Kinetic energy of a hallow sphere

416
0
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
 
416
0
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
 
teng125 said:
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
[tex] K_{tot}=K_{lin} + K_{rot} [/tex]

Since you have rolling without slipping (I would assume) then [tex] v = R \omega [/tex].

-Dan
 
416
0
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
 
30
0
teng125
Write the equation of rotational kinetic energy about a point where V = 0 m/s. By doing so you can solve for omega.
 
416
0
for v=0?? so, is it just the rotatioanl energy only and without 1/2 (mv^2)??
 
30
0
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.
 
416
0
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
 
416
0
still can't do.pls show me the eqn on how u do it
thanx
 
30
0
Nevermind about finding rotational kinetic energy about a point where V=0 m/s.

Here is another approach.
[tex] I_{c} = 2/3MR^2\ kg*m^2 [/tex]
Where [tex] I_{c} = 0.04\ kg*m^2[/tex]
Using this equation you can solve for the mass of the sphere.
 
Last edited:
416
0
i got 9.51 J but the answer is 8 J
 
30
0
I got 8 J, check your math.
 
416
0
for m = 2.67kg
then 1/2m(v^2) + 2/5I(w^2) =20 then w= 21.81

then how to continue??
 
30
0
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:
416
0
why is the moment of inertia changes from 2/3(mr^2) to 2/5(mr^2) ??
 
30
0
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
 
416
0
oh........okok
 

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top