 Hello everyone, myself Sandeep Javeri, Assistant Professor, Department of Civil Engineering from Valchiyas Drop Technology, Swelapur. In today's session, we are going to discuss rotational movement and kinetic energy of rigid body. The learning outcome at the end of this session, students will be able to derive the expression for the rotational moment and kinetic energy of rigid body. Let us consider a wheel shown in this figure which is rotating about its axis in clockwise direction with an acceleration alpha. So this is a wheel having a axis of rotation through point O. So the angular acceleration with which it rotates is alpha. Let delta m be the mass element and a distance r from the axis of rotation and let delta p be the resulting force which is acting on this element. As we know, the force is nothing but mass into acceleration. So that resulting force can be obtained as delta m into A where A is a tangential acceleration. So we know the relation between linear motion and angular motion that means linear acceleration and angular acceleration as A is equal to r into alpha where alpha is a angular acceleration and A is a linear acceleration. How to get the force which is acting on that element delta m that is delta p? We put the value of A as r into alpha in this expression. So we get delta m delta p is equal to delta m into r into alpha. Now rotational movement delta mt that is for the mass delta m, what is the rotational movement? It is given by delta mt due to the force delta p is given by force into the distance r from the axis of rotation. So we have this is a delta p and this is a distance r from the axis of rotation. So delta mt that is rotational movement is equal to delta m into r into alpha into r. So we are putting the value of delta p as delta m into r into alpha in the expression of delta mt. Now total rotational movement is given by for that we have to do summation of all the rotational movement of this different element various element or masses which is present on the wheel. We have to consider all masses of a wheel. So summation of delta m into r square into alpha. So alpha is a constant term that is angular acceleration. So the rotational movement that is total rotational movement is given by alpha into summation of delta m into r square. Now this delta m summation of delta m into r square is nothing but mass movement of inertia. So put this as i. So we are getting the expression for rotational movement as alpha into i or you can say i into alpha. Thus rotational movement is given by i into alpha and in case of linear motion the force is given as mass into acceleration. So force causes linear motion while rotational movement causes rotation. So rotational movement and angular momentum we can define as rotational movement is the product of mass movement of inertia and angular acceleration while angular momentum is the product of mass movement of inertia and the angular velocity of the rotating body. Now we can define rotational movement as the product of mass movement of inertia and angular acceleration. So rotational movement is equal to I into alpha, its unit is Newton meter and angular momentum is the product of the mass movement of inertia and the angular velocity of the rotating body. So here angular movement is equal to I into omega. So omega is the angular velocity of the body that is rotating body. So here the unit of angular momentum is Newton second. Please remember unit of rotational momentum is Newton meter while angular momentum is Newton second. Now let us derive the expression for kinetic energy of rotating bodies. So same figure we are considering this is a flywheel or a wheel. So the axis of rotation is O at del time with the mass which is present on the wheel and it is acting at it is present at a distance of r from the axis of rotation. Let omega be the angular velocity. So consider the wheel which is rotating about the axis in clockwise direction with an angular velocity omega. Let delta m be the mass of an element acting at a distance r from the axis of rotation. Hence if V be the linear velocity of the element the kinetic energy of the element mass is half into delta m V square. Kinetic energy of the rotating body we can write summation of half into delta m into V square that is equals to summation of half into delta m into r square omega square. So this V can be replaced by r into omega. So expression for velocity is r into omega. In case of linear motion we are considering velocity as small v, rotational motion we are considering the velocity as angular velocity that is omega. So kinetic energy is given by half into omega square as omega is a constant term into summation of delta m into r square. But by definition mass movement of inertia i is equal to summation m r square and kinetic energy is equal to half into i into omega square. So what is unit of this Newton meter or Joule? So how to derive this expression we have to consider first the linear motion concept and by using linear motion concept we know the relation between linear motion and rotational motion that is V is equal to r omega that is the relation between linear velocity and angular velocity. And then we are getting the kinetic energy of this rotating body as half into i omega square and the unit is Newton meter or you can say Joule. Now you are supposed to pause the video and answer this question. So these are the answers as we know or we can recall the rotational moment is given by i into alpha, alpha is the angular acceleration. The expression for kinetic energy for rotating body having angular velocity omega is given by half into i omega square. So b is the correct answer and kinetic energy the equation for angular momentum is given by i into omega. So a is the correct answer. Now these are the references which we are using for the creation of this video. Thank you.