Why JEE Students Fail This Rotation & Center of Mass Question | Hinge Force Mechanics by SBT

追加: 2026-05-30

[00:00:00]This is yet another video to bring conceptual clarity on the chapter of rotational motion. This is a very routine kind of question, but I can assure you that if you watch the video completely, you will learn many key points. So, keep watching. Don't skip.

[00:00:19]This is the question.

[00:00:21]There's a rod of mass m and length l.

[00:00:23]This is a uniform rod.

[00:00:26]And it is hinged at this point O.

[00:00:28]It is just hinged like this, right?

[00:00:30]You keep the rod in horizontal position and release it. It can rotate about this hinged point O like this in vertical plane.

[00:00:37]Uh here, the question asks you two things. The first is immediately after the rod is released, what is the hinge force acting on the rod?

[00:00:47]Just after you release the rod, how much force the hinge, the pin, the support will apply on the rod at point O? That is the first question, and the second question is after the rod has rotated through an angle of 30°, what is the hinge force? So, after the rod has turned through an angle of 30°, at that particular instant, how much force the hinge is applying at point O? The support is applying at point O. So, these are two problems. Let us discuss them one by one.

[00:01:17]First one.

[00:01:18]Uh As soon as you release the rod, as soon as you release the rod, it does not have any speed.

[00:01:25]If a question says that a ball is released from height h, what is the initial velocity of the ball? Just after the release, what is the initial velocity of the ball? Zero.

[00:01:35]What its acceleration is not zero.

[00:01:38]Once you release the ball, immediately after releasing the ball, its velocity is zero, but the velocity has a started to change. There is an acceleration.

[00:01:47]Acceleration is caused by force.

[00:01:51]Forces can be changed instantaneously.

[00:01:55]As soon as I As I release this ball there is a net unbalanced force mg on it.

[00:02:01]So, when I was holding it, net force was zero. mg was being balanced by force applied by my hand. But as soon as I release it, there is an unbalanced force mg. And once there is a force, there is an acceleration.

[00:02:14]But if there is an acceleration, it does not mean velocity of an object will change immediately.

[00:02:20]It will take some time. So, you release the ball, its initial velocity is zero.

[00:02:24]But there is some acceleration. Allow some time and it will gain speed.

[00:02:28]Similarly, when you release this rod, immediately after the release, the rod has not gone anywhere. It is still there.

[00:02:35]So, immediately after the release means uh 0.00001 second after the release. So, the rod is still horizontal.

[00:02:44]Right? It has got no velocity right now.

[00:02:48]There is no angular speed. Omega is zero. It is horizontal.

[00:02:52]Its angular velocity is zero.

[00:02:54]But it will have acceleration because as you release it, then there are unbalanced forces and torques which are going to uh produce acceleration or angular acceleration, whatever.

[00:03:08]So, let us have a look at this situation.

[00:03:11]Immediately after the release, this mg force, which is considered to be acting at the center of mass of the rod, uh there is a force that is going to cause a torque about point two.

[00:03:25]Please understand that the hinge, the pin, the support, whatever it is, whatever you call it, this hinge is also applying some force on the rod.

[00:03:34]I do not know the direction right now. I do not know its magnitude. I'm just showing it like this for just for reference, I'm showing it like this that this is the hinge force. I may be wrong.

[00:03:45]Let us discover it. But let us assume that there is some force applied by the hinge in some unknown direction which I do not know right now.

[00:03:53]Uh this force is not known to me. This is what I have to calculate. mg is the another force, but look at the situation. If I consider torque of all the forces about point O, then hinge force is not relevant.

[00:04:11]Hinge force is not going to produce any torque on the rod about point O.

[00:04:16]It is the mg force that is going to have some torque about point O.

[00:04:21]Obviously, if this length is L by 2, mg force mg, its perpendicular distance from point O is L by 2. So, the torque is mgL by 2. So, immediately after you release the rod about point O, the torque on the rod is mg into L by 2.

[00:04:38]This is the torque.

[00:04:41]This torque will produce an angular acceleration.

[00:04:44]How much is that angular acceleration?

