# Mechanics question

1. Apr 22, 2012

### winichris

2. Apr 22, 2012

### winichris

3. Apr 22, 2012

### tiny-tim

hi winichris!
yes, that looks fine

4. Apr 22, 2012

### winichris

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)

5. Apr 22, 2012

### tiny-tim

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")
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)

6. Apr 23, 2012

### winichris

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?

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

7. Apr 23, 2012

### tiny-tim

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

8. Apr 23, 2012

### winichris

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?

9. Apr 23, 2012

### tiny-tim

yes
(a-A)=Rα

(because a-A is the acceleration relative to the platform)
definitely F+f = mA (or is it F-f = mA ?) …

a free body diagram would show both forces on the platform

10. Apr 23, 2012

### winichris

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

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

11. Apr 23, 2012

### tiny-tim

yup, the sphere will roll backward, and get left behind
friction from the sphere on the platform?

12. Apr 23, 2012

### winichris

13. Apr 23, 2012

### 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

14. Apr 23, 2012

### winichris

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..

15. Apr 23, 2012

### tiny-tim

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)

16. Apr 23, 2012

### winichris

17. Apr 23, 2012

### winichris

18. Apr 23, 2012

### tiny-tim

oooh, sorry … two errors

i] rotational kinetic energy is
either 1/2 Iωc.o.rotation2

or 1/2 Iωc.o.mass2 + 1/2 mvc.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 Iωc.o.rotation2 + 1/2 mvc.o.mass2

19. Apr 24, 2012

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

For time used: it should be 2pi/ω?

Last edited: Apr 24, 2012
20. Apr 24, 2012

### tiny-tim

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