Question

The rotational kinetic energy of a body is e in the absence of external torque if mass of the body is halved and radius of gyration doubled then its rotatioanl kinetic energy will be

Answer 1

Rotational Kinetic energy of an object is given by formula

$K = \frac{1}{2} I\omega^2$

we can also write it in form of angular momentum

as we know that

L = Iw

$K = \frac{L^2}{2I}$

now here moment of inertia is given by

$I = mk^2$

here k = radius of gyration

m = mass

when there is no torque acting on the object then we can say

$L_i = L_f$

angular momentum must remain same always

now the mass is half and radius of gyration is double

then moment of inertia is given as

$I = \frac{m}{2}*(2k)^2$

$I = 2mk^2$

so here moment of inertia is double

so kinetic energy is given by

$K' = \frac{L^2}{2mk^2}$

as we compare it with initial energy

$K' = \frac{E}{2}$

so kinetic energy is half that of initial value

Answer 2

Answer: I=MK^2

Also, L(angular momentum)=Iw then,

w=L/I

Here L(i)=L(f)

E=1/2Iw^2

E=1/2I(L^2/I^2)

E=L^2/2MK^2

When mass is halved and radius of gyration is doubled,

E'=(L^2)/(2M.(2K)^2/2)

E' =(L^2/2MK^2)/2 i.e.

E' =E/2

Answer:: 0.5E

Explanation: