Circular Motion and Static Friction

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Homework Statement

A small metal cylinder of mass m=0.20 kg sits on a rotating turntable. The coefficient of static friction between the metal cylinder and the turntable is us=.08. The cylinder is located 0.15m from the center of the turntable. Find the maximum speed that the cylinder can have without skidding. Acceleration due to gravity equal 9.81 m/s^2.

Homework Equations

F=MA
Fmax=us x normal force
Aradial=V^2/r

The Attempt at a Solution

Fmax=.08 x 9.81
Fmax=.7848

A=F/m
A=.7848/.2
A=3.94

Tangenitial acceleration equals zero so.

3.94=V^2/.15
V=.767 m/s

but i don't think this is correct. What am i doing wrong?

Any help is appreciated.
Thanks

 

[Delphi51]

Likely you just forgot the mass when you did
Fmax=μmg = .08 x 9.81
Just a suggestion from an old timer - it is often helpful to separate the big picture from the details. I would begin this problem by writing the slipping condition:
Fc = Ff
then fill in the details with
m*v^2/r = μmg
cancel the m's and solve for v. Put in the numbers only in the last step. I can tell you that this kind of approach will serve you well when the problems get more complex.

 

[tomcue]

for your question. It seems like you are on the right track with your calculations. However, there are a few things that could be adjusted to ensure accuracy and clarity.

Firstly, when calculating the maximum static friction force, it is important to use the normal force of the object on the surface, which in this case would be the weight of the cylinder. So the calculation would be Fmax=0.08 x 0.20 x 9.81 = 0.15768 N.

Next, when using the equation A=F/m, it is important to use the net force acting on the object, which in this case would be the centripetal force (Fcentripetal=mv^2/r) minus the maximum static friction force. So the equation would be A=(mv^2/r)-Fmax.

Finally, when solving for the maximum speed, it is important to use the correct units for each variable. In this case, the mass should be in kilograms, the radius in meters, and the acceleration in meters per second squared. This will give you a final answer in meters per second, which is the correct unit for speed.

With these adjustments, the correct calculation would be:

A=((0.20 kg)(v^2)/0.15 m)-0.15768 N

Solving for v, we get v=0.767 m/s, which is the same answer you had. So it seems that your calculations were correct, but there were some small errors in units and using the correct force for the net acceleration. I hope this helps clarify things for you. Keep up the good work!