axis. The coefficient of static friction between you and the surface of the disk is 0.400.
1) What is the minimum time for one revolution of the disk if you are not to slide off?

2) Your friend's weight is half yours. If the coefficient of static friction for him is the same as for you, what is the minimum time for one revolution if he is not to slide off?

Answer 1

To solve this problem we will make a free body diagram from which we will obtain the magnitude and equivalence of each of the forces. After that we will find the equivalent expression of time and we will make the pertinent considerations for the second part when the weight is diminished.

According to the free body diagram made the meaning of the variables are

N = Normal Force

W = mg = Weight of the person

$f_s =$ Static Friction

If we do sum of vertical direction we have,

$\sum F_y = 0$

$N-mg = 0$

$N = mg$

Now the maximum static friction is defined as

$f_s = \mu_s N$

$f_s = \mu_s mg$

Since person is moving in the circular path, so, net force would be f_s towards the center of the circular path

$f_s$ = Net Force = Centripetal force

$\mu_s mg = \frac{mv^2}{R}$

Here v is the velocity of the person on disk around the circular path, R is the radius of circular disk

$v = \sqrt{\mu_s gR}$

The distance traveled by the person in one round trip around the circular path is

$\phi= 2\pi R$

Then time taken by the person to move around one round trip is

$t = \frac{d}{v}$

$t = \frac{2\pi R}{\sqrt{\mu_s Rg}}$

PART A)  Replacing with our values we have that the minimum time is

$t = \frac{2\pi (3)}{\sqrt{(0.4)(3)(9.8)}}$

$t = 5.4966s$

PART B) Since the expression does not depends on the weight (mass) so, time interval for one round trip would be same as it was in part A.

$t = 5.4966s$