Question

a disk of diameter 0.2 m and mass 10 kg rotating about its axis at a rate of 600 revolutions per minute.calculate the angular momentum and rotational kinetic energy.

Answer 1

$\huge\underline{\underline{\bf \green{Question-}}}$

A disk of diameter 0.2 m and mass 10 kg rotating about its axis at a rate of 600 revolutions per minute.calculate the angular momentum and rotational kinetic energy.

$\huge\underline{\underline{\bf \green{Solution-}}}$

$\large\underline{\underline{\sf Given:}}$

• Diameter of disc (d ) = 0.2m
• Radius (r) = 0.1m
• Mass (m) = 10kg
• Frequency (f) = 600rpm

$\large\underline{\underline{\sf To\:Find:}}$

• Angular Momentum (L)
• Rotational Kinetic Energy (K)

✪$\large{\boxed{\bf \blue{ Moment\: of\:Interia(I)=MR^2 }}}$

$\implies{\sf I=10×(0.1)^2}$

$\implies{\bf I=0.1\:Kg-m^2 }$

❏ Angular Momentum ➝

✪$\large{\boxed{\bf \blue{ L = I\omega }}}$

$\large{\bf \omega =2πf}$

$\implies{\sf \omega = 2π×\dfrac{600}{60} }$

$\implies{\bf \blue{ \omega = 20π}}$

$\implies{\sf L = 0.1×20π}$

$\implies{\bf \red{L= 2π\:Kg.m^2/s}}$

❏ Rotational Kinetic Energy ➝

✪$\large{\boxed{\sf K = \dfrac{1}{2}I\omega^2 }}$

$\implies{\sf \dfrac{1}{2}×0.1×(20π)^2}$

$\implies{\sf \dfrac{1}{2}×0.1×400π}$

$\implies{\bf \red{K= 98.5\:J }}$

$\huge\underline{\underline{\bf \green{Answer-}}}$

Angular Momentum ${\bf \red{(L)=2π\:Kg.m^2/s}}$

Rotational Kinetic Energy ${\bf \red{K=98.5\:J}}$