AP®︎/College Physics 1
- AP Physics 1 review of 1D motion
- AP Physics 1 review of 2D motion and vectors
- AP Physics 1 review of Forces and Newton's Laws
- AP Physics 1 review of Centripetal Forces
- AP Physics 1 review of Energy and Work
- AP Physics 1 review of Momentum and Impulse
- AP Physics 1 review of Torque and Angular momentum
- AP Physics 1 review of Waves and Harmonic motion
- AP Physics 1 Review of Charge and Circuits
In this video, David quickly explains each torque and angular concept and does a sample question for each one. Created by David SantoPietro.
Want to join the conversation?
- At15:34,why does the clay have an ANGULAR momentum before the collision? I thought it was moving translationally?Thanks(8 votes)
- Think about it this way: Once the translationally moving mass comes into contact with the rod and sticks to it, its translational momentum will become angular momentum with the same magnitude, so what you are calculating is not it's angular momentum while flying in a straight line, but it's "potential angular momentum" that it would have if it started moving in a circle.(12 votes)
- At16:18, why did David use the rotational inertia of the clay sphere to be ML^2 when the formula of the rotational inertia of a sphere is (2/5)(M)(L^2)?(5 votes)
- So does rotational kinetic energy affect mechanical energy or just translational kinetic energy?(1 vote)
- Mechanical energy is the sum of KINETIC and gravitational energy so I believe that rotational KINETIC energy would affect mechanical energy.(4 votes)
- What is the right hand rule?(2 votes)
- Given that you're on the video for rotational motion, the right-hand rule (for rotational motion) applies to vector quantities (angular velocity, angular acceleration, torque, etc.); it basically states that if you curl your right hand in the direction of the rotation, the vector's direction is the direction of your thumb.(1 vote)
- Can you please specify the differences/relationships between angular acceleration, tangential acceleration, radial acceleration and centripetal acceleration? I keep mixing up the appropriate equations/units. I feel that is because I do not fully understand the subtleties of these terms.(1 vote)
- Angular acceleration α = dω/dt = d²θ/dt² [rads/sec²]
Angular velocity ω = dθ/dt [rads/sec]
Tangential acceleration a = rα [meters/sec²]
Tangential velocity v = rω [meters/sec]
Tangential displacement s = rθ [meters]
Centripetal acceleration (same thing as radial acc.) a = v²/r = rω² [meters/sec²]
Note that radians are a unitless unit in that they don't have any specific dimension. This is because the radian is defined as the ratio of the circle arc length/radius, so [distance/distance] is unitless.(3 votes)
- At10:27, shouldn't we use the formula of I=mr^2 to culcuate the inertia of those 2 objects? ( since B is a hoop, we can use that formula) If we use that formula, then the answer will be C.(2 votes)
- Your time stamp doesn't direct to the problem you're asking about. Could you recheck it? Or are you referring to a different video altogether?(1 vote)
- At10:52, shouldn't the torque with 1 newton of force be positive since it is counter-clock wise and the torque with the 4 N force be negative since it is pushing clockwise? The final answer would be the same because the answer asks for the magnitude, but just wondering.(1 vote)
- (v^2)/r is centripetal acceleration and so is (omega)^2(r). when exactly do we use the latter one? I was told you only use that when the velocity is changing(1 vote)
- Actually, they are equivalent.
recall: v= r omega
You would simply use them when the data or analysis best suits.
