Question:
A ring rolls down an inclined plane. The ratio of rotational kinetic energy to translational kinetic energy is
A ring rolls down an inclined plane. The ratio of rotational kinetic energy to translational kinetic energy is
Updated On: May 12, 2024
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The Correct Option is B
Solution and Explanation
Translational kinetic energy,
$K_T = \frac{1}{2} mv^2$
where $m$ = mass of the ring,
$v$ = speed of the centre of mass of the ring
Rotational kinetic energy,
$K_R = \frac{1}{2} I \omega^2$
Here, $I = mR^2 , \omega = \frac{v}{R}$
So, $K_R = \frac{1}{2} ( mR^2 ) \left(\frac{v}{R} \right)^2 = \frac{1}{2} mv^2$
$ \therefore\:\:\:\: \frac{K_R}{K_T} = \frac{1}{1}$
$K_T = \frac{1}{2} mv^2$
where $m$ = mass of the ring,
$v$ = speed of the centre of mass of the ring
Rotational kinetic energy,
$K_R = \frac{1}{2} I \omega^2$
Here, $I = mR^2 , \omega = \frac{v}{R}$
So, $K_R = \frac{1}{2} ( mR^2 ) \left(\frac{v}{R} \right)^2 = \frac{1}{2} mv^2$
$ \therefore\:\:\:\: \frac{K_R}{K_T} = \frac{1}{1}$
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