You visited us 1 times! Enjoying our articles? Unlock Full Access!

Moment of Inertia of a Rod

The diameter ...

Question

The diameter of a thin circular disc of mass $2 k g$ is $0.2 m$ . Its moment of inertia about an axis passing through the edge of perpendicular to plane of the disc is

0.01 k g - m^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.02 k g - m^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0.03 k g - m^{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

0.04 k g - m^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $0.03 k g - m^{2}$
R.E.F image
$I_{1} = I_{c m} + m π^{2}$ [ $I_{c m} =$ In about center of mass]
$I_{1} = \frac{1}{2} m R^{2} + m R^{2}$
$= \frac{3}{2} m R^{2} = \frac{3}{2} (2) (0.1)^{2}$ $[R = 0.1 m$ $m = 2 k g]$
$I = 0.03 k g - m^{2}$

Suggest Corrections

Similar questions

2 k g

0.2 m

k g - m^{2}

2 m

1 k g

2 k g - m^{2}

k g - m^{2}

1 m

C O M

I_{c o m} = {2 kg-m}^{2}

I_{d i a}

(I_{c o m} - I_{d i a})

1 m

C O M

I_{c o m} = {2 kg-m}^{2}

I_{d i a}

(I_{c o m} - I_{d i a})

1 {kg/m}^{2}

2 {kg/m}^{2}

r = 1 m

MI

Join BYJU'S Learning Program