• Support PF! Buy your school textbooks, materials and every day products Here!

Angular Speed of a reel

  • Thread starter yb1013
  • Start date
  • #1
56
0

Homework Statement



The reel shown below has radius R and moment of inertia I. One end of the block of mass m 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.

http://s721.photobucket.com/albums/ww212/yb1013/?action=view&current=k.gif&newest=1

(a) Find the angular speed of the reel when the spring is again unstretched. (Answer using theta for θ, g for the acceleration due to gravity, and R, I, m, k, and d, as necessary.)

(b) Evaluate the angular speed numerically at this point if I = 0.50 kg·m2, R = 0.300 m, k = 50.0 N/m, m = 0.550 kg, d = 0.200 m, and θ = 37.0°.
ω = ________ rad/s

Homework Equations



F = ma
t = Ia


The Attempt at a Solution



I got through all the other questions on my hw but this one is just giving me a rough time. I really dont know where to start on this question, could somebody please help?
 

Answers and Replies

  • #2
The angular speed of the reel is related to the linear speed of the block. Conservation of mechanical energy might tell you the total kinetic energy of the system, from which you can get the speeds.
 
  • #3
56
0
yeaa, I kind of understand how the system works and what you need for the angular speed, but my real problem comes about on part a. I have no idea how to state it using all those variables..
 
  • #4
Well, I would start by solving for the linear velocity of the block in terms of omega and the vertical distance the block moves in terms of d and theta.

Then I'd set up a conservation of energy equation: (rotational kinetic energy of reel in terms of I and omega) + (linear kinetic energy of block in terms of m, R and omega) = (change in spring potential energy from spring contraction in terms of k and d) + (change in gravitational potential energy from loss of height in terms of m, d and theta).

The above equation should have all the right variables in it; all that should be left for you to do is some algebraic manipulation (I believe you'll be able to factor out an omega^2 from the left side, and from there it should be easy).
 
  • #5
56
0
Well, I would start by solving for the linear velocity of the block in terms of omega and the vertical distance the block moves in terms of d and theta.

Then I'd set up a conservation of energy equation: (rotational kinetic energy of reel in terms of I and omega) + (linear kinetic energy of block in terms of m, R and omega) = (change in spring potential energy from spring contraction in terms of k and d) + (change in gravitational potential energy from loss of height in terms of m, d and theta).

The above equation should have all the right variables in it; all that should be left for you to do is some algebraic manipulation (I believe you'll be able to factor out an omega^2 from the left side, and from there it should be easy).
this is very frustrating... lol
when you explain it, the problem sounds so simple, but when I actually try to put it together, I keep getting the wrong answer and I know Im just doing something wrong..

when your supposed to put equations together while only using the variables, I always get extremely confused... I get the first KEr, but then i dont get how to put KEl in terms of m, R, omega and so on.. I think i might be helpless on this problem..
 
  • #6
I'm sure you can get it. Have you figured out v in terms of r and omega? Think about it: the block is moving at the same speed as the string holding it, which is moving at the same speed as the outer edge of the reel. The reel is spinning at omega radians per second, and the outer edge is moving a linear distance of r meters per radian - some dimensional analysis gives you v = r*omega meters per second.

So you plug that formula for v into the equation for linear KE: KEl=1/2 m*v^2 = 1/2 m*(r*omega)^2 = 1/2 m*r^2*omega^2.

You should already have KEr=1/2 I*omega^2. So, factoring out an omega^2, the left side of the equation should look like:
omega^2 (1/2 I + 1/2 m*r^2)

Any progress on the right side (the potential energies)?
 
  • #7
56
0
ooo ok, i think i've figured it out, that definitely cleared it up alot more.

thanks alot man!
 

Related Threads for: Angular Speed of a reel

Replies
8
Views
4K
  • Last Post
Replies
1
Views
3K
Replies
4
Views
559
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
5K
  • Last Post
Replies
4
Views
448
Replies
3
Views
7K
Top