What Mechanics Questions Are Urgently on Your Mind?

  • Thread starter Thread starter winichris
  • Start date Start date
  • Tags Tags
    Mechanics
AI Thread Summary
The discussion centers around a complex mechanics problem involving conservation of momentum and energy related to a rod and a ring system. Participants emphasize the importance of correctly identifying the velocities of the center of mass and the rod, clarifying that the rod does not have an ordinary velocity but rather an angular velocity. There is confusion regarding the tangential and radial velocities, especially at the highest point of the rod's swing, with participants seeking to establish the relationships between these velocities. The conversation also touches on calculating the moment of inertia and kinetic energy, highlighting the need for careful application of formulas depending on the system's configuration. Overall, the thread reflects a collaborative effort to solve a challenging mechanics question.
winichris
Messages
38
Reaction score
0
Physics news on Phys.org
welcome to pf!

hi winichris! welcome to pf! :wink:

show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 


tiny-tim said:
hi winichris! welcome to pf! :wink:

show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:

My teacher taught some very easy notes.
But he gave us this kind of so difficult problems.
I don't have ideas about that.

For Q1,
velocity of mass m = 2a x (g/a)^0.5

By conservation of momentum, m x 2a x (g/a)^0.5 = m x (velocity of ring)
velocity of ring will be = 2a x (g/a)^0.5

But I think this solution is wrong since by using conservation of energy:

0.5 x m x (velocity of mass)^2 = 0.5 x m x (velocity of ring)^2 + mgh

This will make h=0

So can you help me to think of this question?
 
you need to start by finding expressions for the momentum and the energy at a general speed v (of the ring) and angle θ

for horizontal momentum of the rod, you must use the velocity of the centre of mass of the rod

for kinetic energy of the rod, you must add the rotational kinetic energy (using moment of inertia about the centre of mass) to the "translational" kinetic energy of the centre of mass

start again! :smile:​
 


tiny-tim said:
you need to start by finding expressions for the momentum and the energy at a general speed v (of the ring) and angle θ

for horizontal momentum of the rod, you must use the velocity of the centre of mass of the rod

for kineticenergy of the rod, you must add the rotational kinetic energy (using moment of inertia about the centre of mass) to the "translational" kinetic energy of the centre of mass

start again! :smile:​
When the mass is at its highest point, its velocity and angular velocity are both zero?
 
winichris said:
When the mass is at its highest point, its velocity and angular velocity are both zero?

when the rod is at its greatest angle, its angular velocity is obviously zero

but why would its ordinary velocity also be zero? :confused:
 


tiny-tim said:
when the rod is at its greatest angle, its angular velocity is obviously zero

but why would its ordinary velocity also be zero? :confused:
http://imageshack.us/photo/my-images/823/phy.png/
When it is in the highest point, is the tangential velocity is zero? because v=rw, where w=0. (red blanket in the picture)

As the whole system is moving to right, so there should be a radial velocity of the mass ??(blue blanket in the picture)

But there is a question, there are 3 unknowns (angle theta, velocity of mass and velocity of ring) but there are only 2 equations.
So whether the radial velocity of the mass is equal to initial velocity of the mass?

Thanks a lot!
 
winichris said:
When it is in the highest point, is the tangential velocity is zero? because v=rw, where w=0. (red blanket in the picture)

As the whole system is moving to right, so there should be a radial velocity of the mass ??(blue blanket in the picture)

why are you using "radial" and "tangential"? there's no origin here (the ring is moving) :confused:
But there is a question, there are 3 unknowns (angle theta, velocity of mass and velocity of ring) but there are only 2 equations.

no, there are only 2 unknowns … position of ring, and angle of rod
 


tiny-tim said:
why are you using "radial" and "tangential"? there's no origin here (the ring is moving) :confused:


no, there are only 2 unknowns … position of ring, and angle of rod

radial and tangential is for the rod (not the ring above).
Because when it is at highest point, w=0, can i conclude that tangential velocity=0?

Position of ring means velocity of ring?
 
