Answer 2nd Question Part 1: Mechanics Ques.

[winichris]

http://imageshack.us/photo/my-images/152/12617395.png/
The above link is the question.

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

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

 

[winichris]

need your help tiny-tim pleaseeee:)

 

[tiny-tim]

hi winichris!

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

yes, that looks fine

 

[winichris]

hi winichris!


yes, that looks fine

THANKS.

For the part 2bi:

let f be the frictional force acting on the sphere pointing towards right.

Setting the below 3 equations, then i can solve for a,α and f right?
fR=Iα
a=Rα
f=Ma

But the moment of inertia found from part a is about diameter.
In part b, the moment of inertia should be in the center, so I need the use the part a result and times 2? (because i rarely remember that moment of inertia about diameter = moment of inertia about center /2)

 

[tiny-tim]

hi winichris!

let f be the frictional force acting on the sphere pointing towards right.

Setting the below 3 equations, then i can solve for a,α and f right?
fR=Iα
a=Rα
f=Ma

those three equations would be fine if the platform was not accelerating

you need to adjust a=Rα, and you need a fourth equation, F=ma for the platform

(call the acceleration of the platform "A")

But the moment of inertia found from part a is about diameter.
In part b, the moment of inertia should be in the center, so I need the use the part a result and times 2? (because i rarely remember that moment of inertia about diameter = moment of inertia about center /2)

no, you're rambling

this is a (hollow) sphere, and the diameter is through the centre, isn't it?

(you're thinking about the perpendicular axis theorem, it only applies to "2D" bodies, eg a disc)

 

[winichris]

hi winichris!


those three equations would be fine if the platform was not accelerating

you need to adjust a=Rα, and you need a fourth equation, F=ma for the platform

(call the acceleration of the platform "A")

My professor said the velocity of the platform and the velocity of the sphere are the same.
So I think their acceleration will be the same, right?

no, you're rambling

this is a (hollow) sphere, and the diameter is through the centre, isn't it?

(you're thinking about the perpendicular axis theorem, it only applies to "2D" bodies, eg a disc)

Ok then i can just use the part a result! thx

 

[tiny-tim]

My professor said the velocity of the platform and the velocity of the sphere are the same.

no, that's rubbish …

the sphere will roll backwards relative to the platform: its velocity and acceleration will be less than that of the platform

 

[winichris]

no, that's rubbish …

the sphere will roll backwards relative to the platform: its velocity and acceleration will be less than that of the platform

Let A be acceleration of platform, a be acceleration of sphere
M be mass of sphere, m be mass of platform

4 equations:
fR=Iα
(A-a)=Rα<--i am not sure whether this adjustment is correct or not..
f=Ma
F=mA<--use F=mA or F+f=mA?

 

[tiny-tim]

fR=Iα
f=Ma

yes

(A-a)=Rα<--i am not sure whether this adjustment is correct or not..

(a-A)=Rα

(because a-A is the acceleration relative to the platform)

F=mA<--use F=mA or F+f=mA?

definitely F+f = mA (or is it F-f = mA ?) …

a free body diagram would show both forces on the platform ​

 

[winichris]

yes


(a-A)=Rα

(because a-A is the acceleration relative to the platform)

As you mentioned before, A>a, so α will be negative?

definitely F+f = mA (or is it F-f = mA ?) …

a free body diagram would show both forces on the platform ​

friction should be point towards right, so should be F+f=mA instead of F-f=mA?

 

[tiny-tim]

As you mentioned before, A>a, so α will be negative?

yup, the sphere will roll backward, and get left behind

friction should be point towards right

friction from the sphere on the platform?

 

[winichris]

yup, the sphere will roll backward, and get left behind


friction from the sphere on the platform?

http://imageshack.us/photo/my-images/821/62646872.png/

The free body diagram is like above?

So for platfrom: F+f=mA?

