Find angular velocity of the system

[huybinhs]

Homework Statement

I did:

Center of mass = m(R+r) / M+m = 0.0708*0.0956 / (0.123+0.0708) = 0.034925 m

I did:

sys = m1, m2

Pf = Pi + Isystem => (M+m) v = mvi => v = m*vi / (M+m) = 0.533375 m/s

How to get start on this part? Please help!
Please check if part a and b are correct! Thanks!

 

[thepatient]

For parts a and b, those look good. In part c though, remember angular momentum is equal to the cross product between the position vector and momentum vector. The magnitude would be: L = |r| * |mv| * sinß. (I couldn't find theta so I used ß)

 

[huybinhs]

Which v should I use and how can I find theta?

 

[gneill]

For parts a and b, those look good. In part c though, remember angular momentum is equal to the cross product between the position vector and momentum vector. The magnitude would be: L = |r| * |mv| * sinß. (I couldn't find theta so I used ß)

Keep in mind that the disk is not a point mass. Different parts of it will be at different distances from the center of mass.

The instant before the collision, what will be the geometry? Where's the COM? How can you calculate the moment of inertia of the moving disk about the center of mass?

 

[thepatient]

Keep in mind that the disk is not a point mass. Different parts of it will be at different distances from the center of mass.

The instant before the collision, what will be the geometry? Where's the COM? How can you calculate the moment of inertia of the moving disk about the center of mass?

Yes, I agree.

You will want to use the formula that includes the moment of inertia and the angular velocity.

 

[thepatient]

The object will have rotational and translational velocities.

There will also be conservation of angular momentum, so calculate the moment of inertia about the new center of mass using the parallel axis theorem and use the conservation of momentum to find the final angular velocity.

Actually using angular momentum initial = angular momentum final is a bad idea since initially none of the objects are rotating. You can use kinetic energy though. KE = .5mv^2 + .5Iw^2

 

[huybinhs]

The object will have rotational and translational velocities.

There will also be conservation of angular momentum, so calculate the moment of inertia about the new center of mass using the parallel axis theorem and use the conservation of momentum to find the final angular velocity.

Actually using angular momentum initial = angular momentum final is a bad idea since initially none of the objects are rotating. You can use kinetic energy though. KE = .5mv^2 + .5Iw^2

So because this is inelastic collision, we have:

m1v1+m2v2 = (m1+m2)v ,correct? then what should i do next?

 

[thepatient]

So because this is inelastic collision, we have:

m1v1+m2v2 = (m1+m2)v ,correct? then what should i do next?

This actually is a second way to calculate final translational velocity. If you plug in all the values, for final velocity you will get 0.533m/s.

I meant to say you can also use that kinetic energy is also conserved, since there is no friction. You can use the total initial kinetic energy of the system and set it equal to the total final kinetic energy.

 

[huybinhs]

This actually is a second way to calculate final translational velocity. If you plug in all the values, for final velocity you will get 0.533m/s.

I meant to say you can also use that kinetic energy is also conserved, since there is no friction. You can use the total initial kinetic energy of the system and set it equal to the total final kinetic energy.

If the answer is 0.533 m/s, this is not right because this is the answer for part b already, right? I'm confused :(

 

[huybinhs]

The object will have rotational and translational velocities.

There will also be conservation of angular momentum, so calculate the moment of inertia about the new center of mass using the parallel axis theorem and use the conservation of momentum to find the final angular velocity.

Actually using angular momentum initial = angular momentum final is a bad idea since initially none of the objects are rotating. You can use kinetic energy though. KE = .5mv^2 + .5Iw^2

If using this, I have:

Isystem = 1/2 m r^2 + 1/2 MR^2

then how can I get KE in order to find w (angular velocity)?

 

[thepatient]

Rotational kinetic energy is 1/2 Iw^2

Translational kinetic energy is 1/2 mv^2

The sum of those two will give you the total kinetic energy.

Sum up the total kinetic energy before the collision set it equal to the total kinetic energy after the collision.

You have everything you need to calculate this, all you need is the moment of inertia of after the collision. For that you will need to use the parallel axis theorem. Once you have that you can just plug in and solve for final angular velocity.

 

[huybinhs]

Rotational kinetic energy is 1/2 Iw^2

Translational kinetic energy is 1/2 mv^2

The sum of those two will give you the total kinetic energy.

Sum up the total kinetic energy before the collision set it equal to the total kinetic energy after the collision.

You have everything you need to calculate this, all you need is the moment of inertia of after the collision. For that you will need to use the parallel axis theorem. Once you have that you can just plug in and solve for final angular velocity.

so KEafter collision = .5mv2^2 + .5Iw^2

and KEbefore collison = .5mv1^2 + .5Iw^2

then set it to equal:

.5mv2^2 + .5Iw^2 = .5mv1^2 + .5Iw^2 (why are they equal?)

with I = 1/2 m r^2 + 1/2 MR^2

Am I on the right track?

Thanks so much for your patience ;)

 

[thepatient]

Conservation of kinetic energy. Just remember that I final will be the total moment of inertia of both discs about the new center of mass.

 

[huybinhs]

Conservation of kinetic energy. Just remember that I final will be the total moment of inertia of both discs about the new center of mass.

Sorry! What do u mean by that?

Are those equation set up right?