[00:04:47]Answer is torque divided by moment of inertia.

[00:04:50]Torque is equal to I alpha. So, about point O, in fact, about this axis, when is the axis? About this axis, mg has a torque. This hinge force does not have torque because it is passing through this point O.

[00:05:03]So, such a force cannot cause any rotation about point O. So, mg into L by 2 is the torque, and the moment of inertia of the rod about this axis, all of you must be knowing, is 1 by 3 ml squared. So, torque divided by moment of inertia gives me the angular acceleration.

[00:05:21]This is the angular acceleration of the rod immediately after it is released.

[00:05:27]Angular acceleration means an acceleration which causes the angular speed to change. Initially, the there is no angular speed. The rod is not having any speed, but there is an angular acceleration means you allow some time, and now it will have some speed. Radian per second value speed. 1 second may kitne radians se ghum gaya usko omega kehte hain. And if omega is changing, we say there is some angular acceleration.

[00:05:52]So, this is angular acceleration. Its unit is radian per second square SI unit.

[00:05:58]Okay, fine.

[00:06:00]Now, this is the angular acceleration.

[00:06:03]Now, let us think about the acceleration of center of mass of the rod.

[00:06:09]Why center of mass? Why I have suddenly moved to center of mass? There is no reference to center of mass in this question. Why are you talking about center of mass? Because Newton's second law of motion is correctly read in this way.

[00:06:26]This is the correct way of writing Newton's second law of motion.

[00:06:29]External force acting on a body is mass times acceleration of its center of mass.

[00:06:35]Students, if you know the acceleration of center of mass of a body, then you know the total external force acting on the body.

[00:06:44]So, I have been asked to find the hinge force. I have been asked to find some force that is acting on this particular rod.

[00:06:53]So, that force, external force, can be calculated, there is a possibility, if I know the acceleration of center of mass.

[00:07:02]So, let us focus on motion of center of mass. Forget about everything else.

[00:07:06]This rod, as soon as it is left, each and every point of this rod is going to move in a circle, right? The center of mass will move in a circle of radius L by 2.

[00:07:19]But, immediately after it is released, there is no angular speed. There is no speed of center of mass.

[00:07:25]It is going to move in this circle, but right now it is not having any speed.

[00:07:31]But, its speed is going to increase because this thing is having some angular acceleration.

[00:07:38]The situation is somewhat like this.

[00:07:40]So, assume yourself to be uh on a racing track. A motorcycle race 1 2 3 go.

[00:07:51]What is your acceleration immediately after I say go? What is the direction of your acceleration? Can you guess that?

[00:07:57]Immediately after I say go, your speed is still zero.

[00:08:01]So, you are going to move in this circle. You are going to move in this circle, but right now your speed is zero.

[00:08:07]So, if you are not having any speed, there is no radial acceleration.

[00:08:12]V squared by R is zero right now. There is no centripetal acceleration.

[00:08:17]So, whatever acceleration you are having right now is in the direction of tangent.

[00:08:22]You are having a tangential acceleration. Immediately after you you hear the word go, there is a tangential acceleration. This tangential acceleration causes your speed to change. Your speed starts increasing.

[00:08:34]And once you gain some speed, then obviously there is a radial acceleration if you are moving in a circle, radial acceleration will be V squared by R.

[00:08:43]But immediately after I say go, there is no acceleration in this direction.

[00:08:47]Acceleration is only in tangential direction. And this tangential acceleration is actually equal to R into alpha, where R is this radius of this circular path, and alpha is your angular acceleration.

[00:08:58]So, the tangential acceleration and angular acceleration are related through this relation.

[00:09:03]The same is the situation of center of mass of the rod. As soon as I release the rod, the center of mass is not having any speed. It is going to move in this circle, but right now it is not having any speed. Because it is not having any speed, therefore it is having no radial acceleration.

[00:09:20]The acceleration of center of mass is not along the radius of the circular path that it is going to follow right now.

[00:09:26]So, the only acceleration that center of mass is having right now is tangential.

[00:09:31]So, in the direction of tangent, which is of course vertical, this is the acceleration of center of mass, and its acceleration is nothing but R alpha. R means this radius L by 2. L by 2 into alpha. What is alpha? Alpha is this number, this number.

