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Sati [7]
3 years ago
15

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

Physics
1 answer:
ollegr [7]3 years ago
7 0

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}

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Replacing,

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PART B) To solve this section it is necessary to apply the concepts related to the conservation of energy both potential (simple harmonic) and kinetic in the spring.

\frac{1}{2}kA^2 = \frac{1}{2}mv^2 + \frac{1}{2} kY^2

Where,

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This expression is equivalent to,

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E = \frac{1}{2}kA^2

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