They also showed me that math is all around us and is used for more things than we can imagine. The most important thing was that they made me excited and curious to learn more. I discovered several channels on YouTube where they would explain science concepts in a fun and interesting way. We were simply told to memorize a bunch of formulas and use them without knowing how or why they worked.īut one day everything changed. In school, science didn’t interest me because the lessons were not exciting.
I love learning about new concepts from science, math and technology, although it was not always like that. At the moment I’m finishing my Master’s and after that I would like to start a PhD, as I also like research and teaching. This equation is equivalent to I D 4 / 64 when we express it taking the diameter (D) of the circle. Here, R is the radius and the axis is passing through the centre. Apart from my passion for science, I also enjoy learning new languages, and so I decided to move to Austria to improve my German while studying Engineering. Moment of inertia of a circle or the second-moment area of a circle is usually determined using the following expression I R 4 / 4. In 2015 I moved to Austria to start studying Mechanical Engineering, as I had always wanted to study abroad. I.My name is Alvaro Nebreda and I was born in Spain in 1996. of a disc of mass 0.5 kg and radius 10 cmĪbout an axis passing through its centre and at right angles to its plane. Moment of inertia about an axis tangent to ring and perpendicular to its plane is 0.0625 kgm 2Ĭalculate the M.I. Moment of inertia about an axis tangent to ring andĪns: Moment of inertia about its diameter is 0.0625 kgm 2. Given: Mass of Ring = M = 500 g = 0.5 kg, Radius of ring = R Radius 0.5 m about an axis of rotation coinciding with its diameter and tangent Moment of inertia about an axis tangent to the ring and its plane is 0.09375 kgm 2Ĭalculate the moment of inertia of a ring of mass 500 g and Moment of inertia about its diameter is 0.03125 kgm 2. Moment of inertia about a transverse axis passing through its circumference is 0.125 kgm 2. I = 3/2MR 2 = 3/2 x (0.25 x (0.5) 2) = 3/2Īns: Moment of inertia about a transverse axis passing through its centre is 0.0625 kgm 2. Moment of inertia about an axis tangent to the ring and its Moment of inertia about its diameter is given by Moment of inertia about a transverse axis passing through
Perpendicular to its plane iii) diameter. To the plane and ii) an axis passing through a point on its circumference, Moment of inertia about i) an axis passing through its centre and perpendicular
Point 20 cm from one end is 4.16 x 10-3 kg m 2Ī thin ring has mass 0.25 kg and radius 0.5 m. Of inertia of the thin uniform rod about a transverse axis passing through a Uniform rod about a transverse axis passing through its end is 0.008 kg m 2, Moment Of inertia of the thin uniform rod about a transverse axis passing through itsĬentre is 0.002 kg m 2, Moment of inertia of the thin Given: Mass of rod = M = 0.024 kg, length of rod = l = 1 m. of a thin uniform rod 1 m long and weighingĠ.024 kg about a transverse axis passing through (1) its centre (2) one end (3) Thin uniform rod about a transverse axis passing through its end is 0.012 kg m 2.įind the M. Inertia of the thin uniform rod about a transverse axis passing through its endĬentre is 0.003 kg m 2 and the moment of inertia of the Inertia of the thin uniform rod about a transverse axis passing through its Given: Mass of rod = M = 100 g = 0.1 kg, length of rod = l Length 60 cm about an axis perpendicular to its length and passing through (1) The required axis is 0.1458 kg m 2, and radius of gyration is0.3818 mĬalculate the M.