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

 

[asteriatic]

I would approach this question by first considering the forces acting on the cylinder and using Newton's laws of motion to determine the maximum speed at which the cylinder can move without slipping off the turntable.

Firstly, the weight of the cylinder (mg) and the normal force from the turntable (N) are balanced, so the net force acting on the cylinder in the radial direction is zero. This means that the maximum speed of the cylinder can be calculated using the equation for centripetal force:

F_{centripetal} = \frac{mv^2}{r}

Where m is the mass of the cylinder, v is the speed, and r is the distance from the center of the turntable to the cylinder.

Next, we need to consider the maximum frictional force that can act on the cylinder without it slipping off the turntable. This can be calculated using the coefficient of static friction (μ) and the normal force:

F_{friction,max} = μN

Substituting this into the equation for centripetal force, we get:

\frac{mv^2}{r} = μN

We can then substitute the values given in the question to solve for the maximum speed:

v_{max} = \sqrt{\frac{μN}{m}} = \sqrt{\frac{(0.080)(mg)}{m}} = \sqrt{0.080(9.81)(0.20)} = \sqrt{0.156} = 0.40 \rm m/s

Therefore, the maximum speed at which the cylinder can move along its circular path without slipping off the turntable is 0.40 m/s.