 

[thepatient]

For example, let's say you have an object that has an unevenly distributed mass. A hammer for example where the head is much heavier than the wooden handle. The center of mass would be very close to the neck of the head. If you were to spray paint a dot on this center of mass red, and toss it at a distance, the object will always rotate about this red dot. The object may be rotating a lot, but if you were to plot out a x-y graph of just the dots motion, it will be a perfect parabolic shape (depending if there is no air resistance, etc).

Similarly in this problem, once the objects collide, there will be a rotation about the new center of mass. So when finding I total, you will use the parallel axis theorem I = Icm + m(d)^2; d being the new distance between the old center of mass and the new center of mass.

This would be done for both discs and added together to get the moment of inertia total.

 

[huybinhs]

For example, let's say you have an object that has an unevenly distributed mass. A hammer for example where the head is much heavier than the wooden handle. The center of mass would be very close to the neck of the head. If you were to spray paint a dot on this center of mass red, and toss it at a distance, the object will always rotate about this red dot. The object may be rotating a lot, but if you were to plot out a x-y graph of just the dots motion, it will be a perfect parabolic shape (depending if there is no air resistance, etc).

Similarly in this problem, once the objects collide, there will be a rotation about the new center of mass. So when finding I total, you will use the parallel axis theorem I = Icm + m(d)^2; d being the new distance between the old center of mass and the new center of mass.

This would be done for both discs and added together to get the moment of inertia total.

Ok, so:

I before collision = (1/2 m r^2 + 1/2 MR^2) + [(m+M) * (R+r)]

then

I after collision = (1/2 m r^2 + 1/2 MR^2) + [(m+M) * 0.034925 m]

which 0.034925 m is the center of mass of part a, am I right?

 

[thepatient]

Initially none of the objects are rotating, not really given in the problem. The only thing you really need is the moment of inertia after the collision, since if you were to calculate I's for the initial part, they are just going to cancel out with the w = 0.

In the final I, it's best to draw a picture. Draw both circles stuck together, maybe small circle above the large circle. Calculating I about the individual center of masses is easy, the tricky part is over the new axis of rotation.

The axis of rotation is .034925m above the center of the larger circle towards the center of the smaller circle. So individually:
I total = I (smaller circle) + I(larger circle) = (.5mr^2 +m(r+.034925m)^2) + (.5MR^2 +M(.034925)^2.)

Drawing a picture makes it much easier to see what's going on. Physics is fun, review your notes on the parallel axis theorem. XD

 

[huybinhs]

Initially none of the objects are rotating, not really given in the problem. The only thing you really need is the moment of inertia after the collision, since if you were to calculate I's for the initial part, they are just going to cancel out with the w = 0.

In the final I, it's best to draw a picture. Draw both circles stuck together, maybe small circle above the large circle. Calculating I about the individual center of masses is easy, the tricky part is over the new axis of rotation.

The axis of rotation is .034925m above the center of the larger circle towards the center of the smaller circle. So individually:
I total = I (smaller circle) + I(larger circle) = (.5mr^2 +m(r+.034925m)^2) + (.5MR^2 +M(.034925)^2.)

Drawing a picture makes it much easier to see what's going on. Physics is fun, review your notes on the parallel axis theorem. XD

Great! Then I just plug it in:

.5mv2^2 + .5Iw^2 = .5mv1^2 + .5Iw^2 to find w, correct?

 

[thepatient]

Yup yup.

 

[huybinhs]

Yup yup.

LOL

.5mv2^2 + .5Iw^2 = .5mv1^2 + .5Iw^2

where v1 is initial given, v2 is the part b answer, correct? ;)

 

[thepatient]

Thanks so much for your patience ;)

I chose my log in name as a reference to a song by the band Tool; has nothing to do with me. XD

 

[thepatient]

LOL

.5mv2^2 + .5Iw^2 = .5mv1^2 + .5Iw^2

where v1 is initial given, v2 is the part b answer, correct? ;)

Yea.

 

[huybinhs]

I chose my log in name as a reference to a song by the band Tool; has nothing to do with me. XD

Now I can see that it fits you perfectly ;)

 

[huybinhs]

Yea.

So, the I on both side will be the same as I(total) , right?

 

[huybinhs]

LOL

.5mv2^2 + .5Iw^2 = .5mv1^2 + .5Iw^2

where v1 is initial given, v2 is the part b answer, correct? ;)

I think the I on the right side will = 0,

so the final equation is:

.5mv2^2 + .5Iw^2 = .5mv1^2

Right?

 

[thepatient]

Yea. What did your answer come out to?

 

[huybinhs]

Yea. What did your answer come out to?

Hang on. Let me calculate it now ;)

 

[huybinhs]

Yea. What did your answer come out to?

.5mv2^2 + .5Iw^2 = .5mv1^2

I have some problem on this equation. It's about the m, is it? :

.5(m+M)v2^2 + .5Iw^2 = .5mv1^2 , correct?

 

[huybinhs]

.5mv2^2 + .5Iw^2 = .5mv1^2

I have some problem on this equation. It's about the m, is it? :

.5(m+M)v2^2 + .5Iw^2 = .5mv1^2 , correct?

If so, I calculated I(total) = 8.00321 * 10^-4
=> w = 14.32 rad/s

Could u check whether it's correct or not, and let me know, k?

Thanks ;)

 

[thepatient]

Seems about ok. I got 12.1 rad/s but I used different amount of significant figures. What does the answer on the back of the book say?

 

[huybinhs]

Seems about ok. I got 12.1 rad/s but I used different amount of significant figures. What does the answer on the back of the book say?

It's on a paper exam, so not sure about the answer yet :(