  • #10
winichris said:
radial and tangential is for the rod (not the ring above).
Because when it is at highest point, w=0, can i conclude that tangential velocity=0?

yes of course :smile:
Position of ring means velocity of ring?

you can use either
 
  • #11


tiny-tim said:
yes of course :smile:


you can use either

So with tangential velocity=0, the rod only consist of radial velocity.
Is the radial velocity (blue blanket in the last photo) = initial velocity of rod ( radius x angular velocity)
 
  • #12
can you please type your equations directly into the post, instead of photocopying them? :redface:

and what is y? and remember that there's no mass at the free end of the rod
 
  • #13


tiny-tim said:
can you please type your equations directly into the post, instead of photocopying them? :redface:

and what is y? and remember that there's no mass at the free end of the rod

Let y be the radial velocity of the rod at highest point, x be the velocity of the ring when rod at highest point, v be initial velocity of the rod.
v=rw=2a(g/a)^0.5, right?
Will y and v be equal?

Momentum: mv = my cos(theta) + mx
Energy: 0.5mv^2 + 0.5Iw^2 = mg2a(1-cos theta) + 0.5mx^2 + 0.5my^2
 
  • #14
winichris said:
Let y be the radial velocity of the rod at highest point

the radial velocity is always 0 :confused:
, x be the velocity of the ring when rod at highest point, v be initial velocity of the rod.

the rod doesn't have a velocity, it only has an angular velocity
Energy: 0.5mv^2 + 0.5Iw^2 = mg2a(1-cos theta) + 0.5mx^2 + 0.5my^2

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

kinetic energy = 0.5mv2 for the ring + 0.5mv2 for the centre of mass of the rod + 0.5Iω2 for the rod

(and at the top of the swing, ω = 0)
 
  • #15


tiny-tim said:
the radial velocity is always 0 :confused:the rod doesn't have a velocity, it only has an angular velocity

"when the rod is at its greatest angle, its angular velocity is obviously zero
but why would its ordinary velocity also be zero? "<-- you said that the rod should have zero velocity before.
Or maybe i misunderstand your meaning.
So the ordinary velocity of the rod should be zero at highest point?
 
  • #16
winichris said:
"when the rod is at its greatest angle, its angular velocity is obviously zero
but why would its ordinary velocity also be zero? "<-- you said that the rod should have zero velocity before.

no, i said the opposite ...
tiny-tim said:
when the rod is at its greatest angle, its angular velocity is obviously zero

but why would its ordinary velocity also be zero? :confused:
 
  • #17


tiny-tim said:
no, i said the opposite ...

tiny-tim said:
the radial velocity is always 0 :confused:


the rod doesn't have a velocity, it only has an angular velocity
Sorry i type wrongly.
You say the rod does not have an ordinary velocity here..
 
  • #18


Or maybe can you just tell me what will be the ordinary velocity of the rod when it is at highest point? thanks a lot, you are so helpful:)
 
  • #19
winichris said:
Or maybe can you just tell me what will be the ordinary velocity of the rod when it is at highest point? thanks a lot, you are so helpful:)

the rod does not have an ordinary velocity, the rod has an angular velocity, and the centre of mass of the rod has an ordinary velocity

at its highest point, the angular velocity is 0, and therefore the velocity of the centre of mass of the rod is the same as the velocity of the ring
 
  • #20


tiny-tim said:
the rod does not have an ordinary velocity, the rod has an angular velocity, and the centre of mass of the rod has an ordinary velocity

at its highest point, the angular velocity is 0, and therefore the velocity of the centre of mass of the rod is the same as the velocity of the ring

At the highest point, the direction of the velocity of the center of rod is towards right or towards right upwards direction?

If the direction is towards right,

m(2aw) = m(velocity of ring) + m(velocity of center of mass of rod)
As velocity of ring and velocity of center of mass of rod is same,
so velocity of ring= aw

right?
 
