KE Conservation: Calculating Final Velocities

[ehild]

ok, I'm so lost right now.

lets try to clear things up a bit.

the block is moving, hits into the beam.

I'm guessing the beam makes a full revolution and comes back down to the initial position?

then the block moves backwards with a velocity that was less than what it initially was. and since it was going backwards, the velocity is negative.

is everything correct so far?

No. You need to consider the situation just before and after collision. The block had some velocity initially, the rod was in rest.
The block has some angular momentum both before and after the collision. The beam has angular momentum after the collision. The angular momentum before collision is the same as the angular momentum after it.
Also, the initial KE of the block is the same as the KE of the block plus the rotational energy of the beam just after collision.
You do not know the direction of the final velocity of the block. Consider it positive. If it comes out to be negative at the end, it would mean that the block moves backward after the collision.

 

[goonking]

it's a bit hard trying to grasp how an object moving linearly can have an angular momentum. anyways:

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)

 

[ehild]

it's a bit hard trying to grasp how an object moving linearly can have an angular momentum. anyways:

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)

It is right now. Solve for vf.

 

[goonking]

It is right now. Solve for vf.

is this just all algebra now?

how do we eliminate ω?

and D is unknown.

 

[goonking]

It is right now. Solve for vf.

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

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

?

 

[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.