The velocity (strictly speaking) is always changing because the direction is changing
if the speed is changing then you would need to take into account the tangential acceleration (alpha usually) which is different to the centripetal acceleration
- I don't understand how the problem was approached for15:26. The sign for length (L) and the sign for angular momentum (L) are the same and are treated as the same thing. Why? Is angular momentum equal to the distance between the axis of rotation and the collision point?(1 vote)
- No. Notice that when he refers to angular momentum he writes Li, not just L. The i is for inertial. The problem stipulates that the length of the rod is L. It's confusing since we often use L for angular momentum. Here he tried to resolve that confusion by using Li but that was not a good way to do it because the i is hard to even notice. He should have just given the angular momentum a completely different variable name for this problem.(1 vote)
- At16:51, it says that the gravitational potential energy can be determined using Gm1m2/d. I thought that the formula was Gm1m2/d^2... Or is gravitational attraction different from gravitational potential energy?(1 vote)
- [Instructor] The rotational kinematic formulas allow us to relate the five different rotational motion variables and they look just like the regular kinematic formulas except instead of displacement, there's angular displacement. Instead of initial velocity there's initial angular velocity. Instead of final velocity there's final angular velocity. Instead of acceleration there's angular acceleration and the time is still just the time. You only get the first two of these on the AP exam formula sheet. You do not get three and four. And just like the regular kinematic formulas, these rotational kinematic formulas are only true if the angular acceleration is constant. What do each of these rotational variables mean? Well, the angular displacement is the amount of angle the object has rotated through in the certain amount of time t. Angular velocity is defined to be the amount of angle you rotated through per time just like regular velocity is the displacement per time, and the angular acceleration is defined to be the amount of change in angular velocity per time. Just like regular acceleration is the change in regular velocity per time. For something rotating in a circle, technically the angular velocity points perpendicular to that plane of rotation but it's easiest to just think of omega as being counterclockwise or clockwise. And there's relationships between these angular variables and their linear counterparts. To get the arc length s the object has traveled through you just multiply the radius of the path by the amount of angular displacement. To get the speed of the object just multiply the radius of the path by the angular speed of the object. To get the tangential acceleration multiply the radius of the path by the angular acceleration. Note that this is the tangential acceleration. This acceleration causes the object to speed up or slow down. It's the centripetal component of the acceleration that causes the object to change directions and the formula for that is still just v squared over r. If an object's moving in a circle it must have centripetal acceleration because it's changing directions but only if it's speeding up or slowing down will it have tangential acceleration and angular acceleration. What's an example problem involving the angular motion variables look like? Let's say an object is rotating in a circle at a constant rate. Which would best describe the three different types of accelerations of the object? Well, if an object's moving in a circle at all there has to be centripetal acceleration so that's got to be non-zero. And if it's rotating at a constant rate there's no change in omega and that's means the angular acceleration is gonna be zero. If the angular acceleration is zero the tangential acceleration would also be zero. Only when the object is speeding up or slowing down do you have angular acceleration and tangential acceleration. This change the speed and centripetal acceleration changes the direction. What does torque mean? Just like force is what causes acceleration, torque is what causes angular acceleration. In order for an object to speed up or slow down in its angular motion, there's got to be a net torque on the object. What causes a torque? Forces cause torque. In order to have a torque you have to have a force but the same force could exert a different torque depending on where that force is exerted. If the force is exerted far from the axis of rotation, you'll get more torque for that given amount of force compared to forces that are exerted near the axis of rotation. This r represents how far that force is applied from the axis. And to maximize that force you would actually wanna point it perpendicular to this r since sine of 90 degrees is equal to one. In other words, to maximize the amount of torque you get exert the force as far away as possible from the axis and exert that force perpendicular to the line from the axis to that force. You might get many angles in a problem but this angle here is always the angle between the r and the F. And just like an object is in translational equilibrium if the net force is zero, we say that an object is in rotational equilibrium if the net torque is zero. This would cause the angular acceleration to be zero just like translational equilibrium causes the