KE Conservation: Calculating Final Velocities

[goonking]

Homework Statement

Homework Equations

KEi = KEf

The Attempt at a Solution

since it is perfectly elastic, kinetic energy before = kinetic energy after

(1/2) (m) (20m/s^2) = (1/2) (2m) (VBeamf^2) + (1/2) (m) (VBlockf^2)

is this the right approach? [/B]

 

[ehild]

Homework Statement

Homework Equations

KEi = KEf

The Attempt at a Solution

since it is perfectly elastic, kinetic energy before = kinetic energy after

(1/2) (m) (20m/s^2) = (1/2) (2m) (VBeamf^2) + (1/2) (m) (VBlockf^2)

is this the right approach? [/B]

Not quite. The beam can only rotate about the axis. What is its kinetic energy?

 

[goonking]

Not quite. The beam can only rotate about the axis. What is its kinetic energy?

(1/2) (I) (ω^2) ?

 

[ehild]

(1/2) (I) (ω^2) ?

Yes. And you need one more equation. What else is conserved?

 

[goonking]

Yes. And you need one more equation. What else is conserved?

i have no idea, momentum?

 

[ehild]

Momentum is conserved when translation is involved. But the beam can not translate. What can be conserved?

 

[goonking]

Momentum is conserved when translation is involved. But the beam can not translate. What can be conserved?

I can't think of anything else besides momentum and kinetic energy. any hints?

 

[ehild]

For rotation, you have quantities analogous to translational ones. For position, it is the angle of rotation. For velocity, it is angular velocity. For mass, it is moment of inertia. What do you have for momentum?

 

[goonking]

For rotation, you have quantities analogous to translational ones. For position, it is the angle of rotation. For velocity, it is angular velocity. For mass, it is moment of inertia. What do you have for momentum?

angular momentum?

but if the initial angular momentum is 0, then the final has to be 0 too? since it was at rest

 

[ehild]

The angular momentum of the beam is zero, but the little mass has initial and final angular momentums with respect to the axis. How is the angular momentum defined for a point-like body, moving in the plane normal to an axis, traveling with velocity v, at distance D from an axis?

 

[goonking]

The angular momentum of the beam is zero, but the little mass has initial and final angular momentums with respect to the axis. How is the angular momentum defined for a point-like body, moving in the plane normal to an axis, traveling with velocity v, at distance D from an axis?

L = mvr ?

 

[ehild]

L = mvr ?

yes. What is r in this problem?

 

[goonking]

yes. What is r in this problem?

distance traveled before the block hit the beam? I have no idea

 

[ehild]

distance traveled before the block hit the beam? I have no idea

No, it is the distance of the line of track of the little mass from the axis of rotation. It is the same as the distance from the axis of the point where the small mass hits the beam .

 

[goonking]

No, it is the distance of the line of track of the little mass from the axis of rotation. It is the same as the distance from the axis of the point where the small mass hits the beam .

so it is just D/2?

 

[ehild]

so it is just D/2?

Yes. Now you can write the equation for conservation of angular momentum.

 

[goonking]

Yes. Now you can write the equation for conservation of angular momentum.

i don't understand the use of this equation here.

m vi D/2 = m vf D/2 ?

 

[ehild]

The beam also has got angular momentum, you need to include it.

 

[goonking]

The beam also has got angular momentum, you need to include it.

m vi D/2 = (m vf D/2) + ( I ω)

?

 

[ehild]

m vi D/2 = (m vf D/2) + ( I ω)

?

Yes, it is. You know the expression for the moment of inertia for the beam. And you have the equation for the energies.

Eliminate ω and solve for vf.

 

[goonking]

Yes, it is. You know the expression for the moment of inertia for the beam. And you have the equation for the energies.

Eliminate ω and solve for vf.

what is the moment of inertia for the beam? isn't there a whole list of moment of inertia for each shape?

 

[ehild]

You can use the formula valid for a thin rod of length D wit respect of the centre.

 

[ehild]

what is the moment of inertia for the beam? isn't there a whole list of moment of inertia for each shape?

You can use the formula valid for a thin rod of length D with respect of the centre.

 

[goonking]

You can use the formula valid for a thin rod of length D with respect of the centre.

hmm, i didn't know you could do that, since the beam doesn't look like a thin rod.

anyways, it is 1/12*M* L^2

L being D/2 correct?

 

[ehild]

hmm, i didn't know you could do that, since the beam doesn't look like a thin rod.

anyways, it is 1/12*M* L^2

L being D/2 correct?

No the length L is D.
It does not look like a thin rod, but nothing is given about its width.

 

[goonking]

No the length L is D.
It does not look like a thin rod, but nothing is given about its width.

so basically, we have 2 equations.

the kinetic energy and the angular momentum.

do we equate them?

 

[ehild]

You can not equate angular momentum with energy. You have a system of two equations for the unknowns, final velocity vf and angular velocity ω. Solve.

 

[goonking]

1/2 mv^2 = (1/2) (I) (ω^2)

correct?

we use that to solve for ω?

 

[ehild]

1/2 mv^2 = (1/2) (I) (ω^2)

correct?

we use that to solve for ω?

It is wrong.

 

[goonking]

It is wrong.

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?

 

[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.