- #1

Like Tony Stark

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- Homework Statement
- There's an arm which is ##2R## long and rotates with constant angular velocity on a horizontal plane. There's also a slider block with mass ##0.4 kg##. This slider is pushed against a support ##A## by a spring of constant ##2 N/m## and natural length ##2R##.

A) Suppose that the force exerted by the support is half the force exerted by the elastic force, find the angular velocity of the system.

B) Find the minimum angular velocity so that the slider doesn't interact with the support.

C) If the angular velocity found previously is doubled, which would be the equilibrium position with respect to the arm?

D) Which would be the difference if the angular velocity is now rotating the other way?

- Relevant Equations
- Newton's equations

A) The slider experiments three forces, all of them are on the ##x## axis (considering ##x## axis as the axis aligned with the arm): Normal force (exerted by the support), elastic force and centrifugal force, which is ##m.(\omega^2 r)##

Elastic force is equal to

##Fe=-k \delta =-2 (R-2R)=2R## and therefore ##N=R##. Then, Newton equations would be

##N-Fe+m (\omega^2 R)=0##. So, solving for ##\omega## I get:

##m.\omega^2 R=R##, and then, replacing the value of the mass we get ##\omega = \sqrt{2.5}##

B) For B we solve the same equation but set ##N=0##.

C) For C, we solve

##k(R_{eq} -2R)+m\omega^2 R_{eq} =0##, where ##R_{eq}## is the radius for which the system is in equilibrium.

D) The angular velocity affect the centrifugal force, which is ##m.(\omega^2 r)##. As ##\omega## is squared, the sign doesn't change anything.

Have I solved the problem correctly? I don't know if my concepts are correct

Elastic force is equal to

##Fe=-k \delta =-2 (R-2R)=2R## and therefore ##N=R##. Then, Newton equations would be

##N-Fe+m (\omega^2 R)=0##. So, solving for ##\omega## I get:

##m.\omega^2 R=R##, and then, replacing the value of the mass we get ##\omega = \sqrt{2.5}##

B) For B we solve the same equation but set ##N=0##.

C) For C, we solve

##k(R_{eq} -2R)+m\omega^2 R_{eq} =0##, where ##R_{eq}## is the radius for which the system is in equilibrium.

D) The angular velocity affect the centrifugal force, which is ##m.(\omega^2 r)##. As ##\omega## is squared, the sign doesn't change anything.

Have I solved the problem correctly? I don't know if my concepts are correct