acceleration to be zero. Torque is a vector so it has a direction. Typically it's easiest to think of the direction as just being counterclockwise or clockwise based on which way that force would cause the object to rotate. And since torque is r times F, the units are gonna be meters times newton or newton meters. What's an example problem involving torque look like? Let's say you had this rod with this axis here and there were forces applied as shown. We wanna know how large would the force F have to be in order for this rod to be in rotational equilibrium. Remember, rotational equilibrium means that the net torque is equal to zero. In other words, all the torque that's pointing clockwise would have to equal all the torque that points counterclockwise in order for the system to be balanced. The three newton force and the one newton force, you're trying to rotate this system clockwise and the unknown force F is trying to rotate the system counterclockwise. This green one newton force isn't actually exerting any torque even though the r value is not zero. The angle between the force and the r value is gonna be 180 degrees and the sine of 180 degrees is zero. Which makes sense since this force isn't actually causing this rod to rotate clockwise or counterclockwise. The torque and the clockwise direction would be one meter times three newtons to find the torque from the three newton force. Plus you wouldn't use two meters for the r of the one newton force. You have to find the r from the axis which is gonna be three meters times one newton force which gives a total clockwise torque of six newton meters, and we can write the torque applied by the unknown force F as one meter times F. In order for six newton meters to equal one times F, the force F just has to equal six newtons. What's rotational inertia mean? Well, an object with a large rotational inertia will be hard to get rotating and harder to stop rotating. Basically the rotational inertia tells you how much an object will resist angular acceleration. Just like regular inertia tells you how much an object will resist regular acceleration. And this rotational inertia is often referred to as the Moment of inertia. How do you make the rotational inertia large? Well you can increase the rotational inertia if you place the mass far from the axis of rotation and you can make the rotational inertia smaller if you place the mass close to the axis of rotation. In other words if you could push the mass closer to the axis of rotation which is the point about which the object rotates, you can make the moment of inertia smaller and smaller. To find the moment of inertia or rotational inertia of an object whose entire mass rotates at the same radius r, you can just use the formula I equals the mass that's rotating, times how far it's rotating from the axis squared. This formula's not given, you have to memorize it. I equals mr squared. And if you had many masses rotating at different r's you could just add up all the contributions from each single mass. Now if you have a continuous object whose mass is not all at the same radius from the axis, the formulas are a little more complicated. For a rod rotating about its center, the moment of inertia would be 1/12 the mass of the rod times the entire length of the rod squared. For rod rotating about one end, the moment of inertia is gonna be larger since more mass is distributed farther from the axis and this formula is 1/3 the mass of the rod times the entire length of the rod squared. The rotational inertia of a sphere rotating about an axis through its center would be 2/5 the mass of the sphere times the radius of the sphere squared. And the rotational inertia of a cylinder or a disk rotating about an axis through its center would be 1/2 the mass of the disk times the radius of that disk squared. Another example that comes up a lot that you wouldn't be given the formula for is a hoop. That is to say where all the mass is distributed around the center point with a hollow center. Since all the mass is at the same radius r, the formula for the rotational inertia of a hoop is just the same as the formula for the rotational inertia of a single mass rotating at radius r. The fact that the mass is distributed in a circle doesn't actually matter since the mass still stayed at the same radius r away. Rotational inertia is not a vector so it's always positive or zero and the units since it's mr squared would be kilograms times meter squared. What's an example problem involving rotational inertia look like? Let's say two cylinders are allowed to roll without slipping from rest down a hill. The mass of cylinder A is distributed evenly throughout the entire cylinder. Cylinder B is made from a more dense material and it has a hollow center with the mass distributed around that hollow center. If the masses and radii of the cylinders are the same, which cylinder would reach the bottom of the hill first? To figure out which cylinder gets to the bottom of the hill first we have to ask which one would roll more readily. The cylinder with the least moment of inertia is gonna be easier to rotate. That means it would roll more readily and get to the bottom of the hill faster. Whenever the mass is distributed farther away from the axis, the object's gonna have a larger moment of inertia so since the mass of cylinder B overall is farther away from the axis compared to cylinder A, cylinder B has a larger moment of inertia, that means it's harder to rotate. It will take