Question

Can someone help me with this, please? I have the answer but I need help with why the answer is that :(

A block of mass m is placed against the inner wall of a hollow cylinder of radius R that rotates about a vertical axis with a constant angular velocity, ω, as shown above. In order for friction to prevent the mass from sliding down the wall, the coefficient of static friction, μ, between the mass and the wall must satisfy which of the following inequalities? (Ans: μ is greater than or equal to g/(w^2r)

Answer 1

Answer:

See explanation below

Explanation:

Recall that the force of static friction is the product of the coefficient of static friction between surface and object, times the force applied perpendicular to the surface.

In our case, that perpendicular force has magnitude equal to the centripetal acceleration times the mass m of the block. The centripetal acceleration is defined as the quotient between the square of the tangential velocity divided by the radius at which the rotating mass is located (in our case R).

In order to estimate the tangential velocity given the angular velocity ω, we simply multiply it by R:  

$v_t=\omega\,R$

and therefore the force acting on the surface of the cylinder is:

$F=m\,*\,a_c=m\,*\,\frac{v_t^2}{R} =m\,*\, \frac{\omega^2\,R^2}{R} =m\,*\,\omega^2\,*R$

The force of friction between the block and the inner wall of the hollow cylinder will then be:  $f=\mu\, *\,m\,*\omega^2\,*\,R$

In order for this force of static friction to be able to prevent the mass to slide down due to gravity, we need that the friction force is at least equal to the gravitational force. That is:

$f\geq F_g\\f\geq m\,*\,g\\\mu\,*\,m\,*\,\omega^2\,*\,R\geq m\,*\,g\\\mu\geq \frac{g}{\omega^2\,*\,R}$

Answer 2

The force of static friction to prevent the mass to slide down due to gravity, the friction force is must be equal or greater than to the gravitational force.

The force of static friction is the product of the coefficient of static friction between surface,and force applied perpendicular to the surface.

 

The tangential velocity,

$\bold {V_t = \omega R}$

So,  the force acting on the surface of the cylinder,

$\bold {F= m\times a_c}\\\\\bold {F= m \times \dfrac {V_t^2 }{R}}\\\\\bold {F = m \times \dfrac {\omega ^2R2}{R}}}\\\\\bold {F = m \times \omega ^2 \times R}$

The force of friction between the block and the inner wall of the hollow cylinder,

$\bold {F_f = \mu \times m \times \omega ^2 \times R}$

Therefore, the force of static friction to prevent the mass to slide down due to gravity, the friction force is must be equal or grater than to the gravitational force.

To know more about static friction,

https://brainly.com/question/8790433