Answer:
[tex]\omega' = 0.89\omega[/tex]
Explanation:
Rotational inertia of uniform solid sphere is given as
[tex]I = \frac{2}{5}MR^2[/tex]
now we have its angular speed given as
angular speed = [tex]\omega[/tex]
now we have its final rotational kinetic energy as
[tex]KE = \frac{1}{2}(\frac{2}{5}MR^2)\omega^2[/tex]
now the rotational inertia of solid cylinder about its axis is given by
[tex]I = \frac{1}{2}MR^2[/tex]
now let say its angular speed is given as
angular speed = [tex]\omega'[/tex]
now its rotational kinetic energy is given by
[tex]KE = \frac{1}{2}(\frac{1}{2}MR^2)\omega'^2[/tex]
now if rotational kinetic energy of solid sphere is same as rotational kinetic energy of solid sphere then
[tex]\frac{1}{2}(\frac{2}{5}MR^2)\omega^2 = \frac{1}{2}(\frac{1}{2}MR^2)\omega'^2[/tex]
[tex]\frac{2}{5}\omega^2 = \frac{1}{2}\omega'^2[/tex]
[tex]\omega' = 0.89\omega[/tex]
Answer:
w_cyl = ±√(4/5) ω
Explanation:
Kinetic energy
E = (1/2)Iw²
where I is the moment of inertia and w the angular frequency of rotation.
The moment of inertia of a solid sphere of mass M and radius R is:
I = (2/5)MR²,
Solid cylinder is of mass M and radius R
I = (1/2)MR²
Equate the energies through
(1/2)×(2/5) M R²× (w_sphere)² = (1/2)× (1/2) MR² × (w_cyl)²
(w_cyl)² = (4/5)(w_sphere)²
w_cyl = ±√(4/5) ω
The energy of rotation is independent of the direction of rotation