longer to get down the hill and cylinder A is gonna win. What's the angular version of the Newton's Second Law? Well, Newton's Second Law says that the acceleration is equal to the net force divided by the mass and the angular version of Newton's Second Law says that the angular acceleration is equal to the net torque divided by the rotational inertia. M tells you how much an object resist acceleration and the moment of inertia or rotational inertia tells you how much an object resists angular acceleration. Just like when you add up force vectors you have to be careful with positive and negative signs. The same holds true with the torque vectors. You've got to treat either counterclockwise or clockwise as positive and then be consistent with it. What's an example problem involving the angular version of Newton's Second Law look like? Let's say the rod shown below has a rotational inertia of two kilogram meter squared and has the forces acting on it as shown. We wanna know what the magnitude of the angular acceleration is of the rod. We use Newton's Second Law in angular form which says that the angular acceleration is the net torque divided by the rotational inertia. We got the rotational inertia, we just need the net torque. We have to figure out the total torque from all these forces. The torque from the one newton force would be r which is gonna be three meters from the axis to that force one newton. And since it's applied perpendicular, the sine of 90 is gonna be one. The torque from the one newton force would be three newton meters in the counterclockwise direction. And the torque from the four newton force would be one meter since it's applied one meter from the axis times four newtons and we get four newton meters in the clockwise direction. That means the total net torque when you have four newton meters in the clockwise direction and three newton meters in the counterclockwise direction would just be one newton meter in the clockwise direction since four is one unit bigger than three. And now we divide by the rotational inertia which was two which gives us an angular acceleration of 1/2 or 0.5. What's rotational kinetic energy mean? Well if an object is rotating or spinning, we say it has rotational kinetic energy. If the center of mass of an object is moving and the object is rotating, we typically say that object has translational kinetic energy and rotational kinetic energy. They're both kinetic energies, this is just a convenient way to delineate between two types of kinetic energy and a particularly convenient way to find the total kinetic energy for something that's moving and rotating. The formula for rotational kinetic energy is 1/2 times the moment of inertia or the rotational inertia times the angular speed squared. Which makes sense because the formula for regular kinetic energy is 1/2 times the regular inertia, the mass times the regular speed squared. Again, if an object is rotating, it's got rotational kinetic energy. If the center of mass of an object is moving it's got regular translational kinetic energy. And if the center of mass is moving and the object is rotating, then we say that object has both rotational energy and translational energy. Rotational kinetic energy is not a vector. It is always positive or zero and the units can be written as kilogram meter squared per second squared but it's an energy so we know that just has to equal joules. What's an example problem involving rotational kinetic energy look like? Let's say a constant torque is exerted on a cylinder that's initially at rest and can rotate about an axis through its center. Which of these curves would best give the rotational kinetic energy of the cylinder as a function of time? Well if there's a constant amount of torque on an object, that will cause a constant angular acceleration. And if the angular acceleration is constant, we can use the kinematic formulas to figure out the final velocity of this object. The final angular velocity if it's started at rest would just be alpha times t. That means the rotational kinetic energy of this object could be written as 1/2 the moment of inertia which is a constant times omega squared. Which in this case would be 1/2 I times alpha t squared. Since the function for kinetic energy is proportional to the time squared, if you graph kinetic energy as a function of time, it would look like a parabola so the correct answer would be B. What's angular momentum? Well, the reason we care about angular momentum is that it will be conserved for a system if there's no external torque on that system. And just like regular momentum is mass times velocity, angular momentum will be the rotational inertia times the angular velocity. And this is a convenient formula to find the angular momentum of an extended object whose mass is distributed at different points away from the axis of rotation. The strange thing about angular momentum is that even a point mass moving in a straight line can have angular momentum. To find the angular momentum of a point mass moving in a straight line, take the mass of the object times the velocity of that object, and either multiply by how far that object is from the axis times sine of the angle between the velocity vector and that R. Or the easier way to do it is to just multiply by the distance of closest approach which is how close that mass will ever get to or ever has been from the axis. In other words to determine the angular momentum of this mass moving in a straight