I really - angular speed and speed at equilibrium position

In summary: And, if you're really stuck, you could try kinetics.org or hastings' laws.In summary, the angular speed of the reel when the spring is again unstretched is 2.40 revolutions per minute.
  • #1
BananaRed
7
0
I really need help !- angular speed and speed at equilibrium position

Problem 1. A 5.00-kg block free to move on a horizontal, frictionless surface is attached to one end of a light horizontal spring. The other end of the spring is held fixed. The spring is compressed 0.100m from equilibrium and released. The speed of the block is 1.20m/s when it passes the equilibrium position of the spring. The same experiment is now repeated with the frictionless surface replaced by a surface for which the coefficient of kinetic friction is 0.300. Determine the speed of the block at the eqilibrium position of the spring.

The second problem I have is: The reel has radius R and moment of inertia I. One end is connected to a spring of force constant k, and the other end is fastened to a cord wrapped around the reel. The reel axle and the incline are frictionless. The reel is wound counterclockwise so that the spring stretches a distance d from its unstretched position and is then released from rest. Find the angular speed of the reel when the spring is again unstretched.
:confused: :confused: :confused: If someone could be help me I would really appreciate it
 
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  • #2
What have you done? Where are you stuck?
 
  • #3
getting started..I lost my physics book while I was packing and now I'm having trouble starting these two problems. I can remember how to start them.
 
  • #4
BananaRed said:
getting started..I lost my physics book while I was packing and now I'm having trouble starting these two problems. I can remember how to start them.
Is that can or can't?

Some useful facts (not all necessarily needed for these problems):

[tex]x = A\cos(\omega t + \phi)[/tex]
[tex]v = -\omega A \sin(\omega t + \phi)[/tex]
[tex]U = \frac{1}{2}kx^2[/tex]
[tex]K = \frac{1}{2}mv^2[/tex]
[tex]K = \frac{1}{2}I\omega^2[/tex]
[tex]\omega = 2\pi f[/tex]
[tex]s = r\theta[/tex]
[tex]v_t = r\omega[/tex]

A hint for both problems: what can you determine about the total energy of the system?
 
  • #5
:confused: :confused: :confused: :confused:
 
  • #6
In the first setup, there is no friction, and you can ignore gravity since acts perpendicular to the direction of motion. So the only force you need to consider is the spring tension. This is a conservative force, meaning that TOTAL energy is conserved -- i.e. constant. Total energy in this case is kinetic energy plus potential energy.

When the spring has sprung back to its equilibrium position (it is not compressed or stretched), how much potential energy is there? How much kinetic energy? (You are given the mass and the velocity at this point.) Now you know the total energy.

When the spring is at the maximum compression or extension and the object is NOT MOVING, how much kinetic energy does it have? How much potential energy does it have? (You can figure this out since you know the total energy, which is constant as long as there is no friction.) Now, knowing the amount of potential energy, and the amount of displacement (0), you could figure out the spring constant k (but you don't have to to solve this problem). What's important is to know how much energy the system has, and what kind.

But on the second run, there is friction. You can figure out how much work is done by friction as the block slides back towards the equilibrium point, and then ...

(I don't want to do the whole thing for you. Work on it.)
 
  • #7
sorry you're so confused...I'll already figure out the first one.The second is the one I'm having trouble with
 
  • #8
Do you have any numbers for the second problem?

Here's a couple equations that might help your thinking

alpha = a / R, (a is the tangental acceleration, not the radial acceleration)

angular speed = alpha * time

Think about the spring. The spring constant's always a biggy.

U = 1/2*k*x^2 could come in very handy. Like gnome said, energy's handy.
 

What is angular speed?

Angular speed is a measure of how fast an object is rotating or moving in a circular path. It is typically measured in radians per second (rad/s) or degrees per second (deg/s).

What is speed at equilibrium position?

The speed at equilibrium position is the maximum speed an object has when moving in a circular motion. It is achieved when the object is at its furthest distance from the center of rotation and is equal to the angular speed multiplied by the radius of the circular path.

How is angular speed calculated?

Angular speed can be calculated by dividing the change in angular displacement by the change in time. It can also be calculated by dividing the linear speed by the radius of the circular path.

What factors affect angular speed?

The factors that affect angular speed include the radius of the circular path, the magnitude of the centripetal force, and the mass of the object. In addition, the angular speed may also be affected by external forces such as friction or air resistance.

How is angular speed related to linear speed?

Angular speed and linear speed are directly related. As the angular speed increases, the linear speed also increases. This relationship is described by the formula v = rω, where v is the linear speed, r is the radius of the circular path, and ω is the angular speed.

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