How Much of the Initial Kinetic Energy of a Hollow Sphere is Rotational?

[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

 

[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

 

[topsquark]

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

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

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

-Dan

 

[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

 

[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.

 

[teng125]

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

 

[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.

 

[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

 

[teng125]

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

 

[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.

 

[teng125]

i got 9.51 J but the answer is 8 J

 

[Dmitri]

I got 8 J, check your math.

 

[teng125]

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

then how to continue??

 

[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

 

[teng125]

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

 

[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

 

[teng125]

oh...okok