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 1
 Homework Statement

You are sitting on the edge of a horizontal disk (for example, a playground merrygoround) 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 0.280
A) What is the minimum time for one revolution of the disk if you are not to slide off? Express your answer with the appropriate units.
B) 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? Express your answer with the appropriate units.
 Homework Equations
 a = 4*π^2*r / T^2
A) So we are given the radius and the coefficient of static friction as 3.0 m and 0.28 respectively. I know that in the vertical direction the only forces acting are the normal force and the gravitational force. Therefore, the normal force is equal to mg because net force is equal to 0, due to no vertical acceleration.
I can find the force of static friction, F_{S} = F_{N} * μ_{S}
Net force in the horizontal direction, ma = F_{S}, the only force is the force of friction
ma = mg * μ_{S}
a = g * 0.28
a = 2.744 m/s^{2}
I'm not entirely sure where the frictional force would point on a freebody diagram but, I can use the acceleration to solve for the time for one revolution (T) with the equation mentioned above.
2.744 = 4 * π^{2} * (3) / T^{2}
T = √(4 * π^{2} * (3) / 2.744)
T = 6.57 s
B) For this part, I don't think the differences in mass will have an affect on the time for one revolution.
When solving for the acceleration,
0.5m * a = 0.5m * g * μ_{S}
The 0.5m will just cancel out and we'll be left with the same acceleration and the same time for one revolution.
T = 6.57 s
Is my answer correct? I'm not sure if there is anything I missed or if I used my equations incorrectly.
I can find the force of static friction, F_{S} = F_{N} * μ_{S}
Net force in the horizontal direction, ma = F_{S}, the only force is the force of friction
ma = mg * μ_{S}
a = g * 0.28
a = 2.744 m/s^{2}
I'm not entirely sure where the frictional force would point on a freebody diagram but, I can use the acceleration to solve for the time for one revolution (T) with the equation mentioned above.
2.744 = 4 * π^{2} * (3) / T^{2}
T = √(4 * π^{2} * (3) / 2.744)
T = 6.57 s
B) For this part, I don't think the differences in mass will have an affect on the time for one revolution.
When solving for the acceleration,
0.5m * a = 0.5m * g * μ_{S}
The 0.5m will just cancel out and we'll be left with the same acceleration and the same time for one revolution.
T = 6.57 s
Is my answer correct? I'm not sure if there is anything I missed or if I used my equations incorrectly.