Calculate Max Angular Velocity of Spring: Moment of Inertia Homework

[physicsfun_12]

Homework Statement

I linear spring has one end attached to a ceiling, the other end hanging freely. It was noted that the spring was extended by 120mm when a mass of 2.5kg was attached at the bottom.

If the moment of inertia of the mass is 10kg.m² about its vertical axis, calculate the maximum angular velocity of the spring when the system is released from from the 360 degree twisted position. Assume no change in height and the system is conservative.

I am not to sure how to approach this. I think it may have something to so with the following formulas:

Homework Equations

K.E=1/2.K.theta² and K.E=1/2.I.w²

where k=torsional stiffness, I=moment of inertia and w=angular velocity

The Attempt at a Solution

Other than those equations, I'm really not sure.

Any help much appriciated with this, thanks in advance.

Mike

 

[Yitzach]

Wikipedia informs me that $$\kappa$$ is also called the spring constant. The reason for the "2.5 kg mass extends the spring 120 mm" is so you can find it using Hooke's law where k is preferred. You should be able to get much closer to the answer with that.

 

[kerimek]

I would approach this problem by first understanding the physical principles involved. In this case, we are dealing with a system that involves both linear and rotational motion. The linear motion is caused by the weight of the mass attached to the spring, while the rotational motion is caused by the torsional stiffness of the spring.

To calculate the maximum angular velocity of the spring, we need to consider the conservation of energy in the system. When the system is released, the potential energy stored in the spring will be converted into kinetic energy. This kinetic energy will be shared between the linear motion of the mass and the rotational motion of the spring.

Using the equations you mentioned, we can set up the following equation:

1/2*K*theta^2 = 1/2*I*w^2

Where K is the torsional stiffness of the spring, theta is the angle of rotation of the spring, and I is the moment of inertia of the mass. We can rearrange this equation to solve for w, the angular velocity:

w = √(K*theta^2/I)

To find the maximum angular velocity, we need to find the maximum angle of rotation of the spring. This can be found by considering the 120mm extension of the spring. We can use the equation for Hooke's Law (F = -Kx) to relate the extension of the spring (x) to the force applied (F) by the mass:

K*x = F = m*g

Where m is the mass of the object and g is the acceleration due to gravity. Solving for x, we get:

x = m*g/K

Substituting this value for x into our equation for the maximum angle of rotation, we get:

theta = x/R = m*g/(K*R)

Where R is the length of the spring. Now, we can substitute this value for theta into our equation for w to find the maximum angular velocity:

w = √(K*(m*g/(K*R))^2/I)

Simplifying this equation, we get:

w = √(m*g/(K*R*I))

Therefore, the maximum angular velocity of the spring can be calculated by taking the square root of the mass of the object times the acceleration due to gravity, divided by the product of the torsional stiffness, length of the spring, and moment of inertia of the mass.

I hope this helps you to approach this problem and understand the physical principles involved.