# Kinetic energy of a hallow sphere

1. Feb 21, 2006

### teng125

a hollow sphere or radius 0.15m with rotational inertia=0.04 about a line throughits centre of mass,rolls withoutslipping up a surface inclined 30 degree to the horizontal.at a cwrtain initial position,the sphere's total kinetic energy is 20J.
how much of this initial kinetic energy is rotational.
pls help.i can't do it

2. Feb 21, 2006

### teng125

A hollow sphere or radius 0.15m with rotational inertia=0.04 about a line through its centre of mass,rolls without slipping up a surface inclined
30 degree to the horizontal. At a certain initial position, the sphere's total kinetic energy is 20J.

how much of this initial kinetic energy is rotational??

pls help.i can't do it

3. Feb 21, 2006

### topsquark

$$K_{tot}=K_{lin} + K_{rot}$$

Since you have rolling without slipping (I would assume) then $$v = R \omega$$.

-Dan

4. Feb 21, 2006

### teng125

then, 0.5m(r^2)(w^2) + 0.5 I (w^2) =20J
from here,i don't have the w and m
how to continue

5. Feb 21, 2006

### Dmitri

teng125
Write the equation of rotational kinetic energy about a point where V = 0 m/s. By doing so you can solve for omega.

6. Feb 22, 2006

### teng125

for v=0?? so, is it just the rotatioanl energy only and without 1/2 (mv^2)??

7. Feb 22, 2006

### Dmitri

Yes, V=0 m/s is at the point where the sphere touches the surphase. You would have to recalculate moment of inertia at that point and solve for omega. Then use the omega in your first equation to solve for rotational kinetik energy.

8. Feb 22, 2006

### teng125

what u mean again is it 1/2 (2/5) m (r^2) (w^2) = 20J??
but i don't have the mass

pls help

9. Feb 22, 2006

### teng125

still can't do.pls show me the eqn on how u do it
thanx

10. Feb 22, 2006

### Dmitri

Nevermind about finding rotational kinetic energy about a point where V=0 m/s.

Here is another approach.
$$I_{c} = 2/3MR^2\ kg*m^2$$
Where $$I_{c} = 0.04\ kg*m^2$$
Using this equation you can solve for the mass of the sphere.

Last edited: Feb 22, 2006
11. Feb 23, 2006

### teng125

i got 9.51 J but the answer is 8 J

12. Feb 23, 2006

### Dmitri

I got 8 J, check your math.

13. Feb 23, 2006

### teng125

for m = 2.67kg
then 1/2m(v^2) + 2/5I(w^2) =20 then w= 21.81

then how to continue??

14. Feb 23, 2006

### Dmitri

Plug m=(3/2)*Ic/R^2 in 1/2m(w^2)(R^2) + (1/2)I(w^2) =20
You will get w^2 = 16/Ic where Ic=0.04 kg*m^2
Solve for w (20 rad/s)
Plug w in Kr = (1/2)*Ic*w^2

Last edited: Feb 23, 2006
15. Feb 23, 2006

### teng125

why is the moment of inertia changes from 2/3(mr^2) to 2/5(mr^2) ??

16. Feb 23, 2006

### Dmitri

It did not Ic = 2/3(mr^2)
I copied this equation from your message 1/2m(w^2)(R^2) + (2/5)I(w^2) =20. It should be 1/2m(w^2)(r^2) + 1/2Ic(w^2) = 20

17. Feb 23, 2006

### teng125

oh........okok

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