Question

The kinetic energy of a circular disc rotating with a speed of 60 r.p.m. about an axis passing through a point on its circumference and perpendicular to its plane is (mass of circular disc = 5 kg, radius of disc = 1 m) approximately:

Answer 1

We know the equations:

• $\boxed{\sf{I=mr^2}}$

• $\boxed{\sf{\omega=2\pi f}}$

• $\boxed{\sf{KE=\dfrac{1}{2}I\omega^2}}$

Given,

• Mass (m) = 5 kg

• Radius (r) = 1 m

Which gives,

$\sf{\implies K.E=\dfrac{1}{2}\times\dfrac{3}{2}mr^2\times 4\pi^2f^2}$

$\sf{\implies K.E=\dfrac{1}{2}\times\dfrac{3}{2}\times 5\times1\times4\times10\times 1}$

$\sf{\implies K.E=3\times5\times1\times10\times1}$

$\sf{\implies K.E=150 \ J}$

$\boxed{\bold{\sf{\therefore Kinetic \ Energy=150 \ J}}}$

Answer 2

Answer:

answer is 150J

Explanation:

answer is 150J