[00:09:47]When you leave this rod, each and every particle in this rod will have same angular acceleration.

[00:09:53]At any instant of time, they will have same angular speed.

[00:09:57]After some time, they will have same angular displacement. So, theta, omega, alpha is same for each and every particle.

[00:10:03]Each and every particle is not moving with same speed.

[00:10:06]Each and every particle is not having same acceleration. But, each and every particle is having same angular speed at a moment, same angular acceleration at a moment.

[00:10:15]So, the angular acceleration of the whole rod is this much because each and every particle has got same angular acceleration. That's why we say that angular acceleration of the rod is 3 by 2g by L.

[00:10:27]So, the angular acceleration of center of mass is of course this much, and the center of mass is going to move in a circle of radius L by 2. Therefore, its acceleration will be in tangential, that is vertical direction, L by 2 alpha. So, L by 2 alpha L L cancels out 3 by 4g.

[00:10:46]Yes. So, immediately after you release the rod, the acceleration of center of mass is 3 by 4g in downward direction.

[00:10:55]Downward direction.

[00:10:56]Now, look at this equation.

[00:10:59]We know that acceleration of the rod is 3 by 4g in downward direction. 3 by 4g in downward direction.

[00:11:07]What are the forces acting on this rod?

[00:11:09]Immediately after the rod is released, there is an mg force, external force. mg If this is the only force, then acceleration of center of mass will be g downward.

[00:11:19]But, we have calculated that acceleration of center of mass is 3 by 4g downward.

[00:11:23]It is less than g.

[00:11:25]So, there must be an upward force acting on this rod. Yes, there is an upward force acting on this rod, and this force is applied by this hinge, this pin.

[00:11:36]How much is this force? And just you can make elementary calculation that this is mg by 4.

[00:11:42]If the hinge up hinge applies a force of mg by 4, then the net force acting on the rod in downward direction will be mg minus mg by 4.

[00:11:50]Which is 3 by 4 mg.

[00:11:53]If hinge applies mg by 4 force, then the net downward force is mg minus mg by 4.

[00:11:57]Which is 3 by 4 mg.

[00:12:00]So, if 3 by 4 mg is the net downward force, the acceleration of center of mass will be 3 by 4 g.

[00:12:08]3 by 4 mg is the force divided by mass.

[00:12:11]So, this is the acceleration of center of mass. This is possible only when the hinge applies an upward force of mg by 4. There is no question of any horizontal force right now acting on the rod because there is no horizontal acceleration of center of mass.

[00:12:27]Had there been a horizontal acceleration of center of mass, then it would mean that hinge is also applying some horizontal force.

[00:12:36]But right now, the acceleration is only vertical, and therefore hinge force has to be vertical. It cannot have any horizontal component.

[00:12:44]So, we have got the our answer. It is mg by 4 in upward direction, the hinge force in first case.

[00:12:51]In the second case, after the rod has rotated through some angle, uh I have given you all the fundamentals, but still I'll do few steps so that uh you don't find it really very difficult.

[00:13:07]So, this rod has already rotated through some angle, theta. This is what the question says.

[00:13:12]Uh 30 degree, yes.

[00:13:14]So, this is 30 degree.

[00:13:18]Okay.

[00:13:19]And in this particular position, I have to find what is the hinge force.

[00:13:25]Now, the rod is rotating. It is having some angular speed. Center of mass is also having some speed. So, right now, center of mass is going in a circle, no doubt about it.

[00:13:38]And this center of mass is having some radial as well as tangential acceleration.

[00:13:43]So, again, the approach is same. I need to figure out what is the acceleration of center of mass of this rod right now.

[00:13:51]I need to understand that because center of mass is moving, it is having some speed.

[00:13:57]It is going in a circle and it is performing non-uniform circular motion. Therefore, the center of mass will have some radial acceleration as well as some tangential acceleration.

[00:14:09]So, in order to get the acceleration of center of mass, I need to know two things. First is speed of center of mass and second, the angular acceleration of the rod. So, let us figure out these two things.

