 Hello everyone, Myself Sandeep Javeri, Assistant Professor, Department of Civil Engineering from Waltz Institute of Technology Shwilaapur. In today's session, we are going to discuss problem on rotating rigid body. The learning outcome at the end of this session, students will be able to solve problem on the rotational rigid body. Now let us consider the problem of flywheel having weight 50 kiloton and having radius of variation 1 meter loses its speed from 400 rpm to 280 rpm in 2 minutes. So we are supposed to calculate the retarding torque which is acting on it and also changes in its kinetic energy during the above period and change in its angular momentum during the same period. Now before solving this we should first understand what is rpm, rpm is nothing but revolution per minute. So the relation between omega and rpm that is f n is a rpm so we have omega is equal to 2 pi n. So if you are writing the speed in rpm so the angular velocity can be written as omega is equal to 2 pi n where n is a speed in rpm that is revolution per minute. So let us consider this is a flywheel which is rotating with a velocity that is omega. So let us consider initial velocity as omega naught and the final velocity is omega. So initial speed is given as 400 rpm and then real speed is given as 280 rpm. Now let us consider the expression omega is equal to 2 pi n where n is a rpm that is speed is given in rpm. Now 400 rpm that can be converted into radian per second that is angular velocity as 400 into 2 pi. So this is divided by 60 that is we have to write in a second so it is 2 pi into 400 divided by 60 that is equals to 41.88 radian per second. We have to convert into second that is why we have divided by 60. Now the time is given for the flywheel to come from speed of 400 rpm to 280 rpm as 2 minutes. So let us t is equal to 2 minute that is equals to 120 second. If you multiply 2 by 60 you are getting it is equal to 120 seconds. As we have expression for rotational motion as omega is equal to omega naught plus alpha t that means we can find out the final velocity expression for the rotational motion as omega is equal to omega naught plus alpha t alpha is the angular acceleration and omega naught is the initial angular velocity. So alpha is equal to omega minus omega naught divided by t that is final angular velocity minus initial angular velocity divided by the time. For omega we can work out by using the expression that is 2 pi n. So that is n is nothing but 280 rpm. Now after putting the value of omega and omega naught and t we are getting the value of angular acceleration as minus 0.1047 radian per second square. This negative sign indicates that there is a retardation. So radius of gyration k is equal to 1 meter. So i is equal to mk square which is equals to 50,000 divided by 9.81 into 1 square. So as we know the weight is given as 50 kiloton we have to write the mass in kg. So this 50 kiloton is multiplied by 1000 so we are getting 50,000 and g as 9.81. So convert it into kg for that we have to use the expression m mass is equal to weight divided by acceleration due to gravity that is g. So by putting this and multiplying this by k that is 1 so we have i that is mass movement of inertia is equal to mass of the wheel into square of the radius of gyration getting mass movement of inertia as 5096.84 kg per kg meter square. Now let us work out one by one retarding torque which is acting on the flywheel that is given by i into alpha. So i is the mass movement of inertia 5096.84 multiplied by 0.1047. So we are getting that retarding torque as 533.63 Newton meter. Similarly to get change in kinetic energy we know we have to subtract final kinetic energy from the initial kinetic energy. So we are getting half into i omega naught square minus half i into omega square. So omega naught is a initial angular velocity, omega is a final angular velocity. Now after putting this value of omega naught and omega we are getting change in kinetic energy as 2278.910 kilo Newton meter. So change in angular momentum the expression for that is i omega naught minus i omega that is initial momentum angular momentum minus final angular momentum. So i is a constant that is 5096.84 which is taken out from the bracket and omega naught is 41.88 and omega that is a final velocity is 29.32. So we are getting change in angular momentum as 64016.31 Newton second. Now let us solve another problem a wheel rotating about axis with a constant acceleration 1 random per second square if the initial and final velocities are 50 rpm and 100 rpm determine the time taken and number of revolutions made during this period. So first of all we have to first convert the velocities in random per second. So what they have given in the problem they have given angular acceleration 1 random per second square initial and final velocity is given in rpm. They ask to find out the time taken and number of revolutions made for the given period. So we have omega naught initial angular velocity of the wheel, we have n naught is initial revolution in rpm that is 50 rpm, we have omega that is final angular velocity of wheel, we have n that is a capital N which is a final revolution in rpm that is 100 rpm. Now t be the time required for to attain angular velocity from initial velocity that is omega naught to final velocity that is omega. So these are the angular velocities. Now omega naught is equal to 2.0 divided by 60 if you are going to convert into second random per second then we have to multiply divided by 60. As we know omega is equal to 2.0 that expression is true but if you want to convert into random per second then we have to divide by 60 and naught is 50. So we are getting the initial angular velocity as 5.236 random per second. Similarly final angular velocity is 2.0 by 60 in random per second. So it is obtained by 2.200 divided by 60 that is equals to 10.47 random per second. So n value is here 100. So we are getting omega that is angular velocity as 10.47 random per second. Now they ask in the question to determine the total time required. Now we have angular acceleration with us. We know the formula how to get the angular acceleration. So it is equals to omega minus omega naught divided by t it is just similar to the we have in linear motion a is equal to v minus u divided by t. So here alpha is equal to omega minus omega naught divided by t. Now put the value of alpha as 1 which is given omega as 10.47 it is in random per second omega naught is 5.236 which is also in random per second. Now find out the value of t we are getting 5.234 second. Now let us determine the number of revolutions required. So for that we consider theta as the angular displacement which is travel during motion of flywheel using the expression omega square minus omega naught square which is equals to 2 times angular acceleration multiplied by the displacement. So it is just similar to the equation what we are using in linear motion that is v square minus u square is equal to 2 As right. So here omega is a final angular velocity omega naught is a initial angular velocity. Now putting this value of omega and omega naught and alpha we are getting the displacement that is angular displacement as 40.476 radians. As we know the 1 revolution is theta is equal to 2 pi right. So for 40.633 radian displacements the number of revolution is 6.44 revolution. So the answer is 6.44 revolution this is the answer. Now you are supposed to pause the video and answer this question. So these are the answers in the first problem. The torque required to stop a wheel of moment of inertia phi into minus 3 kg meter square from a speed of 20 radian per second in 10 second. So we know the torque is equal to i into alpha. Now i is given that is mass moment of inertia is given alpha we can find out as we are going to stop the wheel so the final velocity is 0. Using the expression alpha is equal to alpha minus alpha naught or omega minus omega naught divided by t. So we are getting the value of alpha and we are putting the value of alpha in the expression for rotational moment or the torque that is i into alpha we are getting the value of torque as 0.01 Newton meter. The second question is time required for a flywheel to start from rest to gain an angular speed of 50 radian per second. So given angular acceleration is 2.5 radian per second square. Here we have to find out the time required for a flywheel to start from rest that means initial angular velocity is 0 final angular velocity is given as 50 radian per second. So using the expression alpha is equal to omega minus omega naught divided by t we are getting the value of t as 20 second. So this is the correct answer these are the references that we are using for the creation of this video. Thank you.