Solve Using Conservation of Energy

[joemama69]

Homework Statement

Solve Using Conservation of Energy

The spool has a mas of 50 kg and a radius of gyration k(o) = .280m. If the 20 kg block A is released form rest, determine the distance the block must fall in order for the spool to have an angular velocity w = 5 rad/s. Also what is the tension in the cord while the block is in motion? Neglect the mass of the cord

Homework Equations

The Attempt at a Solution

T₁ + V₁ + T₂ + V₂

Does V₁ = 0 because it is at rest from the start

T₁ = .5mv² + .5Iw² = .5*20*0 + .5*50*.28² = 1.96

V_2g = Wy (y is what I am looking for but how do i calculate work W)
V_2e = .5ks² = .5*.28*s² = .14s²

T₂ = .5*20*v² + .5*50*.28² = 10v² + 1.96

1.96 = Wy + .14s² + 10v² + 1.96



To answer the question "determine the distanc ethe block must fall" means solve for y right

To answer the question "what is the tension of the cord" means solve for s right

How do i find v

 

[LowlyPion]

The distance the block falls is your energy..

What does that go into?

m_b*g*h = 1/2*m_b*V² + 1/2*m_s*R_s²*ω²

V = ω*R_s

So h is what you are solving for and I think that looks like ...

h = 1/2*ω²R_s²*(m_b + m_s)/(g*m_b)

 

[nlightner]

2?

Yes, to solve for the distance the block must fall, you need to solve for y. To solve for the tension in the cord, you need to solve for s.

To find v2, you can use the equation v^2 = u^2 + 2as, where u is the initial velocity (which is 0 in this case), a is the acceleration (which can be found using conservation of energy), and s is the distance the block falls.

So, solving for y, you can use the equation you have already set up:

1.96 = Wy + .14s^2 + 10v2 + 1.96

Since the block is released from rest, the initial velocity (u) is 0, so the equation becomes:

1.96 = Wy + .14s^2 + 10(0) + 1.96

Simplifying, you get:

1.96 = Wy + .14s^2 + 1.96

Subtracting 1.96 from both sides:

Wy + .14s^2 = 0

Now, you can use conservation of energy to solve for a (acceleration):

T1 + V1 = T2 + V2

Since the block is released from rest, V1 = 0. Also, since the block is falling, T2 = 0. So the equation becomes:

T1 = V2

Substituting in the equations for T1 and V2 from the problem statement, you get:

.5mv^2 + .5Iw^2 = Ws

Now, you can solve for a:

a = W/m

Substituting this into the equation for y:

Wy + .14s^2 = 0

You get:

Ws + .14s^2 = 0

Substituting in the value for a:

(W/m)s + .14s^2 = 0

Solving for s, you get:

s = -Wm/0.14

Since we know the values for W and m, we can substitute them in and solve for s:

s = -(50*5)/0.14 = -1785.71

Since distance cannot be negative, the block must fall approximately 1785.71 meters in order for the spool to have an angular velocity