Question

Calculate the M.I. and rotational K.E. of a thin uniform rod of mass \( 10 g \) and length \( 60 cm \), wher it rotates about a transverse axis through its centre at \( 90 rpm \)

Answer 1

Given

L = 60 cm

L = 0.60 m

M = 10 g

M = .010 kg

n = 90 rpm

n = \(\frac{90}{60}\) = 1.5 rps

We know that, moment of inertia thin uniform rod transverse axis through its centre.

I = \(\frac{ML^2}{12}\)

I = \(\frac{.010 \times (0.60)^2}{12}\)

I = 0.0003

I = 3 × 10^-4 kg m² 

Rotational kinetic energy K.E = \(\frac{1}{2}\,I\omega^2\) 

∵ \(\omega = 2\pi n\)

\(=\frac{1}{2} \times (0.0003) \times (2\pi n)^2\)

\(=\frac{1}{2} \times (0.003) \times (2\pi \times 1.5)^2\)

\(=\frac{1}{2} \times 0.003 \times 9\pi^2\)

K.E = 0.133 Joule