KE Conservation: Calculating Final Velocities

[ehild]

shouldn't the M in I =1/12*M* D^2 be 2m?

so
I =1/12*2m* D^2

?

D is unknown, solve in terms of it. And yes, the mass of the beam is M=2m

 

[goonking]

shouldn't the M in I =1/12*M* D^2 be 2m?

D is unknown, solve in terms of it. And yes, the mass of the beam is M=2m

after doing some algebra in the angular momentum section. i get :

10 = (Vf /2) + ( D ω )

seems right so far?

D = (10- (Vf/2)) / ω

 

[ehild]

No, I do not think it is right. And you need vf, not D.

 

[goonking]

after doing some algebra in the angular momentum section. i get :

10 = (Vf * 1/2) + ( D ω )

seems right so far?

No, I do not think it is right. And you need vf, not D.

the D's must cancel out somewhere, I'm still trying to figure out where.

 

[ehild]

the D's must cancel out somewhere, I'm still trying to figure out where.

Work symbolically. Keep D and isolate omega.

 

[goonking]

Work symbolically. Keep D and isolate omega.

ω = (60 D - Vf D) / 2D^2

correct?

 

[goonking]

ω = (60 D - Vf D) / 2D^2

correct?

if my algebra is correct, i can simplify more by canceling out the D's to get :

ω = (60 - Vf) / 2?

 

[ehild]

It is certainly wrong, Check the dimensions.

 

[goonking]

I will continue looking at this problem when i come back from school. hopefully I will see things I didn't see before lol

 

[ehild]

Kinetic energy :
1/2 m vi^2 = (1/2 m vf^2) +(1/2) (I) (ω^2)angular momentum:

m vi D/2 = (m vf D/2) + ( I ω)

I =1/12*M* D^2

=
m vi D/2 = (m vf D/2) + ( (1/12*M* D^2) ω)

These are the correct equations, and M=2m.

From the first equation, you got

ω = (60 D - Vf D) / 2D^2

which is almost correct, but you have an algebraic error.

Substituting $I=\frac{1}{12} (2m) D^2$ into the angular momentum equation (m vi D/2 )= (m vf D/2) + ( I ω), you get

$m v_i \frac{D}{2}= m v_f \frac{D}{2}+ \frac{1}{12} (2m) D^2 \omega$ .

Simplifying, it results in $v_i - v_f = \frac{1}{3} D \omega$ . I think you made the error when multiplying by the constants.

So you get Dω in terms of the velocities. Working with the energy equation, (Dω )² appears. Substitute .

 

[goonking]

These are the correct eq uations, and M=2m.

From the first equation, you got

which is almost correct, but you have an algebraic error.

Substituting $I=\frac{1}{12} (2m) D^2$ into the angular momentum equation (m vi D/2 )= (m vf D/2) + ( I ω), you get

$m v_i \frac{D}{2}= m v_f \frac{D}{2}+ \frac{1}{12} (2m) D^2 \omega$ .

Simplifying, it results in $v_i - v_f = \frac{1}{3} D \omega$ . I think you made the error when multiplying by the constants.

So you get Dω in terms of the velocities. Working with the energy equation, (Dω )² appears. Substitute .

yes, the algebra was very long.

anyways, i got 1/2 m v_o ² = 1/2 m v_f² + 1/2 ((m d²) / 6 ) (3/d (v_o - v_f))²

1/2's and m's cancel out from each side.

v_o² = v_f² + (d²/6) ((9/d² (v_o - v_f)²)

d's cancel out

v_o² = v_f² + (3/2) (v_o- v_f)²

now do should i foil the (v_o- v_f)²?

 

[ehild]

v_o² = v_f² + (3/2) (v_o- v_f)²

now do should i foil the (v_o- v_f)²?

You can substitute the numerical value of v_o now. Expand the square and solve the quadratic equation.