[00:14:22]First, what is the angular speed of the rod right now? When the rod has rotated through an angle of 30°, what is its angular speed?

[00:14:30]To know the angular speed, one can apply law of conservation of mechanical energy.

[00:14:37]You release the rod from this position and there is no friction here. Um hinge does not exert any friction. So, the rod is swinging only under the influence of gravitational force as such, isn't it?

[00:14:51]The gravitational force is the only force which is performing work. I should say it like this.

[00:14:56]So, when gravity is performing work, no other force is performing work, obviously. Well, some of you are getting confused. You hinge force can apply energy conservation.

[00:15:09]Hinge this point before a point fixed.

[00:15:11]The hinge force work make a point.

[00:15:13]Hinge force point of application moving.

[00:15:16]So, hinge force is not doing any work.

[00:15:18]MG is the only force which is performing some work.

[00:15:22]And mg is spring force, electrostatic force is of the work at the time that energy comes up at the end. They are These are conservative forces. So, mechanical energy is conserved.

[00:15:32]Or most of the times energy conservation is something that works quite well. So, this rod is initially having no angular speed when it was released. Now, when it has rotated through an angle of 30°, it is having some angular speed. To get that angular speed, one can apply law of conservation of mechanical energy. For that, we need to figure out what is loss in gravitational potential energy.

[00:16:02]The gravitational potential energy we need to find out by what distance the center of mass of the rod has fallen.

[00:16:10]The center of mass of the rod, which is initially here, is now here.

[00:16:15]By what distance it has fallen? It is very easy to work out. This is L by 2.

[00:16:20]This is the center.

[00:16:21]This is L by 2. So, length of this red line will be L by 2 sin 30°.

[00:16:28]Which is, of course, L by 4. So, the center of the rod has fallen vertically by a distance of L by 4.

[00:16:36]So, the rod has lost a potential energy which is mgL by 4.

[00:16:44]And this should be equal to gain in kinetic energy of the rod.

[00:16:47]So, the right rod right now is in pure rotational motion about this axis. It is having angular speed omega right now. So, its kinetic energy can be written as 1/2 I omega squared.

[00:17:04]The rod is positioned here.

[00:17:06]It's the angular speed omega.

[00:17:08]This is my kinetic energy about this axis. It can be written as 1/2 I omega squared. And this kinetic energy is equal to gain in is equal to loss in gravitational potential energy which is mg times l by 4.

[00:17:22]From here you can work out value of omega.

[00:17:26]So it actually becomes two two cancels out. It is 3 by 2 g remains l remains. I guess this is what you get.

[00:17:36]Please cross check it.

[00:17:39]3 by 2 g by l >> [cough] [clears throat] >> root of that. That is the omega.

[00:17:45]So right now you know the angular speed of the rod. Okay, that is one thing.

[00:17:50]So when I know the angular speed of the rod, each and every point in this rod is having this angular speed. So the center of mass is also going in circle with this angular speed. So I can easily write the radial acceleration of center of mass.

[00:18:02]It is omega squared into r where r is l by 2.

[00:18:06]But then I also need to know the tangential acceleration of center of mass and for finding tangential acceleration, I need to know the angular acceleration of the rod. How do I get angular acceleration? For that I need again need to understand how much torque is acting on the rod.

[00:18:21]This is the mg force.

[00:18:23]Again about this particular point I need not worry about the hinge force actually. Hinge force will produce no torque about point O.

[00:18:31]So what is the torque of mg? Very simple. This is the line on which mg force is acting.

[00:18:36]And this line is at this much distance from point O which is l by 2 cos 30°.

[00:18:42]So the torque due to mg is mg l by 2 cos 30°.

[00:18:49]cos 30°. l by 2 cos 30°. This length into mg this is the torque. And torque divided by moment of inertia 1 by 3 ml squared.

[00:18:59]This is angular acceleration.

[00:19:02]So the angular acceleration of the rod is root 3 3 root 3 3 root 3 by 4 uh g upon l something like this.

[00:19:16]Please cross check. 3 root 3 by 4 g upon l, okay. So, I know omega, I know alpha. Now, what am I going to do with these two quantities?