Last edited:
  • #21
winichris said:
If the direction is towards right,

m(2aw) = m(velocity of ring) + m(velocity of center of mass of rod)
As velocity of ring and velocity of center of mass of rod is same,

yes, your RHS (total final horizontal momentum) is correct :smile:

but your LHS (total initial horizontal momentum) is wrong
 
  • #22


tiny-tim said:
yes, your RHS (total final horizontal momentum) is correct :smile:

but your LHS (total initial horizontal momentum) is wrong

Initially, only the rod is moving while the ring is at rest.
And the velocity of the rod is (radius)x(w), so what's wrong?
 
  • #23
winichris said:
And the velocity of the rod is (radius)x(w), so what's wrong?

you keep saying "the velocity of the rod" ! :frown:

the rod does not have a velocity, its centre of mass has a velocity
 
  • #24


tiny-tim said:
you keep saying "the velocity of the rod" ! :frown:

the rod does not have a velocity, its centre of mass has a velocity

Ok sorry, actually I don't quite understand what is the difference between velocity of rod and center of mass of rod..

So what's wrong of my LHS part(the initial momentum)?
 
  • #25
winichris said:
m(2aw) =

why m? why 2? why a? why w?
 
  • #26


tiny-tim said:
why m? why 2? why a? why w?

momentum = mass x velocity

velocity = radius x angular velocity

mass=m
radius=2a (the length of rod)
angular velocity = w =(g/a)^0.5 (from question)

am i correct
 
  • #27
winichris said:
momentum = mass x velocity

velocity = radius x angular velocity

you need the velocity of the centre of mass
 
  • #28


tiny-tim said:
you need the velocity of the centre of mass

I think I don't know how to find the velocity of center of mass:(

2aw/2??

Really no idea:(
 
  • #29
get some sleep :zzz:
 
  • #30


tiny-tim said:
get some sleep :zzz:

ok, have a good sleep:)

thanks a lot:)

I will think of it, but i think i still need your help since i really do not have ideas about that-_-
 
  • #31


I still cannot get any idea..
How to find the velocity of the center of mass of the rod initially?
 
  • #32
winichris said:
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) :smile:
 
  • #33


tiny-tim said:
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) :smile:

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
 
  • #34


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?
 
  • #35
winichris said:
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! :smile:
winichris said:
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" :rolleyes: …

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

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 :wink:) …

so for the final energy, you'll have to use the more awkward formula 1/2mvc.o.mass2 + 1/2 Ic.o.massω2
 
  • #36


tiny-tim said:
yes! :smile:


yes, "but" :rolleyes: …

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

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 :wink:) …

so for the final energy, you'll have to use the more awkward formula 1/2mvc.o.mass2 + 1/2 Ic.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?
 
  • #37
(try using the X2 button just above the Reply box :wink:)
winichris said:
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 1/2mvc.o.mass2 part of the general formula, since in this case the 1/2 Ic.o.massω2 is 0 :smile:
 
  • #39
hmm … we're on page 3 now :rolleyes: …

could you please start a new thread on the new question? :smile:
​
 
  • #41


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?
 
  • #42
let's see …
winichris said:
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 :wink:​
 
  • #43
  • #44
(see other thread)
 

Similar threads

Answer 2nd Question Part 1: Mechanics Ques.
Replies
22
Views
3K
How do you Calculate Ampacity of a Conductor Geometry?
2
Replies
64
Views
5K
Calculation of Box's Acceleration on a Trailer with Friction
Replies
9
Views
1K
3D statics - looks impossible to solve
Replies
11
Views
4K
Is there something wrong with this basic mechanics question?
Replies
2
Views
2K
Conservation of mechanical energy (no friction) - doesn't work in this case?
Replies
28
Views
5K
Solving for Horizontal Projectile Motion: Mechanics Final Question
Replies
10
Views
3K
Confusing free-body diagrams help
Replies
8
Views
15K
What Went Wrong with My Photoelectric Homework?
Replies
6
Views
2K
Statics Mechanics: Looking for the flaw in my logic
Replies
19
Views
3K

Hot Threads

Recent Insights

Back
Top