Circular Motion Bug Question

[neoking77]

If the coefficient of friction between a bug and the turntable is 0.55 and the bug is 25 cm from the centre, how fast (in RPM) does the turntable have to spin to cause the bug to slide off?

All I know so far is that the frictional force must be 0 (?)
Fc = m4pi^2r/T^2
but this obviously can't work if Ff is 0...so what am I doing wrong?
thanks in advance

 

[neoking77]

hey nvm i got it :)

 

[kyle8921]

I would like to clarify that the frictional force cannot be zero in this situation. The coefficient of friction (μ) is a measure of the resistance to motion between two surfaces in contact. In this case, the bug and the turntable are in contact and therefore, there will always be some frictional force acting on the bug.

To calculate the required RPM for the bug to slide off, we can use the equation Fc = μmg, where Fc is the centripetal force, μ is the coefficient of friction, m is the mass of the bug, and g is the acceleration due to gravity.

Since we know that the bug is 25 cm from the center, we can calculate the radius (r) as 0.25 meters. We can also assume a mass of 0.01 kg for the bug. Plugging these values into the equation, we get Fc = (0.55)(0.01 kg)(9.8 m/s^2) = 0.0539 N.

Now, we can use the equation Fc = m4π^2r/T^2 to calculate the required RPM. Rearranging the equation, we get T = √(m4π^2r/Fc). Plugging in the values, we get T = √((0.01 kg)(4π^2)(0.25 m)/0.0539 N) = 0.371 seconds.

To convert this to RPM, we divide 60 seconds by the time in seconds, so we get 60/0.371 = 161.7 RPM. Therefore, the turntable would need to spin at a speed of approximately 161.7 RPM for the bug to slide off.

I hope this explanation helps. It is important to note that the coefficient of friction may vary depending on the surface and conditions, so this calculation is an approximation. Additionally, other factors such as the shape and size of the bug may also affect the required RPM. Further experimentation and analysis may be necessary for a more accurate result.