[00:19:27]Uh let me change the diagram for the sake of clarity.

[00:19:32]This is our rod.

[00:19:36]And this is having an angular speed omega and angular acceleration alpha, right?

[00:19:44]Now, let us worry about the motion of center of mass. Center of mass is of course going in a circle like this having radius l by 2. Center of mass is an imaginary point which is going in a circle of radius l by 2. Right now, it is having angular speed omega and angular acceleration alpha.

[00:19:59]So, the acceleration radial acceleration of center of mass radial acceleration of center of mass is omega squared r omega squared l by 2 basically.

[00:20:10]So, I know the acceleration of center of mass in this direction.

[00:20:13]Right? Now, center of mass is also having a tangential acceleration. Tangential acceleration, how much is that? That is r alpha. R is l by 2 into alpha. I know alpha, I know omega, so I know these two numbers. I'm not doing any more calculations. You can do it on your own.

[00:20:29]So, center of mass has got two components of acceleration which are known to us, radial and tangential.

[00:20:36]Now, what are the forces acting on this rod? The weight force is there.

[00:20:41]And then there is hinge force.

[00:20:43]Uh just give me a moment which which has done any.

[00:20:46]So, again, you can uh consider two perpendicular directions as horizontal and vertical direction. Or you can just take radial and tangential to be your two perpendicular direction, that is also fine. So, this is what I'm going to consider.

[00:20:59]I'm assuming that hinge is applying a force. The radial component of this hinge force is fr.

[00:21:05]Look at this situation. Radial component of hinge force is FR. Radial means towards point O. Center of mass is going in a circle and center is at O.

[00:21:14]The component of mg in this direction, you can work it out yourself. This 30°.

[00:21:21]Right? This is 60°.

[00:21:23]So, mg cos 60°. Component of mg in this direction is mg / 2.

[00:21:29]Right? So, the net force towards point O acting on this rod is hinge force FR - mg / 2. So, FR - mg / 2 is the net force acting on this rod towards point O. And this should be equal to this should be equal to m omega squared L / 2. What is omega squared L / 2?

[00:21:54]Omega squared L / 2 is the acceleration of center of mass in this direction, radial acceleration.

[00:22:00]So, now you know the value of omega. And this radial acceleration.

[00:22:04]You know this acceleration, you know the value of omega, you can calculate FR.

[00:22:09]FR means the radial component of hinge force has been calculated.

[00:22:14]Right? Now, let us focus on the tangential direction.

[00:22:19]In this particular direction, there is a component of mg.

[00:22:24]This horizontal line, this 30°, this 60°, this 30°. So, component of mg in this direction is mg cos 30°.

[00:22:33]This one.

[00:22:35]And there is a hinge force, of course.

[00:22:38]I'm not sure whether the hinge force is this way or this way. So, I'm taking it down to the up to some extent. May I assume that it rotates in the length of the perpendicular direction? Hinge force, okay.

[00:22:48]So, the net force on the rod in this direction, perpendicular to rod, is mg root 3 / 2 uh the Don't write it as FH, write it as FT, tangential hinge force. FT I hope you are getting it.

[00:23:08]Uh mg root 3 by 2 is the component of mg.

[00:23:12]And then there is a component of hinge force perpendicular to rod.

[00:23:15]Uh so, the net force on this rod in this direction, perpendicular to rod. The direction is perpendicular to rod, tangential direction. That is equal to this minus that.

[00:23:26]And this should be equal to mass of the rod into tangential acceleration of center of mass.

[00:23:36]Right?

[00:23:37]So, I've written these two equations for motion of center of mass. For radial acceleration of center of mass, mass times radial acceleration is net force in this direction.

[00:23:45]Mass times tangential acceleration is net force in this direction.

[00:23:49]So, I know the value of alpha, I can calculate FT.

[00:23:51]Um up to say that calculate they flow. I got FT negative again, I'm going to negative again. It simply means that this force will be in this direction. It has no other meaning. You can work it out on your own.

[00:24:03]So, now you get two components of hinge force, FR and FT. And the resultant hinge force will be under root FR squared plus FT squared.