Help with mechanics problem from images

[winichris]

I still cannot get any idea..
How to find the velocity of the center of mass of the rod initially?

 

[tiny-tim]

How to find the velocity of the center of mass of the rod initially?

initially, the ring is fixed, so the rod's motion is pure rotation

so the speed of any part of the rod is given by the standard formula

v = ωr (strictly, v = ω × r)​

where r is the distance from the centre of rotation (ie from the end of the rod, where the ring is)

 

[winichris]

initially, the ring is fixed, so the rod's motion is pure rotation

so the speed of any part of the rod is given by the standard formula

v = ωr (strictly, v = ω × r)​

where r is the distance from the centre of rotation (ie from the end of the rod, where the ring is)

So the initial momentum= mrω = (m)(a)(g/a)^0.5?

As the distance between center of mass of rod and ring is a?

THX

 

[winichris]

About the moment of inertia of the rod:

As the standard formula of moment of inertia of rod is (1/12)ML^2,
I just need to substitute L=2a such that I=(1/3)Ma^2

Then add Ma^2 to the above (1/3)Ma^2 because the rotation is at the ring,
which is equal to (4/3)Ma^2?

 

[tiny-tim]

So the initial momentum= mrω = (m)(a)(g/a)^0.5?

As the distance between center of mass of rod and ring is a?

THX

yes!

About the moment of inertia of the rod:

As the standard formula of moment of inertia of rod is (1/12)ML^2,
I just need to substitute L=2a such that I=(1/3)Ma^2

Then add Ma^2 to the above (1/3)Ma^2 because the rotation is at the ring,
which is equal to (4/3)Ma^2?

yes, "but" …

yes, you've correctly calculated the moment of inertia about the end of the rod

but, when calculating kinetic energy, you can only use 1/2 Iω² if I is about the centre of rotation

that's ok for the initial energy here (since the ring is fixed),

but it won't work for the final energy (since the ring is moving, and you have no idea where the centre of rotation is ) …

so for the final energy, you'll have to use the more awkward formula 1/2mv_c.o.mass² + 1/2 I_c.o.massω²

 

[winichris]

yes!


yes, "but" …

yes, you've correctly calculated the moment of inertia about the end of the rod

but, when calculating kinetic energy, you can only use 1/2 Iω² if I is about the centre of rotation

that's ok for the initial energy here (since the ring is fixed),

but it won't work for the final energy (since the ring is moving, and you have no idea where the centre of rotation is ) …

so for the final energy, you'll have to use the more awkward formula 1/2mv_c.o.mass² + 1/2 I_c.o.massω²

At the highest point(final stage), ω=0, so I don't need to consider that I?

Momentum: maω=m(velocity of ring)+m(final velocity of center of mass of rod)
Energy: 0.5m(initial velocity of center of mass of rod)^2 + 0.5(4/3ma^2)(ω)^2 = 0.5m(velocity of ring)^2 + 0.5m(final velocity of center of mass of rod)^2

there is no need to consider the moment of inertia of the rod when it is at highest point?

 

[tiny-tim]

(try using the X² button just above the Reply box )

Energy: 0.5m(initial velocity of center of mass of rod)^2 + 0.5(4/3ma^2)(ω)^2 = 0.5m(velocity of ring)^2 + 0.5m(final velocity of center of mass of rod)^2

that's right …

you only need the 1/2mv_c.o.mass² part of the general formula, since in this case the 1/2 I_c.o.massω² is 0

 

[winichris]

Thanks for the 1st question!

http://imageshack.us/photo/my-images/214/39295383.png/

Is this the correct approach of the first part of the 2nd question?

 

[tiny-tim]

hmm … we're on page 3 now …

could you please start a new thread on the new question?
​

 

[winichris]

hmm … we're on page 3 now …

could you please start a new thread on the new question?
​

ok

https://www.physicsforums.com/showthread.php?p=3877422#post3877422

 

[winichris]

For first question, when the rod moves to the highest point, the gain in PE should be??

θ is the angle between rod and vertical line

mg2a-mg2acosθ

or

mga-mgacosθ <-- consider all the mass in the center?

 

[tiny-tim]

let's see …

http://imageshack.us/f/685/19897974.png/

yes, mga-mgacosθ (considering all the mass in the center)

that's why it's called the centre of mass …

it's the average position of all the mass, and you use it in formulas like mgh ​

 

[winichris]

Thanks..!

For the third question, i have done it.
http://imageshack.us/photo/my-images/42/59789974.png/

Can you help me to have a look on this?
This is the first question that can be completed by myself and I am so excited!
But hope it is correct..

 

[tiny-tim]

(see other thread)