line, draw a straight line along its trajectory and ask how close has it gotten or ever will get to the axis? That's the capital R I'm talking about. And if you take that times the mass times velocity, you'll get the angular momentum of that point mass. Angular momentum is a vector and it's easiest to just think about the direction of angular momentum as being either counterclockwise or clockwise depending on which way the object is rotating. And as for the units if you multiply mass of kilograms times meters per second times meters, you'd get kilogram meter squared per seconds as the units of angular momentum. What's an example problem involving angular momentum look like? Let's say a clay sphere of mass M was heading toward a rod of mass three M and length L with a speed v. The rod is free to rotate about an axis around its end. If the clay sticks to the end of the rod, what would be the angular velocity of the rod after the clay sticks to the rod? And we're given that the moment of inertia of a rod about its end is 1/3 ML squared. Since there's gonna be no net external torque on this system, the angular momentum of this system is gonna be conserved. The only object in this system that has angular momentum initially is this clay sphere. Since this is a point mass moving in a straight line we'll use the formula M times the velocity, times the closest it will ever get to the axis which is L, the length of the rod. That's gonna have to equal the final angular momentum which we could write as I times omega. And this I would be the moment of inertia of both the rod and the clay that is now stuck to the end of the rod. We'd have MvL equals the total moment of inertia, moment of inertia of the rod is 1/3 mass of the rod which is three M times the length of the rod squared. Plus the moment of inertia of this piece of clay stuck to the end of the rod rotating in a circle is gonna be the mass of the clay times the radius of the circle that clay traces out which is the length of the rod. In other words, we're using the formula for the moment of inertia of a point mass whose entire mass is rotating at the same radius from the center. And we add that to the moment of inertia of the rod itself. We multiply by omega. The term in brackets comes out to be two ML squared. We can cancel the M's, we can cancel one of the L's and we get that omega is gonna equal v over two L. The last topic I wanna talk about is the more general formula for the gravitational potential energy. Why do we need a more general formula? Well, if you're in a region where the gravitational field little g is constant then you could just use our familiar formula mgh to find the gravitational potential energy. But if you're in a region where the gravitational field is varying then you have to use this more general formula which states that the gravitational potential energy between two masses, m one and m two, is gonna equal negative of the gravitational constant, big G, times the product of the two masses, divided by the center to center distance between the two masses. Note, this is center to center, not surface to surface and it's not squared like it is in the force formula. This one's just the distance. Gravitational potential energy is not a vector but because of this negative sign, the gravitational potential energy is always gonna be negative or zero. It'll only be zero when these spheres become infinitely far apart because then you'd divide by infinity and one over infinity would be zero. Otherwise, it's always negative. But even though this gravitational potential energy is negative, this energy could still get converted into kinetic energy, it's just that in order for this gravitational potential energy to decrease, it would have to become even more negative to convert that energy into kinetic energy. I'm bringing up this topic in this section because oftentimes when planets are orbiting each other in circular orbits, you have to use this formula to determine the gravitational potential energy between them. And since it's an energy, the units are joules, so what's an example problem involving this more general formula for the gravitational potential energy look like? Let's say two spheres of radius R and mass M are falling toward each other due to their gravitational attraction. If the surface to surface distance between them starts off as four R and ends up as two R, how much kinetic energy would be gained by this system? We'll include both masses in our system and that would mean that there's gonna be no external work done so the energy of this system is gonna be conserved. The system's gonna start off with gravitational potential energy negative big G both masses multiplied together which is M squared divided by the distance they start off from each other which is not four R, it's the center to center distance which is gonna six R. We'll assume they start from rest so there'll be no kinetic energy to start with and this is gonna equal the final gravitational potential energy negative big G, both masses multiplied M squared divided by the distance they end up which is not two R, it's center to center distance so that's four R, plus however much potential energy was converted into kinetic energy. If we solve this for kinetic energy we're gonna get negative big G, M squared over six R plus big G, M squared over four R. 1/4 minus 1/6 is gonna be 1/12. The amount of potential energy that was converted into kinetic energy would have been big G, M squared over 12 R.