Max Speed of 0.20 kg Mass on Rotating Turntable

[kenau_reveas]

Information given:

A small metal cylinder rests on a circular turntable that is rotating at a constant speed as illustrated in the diagram View Figure .

The small metal cylinder has a mass of 0.20 \rm kg, the coefficient of static friction between the cylinder and the turntable is 0.080, and the cylinder is located 0.15 \rm m from the center of the turntable.

Take the magnitude of the acceleration due to gravity to be 9.81 \rm m/s^2


Question:

What is the maximum speed v_max that the cylinder can move along its circular path without slipping off the turntable?
Express your answer numerically in meters per second to two significant figures.

 

[Chi Meson]

https://www.physicsforums.com/showthread.php?t=94379

 

[fearthebob]

The centripetal force Fc, for a body of mass m traveling with speed v in a circle of radius r is,

Fc = (mv^2)/r....eqn 1

The object to slip off the turntable if the centripetal force overcomes the maximum force due to static friction F_max:

F_s = (u_s)N = (u_s)mg...eqn 2

u_s is the coefficient of static friction.
g = 9.8 m/s^2
m = mass

Using Eq. 1 and Eq. 2 (Fc=Fs) we solve for the maximum speed vmax.

Thus v_max = sqrt((u_s)mg)

you can do the math.

 

[psysics]

The above process is correct but when u solve eqn1 n 2 u get
Vmax = sqrt(μs*r*g) = =0.34 m/s