 

[tiny-tim]

no, that's not a free body diagram, it's showing two bodies and an internal force between them

free body diagrams don't have internal forces

do a free body diagram for the sphere … which way does the friction go?

then do a free body diagram for the platform

 

[winichris]

no, that's not a free body diagram, it's showing two bodies and an internal force between them

free body diagrams don't have internal forces

do a free body diagram for the sphere … which way does the friction go?

then do a free body diagram for the platform

http://imageshack.us/photo/my-images/839/32859543.png/

Is the above free body diagram of sphere and platform correct?

I am not quite sure about whether the friction on platform should be point to right or left..

 

[tiny-tim]

Is the above free body diagram of sphere and platform correct?

yes!

(very good free body diagrams )

the top diagram shows that the friction from the platform must be to the right

(and, btw, that the rotation must be anti-clockwise)

and so Newton's third law tells us that the reaction force, the friction on the platform, must be to the left

(and so the sphere slows the platform down, exactly as if it wasn't rolling)

 

[winichris]

Sorry, i just find a little problem in question 1
https://www.physicsforums.com/showthread.php?p=3879642#post3879642

About the potential energy..

 

[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]

oooh, sorry … two errors

i] rotational kinetic energy is

either 1/2 Iω_c.o.rotation²

or 1/2 Iω_c.o.mass² + 1/2 mv_c.o.mass²
​

ie if you use the centre of rotation, you can forget about 1/2 mv²

(in the previous question, the end of the rod was attached to a ring which was moving, so it wasn't the centre of rotation, here it's fixed, so it is)

you've done 1/2 Iω_c.o.rotation² + 1/2 mv_c.o.mass²

ii] ω is in radians per second, so time = radians/ω,

but you haven't used radians

 

[winichris]

oooh, sorry … two errors

i] rotational kinetic energy is

either 1/2 Iω_c.o.rotation²

or 1/2 Iω_c.o.mass² + 1/2 mv_c.o.mass²
​

ie if you use the centre of rotation, you can forget about 1/2 mv²

(in the previous question, the end of the rod was attached to a ring which was moving, so it wasn't the centre of rotation, here it's fixed, so it is)

you've done 1/2 Iω_c.o.rotation² + 1/2 mv_c.o.mass²

ii] ω is in radians per second, so time = radians/ω,

but you haven't used radians

So for energy:
mga = 0.5Iω² to find ω is enough?

Then moment of inertia used above should be 1/12m(2a)² or 1/12m(2a)²+ma²?
(As when the hinge breaks, the center of rotation change to center of rod?)

For time used: it should be 2pi/ω?

 

[tiny-tim]

hi winichris!

So for energy:
mga = 0.5Iω² to find ω is enough?

Then moment of inertia used above should be 1/12m(2a)² or 1/12m(2a)²+ma²?

to find the energy for the first part:

yes, mga = 0.5Iω² to find ω is enough (ie no 1/2 mv² is needed),

provided you use the centre of rotation, (ie I = 1/12m(2a)²+ma²)

(As when the hinge breaks, the center of rotation change to center of rod?)

(this is the second part)

why does the centre of rotation matter?

the rotation will be constant in the second part

(and anyway, no, it isn't in the centre of the rod, it probably isn't even in the rod at all)

For time used: it should be 2pi/ω?

well, it's angle/ω, and the angle is 90°, so that's π/2ω

 

[winichris]

hi winichris!


to find the energy for the first part:

yes, mga = 0.5Iω² to find ω is enough (ie no 1/2 mv² is needed),

provided you use the centre of rotation, (ie I = 1/12m(2a)²+ma²)


(this is the second part)

why does the centre of rotation matter?

the rotation will be constant in the second part

(and anyway, no, it isn't in the centre of the rod, it probably isn't even in the rod at all)


well, it's angle/ω, and the angle is 90°, so that's π/2ω

Thanks!
I get 2 last problems about heat, hope you can help me:)

https://www.physicsforums.com/showthread.php?p=3881230#post3881230

 

[tiny-tim]

not really my scene …

i'll let someone else have a go! ​

 

[winichris]

not really my scene …

i'll let someone else have a go! ​

ok.thx anyway, u help me a lot!