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03:02

Averell H.

(II) Superman must stop a 120-km/h train in 150 m to keep it from hitting a stalled car on the tracks. If the train's mass is $3.6 \times 10^5$ kg how much force must he exert? Compare to the weight of the train (give as %). How much force does the train exert on Superman?

03:09

Aditya P.

(I) A 7150-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s. A 3350-kg load, initially at rest, is dropped onto the car. What will be the car's new speed?

00:48

(I) What force is needed to accelerate a sled (mass = 55 kg) at 1.4 m/s$^2$ on horizontal frictionless ice?

0:00

Muhammed S.

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welcome to our third example. Video on two dimensional dynamics in this video will consider a problem that shows up in most physics textbooks. It involves a table I'm attempting to draw here that is frictionless and has a hole in the middle out of the hole. There's a string that is attached to a puck that has a circular motion on the table, and that string comes down and is attached to some mass. So we have mass m one for the puck mass M two for the hanging mass. And I want to figure out how the motion of these two things is related. So if I were to draw the free body diagram for the hanging mass, I would have em to G and force of tension and for the object going around in the circle. Then I would have a centripetal force. Okay, so I would say What's going around in a circle? Have a centripetal force on this thing, and the centripetal forces also force of tension because it's the same string. So we have some trip. It'll force is equal to tension force, which is equal to M v squared over r, and we have f t minus m two g almost forgot to label that one M two g is equal to M to a Let's consider a situation. Remember, this is acceleration in the Y direction. Who are the acceleration in the Y? Direction is zero. Let's not worry about that. And we want to see what happens if I add a Well, let's see. First of all, how the motion of these two things is related. I know the F. T, in the case of a Y equals zero would be equal to M two G, which in that case I have m two g equals m one v squared over R. So the velocity of the puck on the top. So this is the tangential velocity, remember of the puck is gonna be equal to M two over m one times r g square root of the whole thing. So we take the ratio of these two masses. If I were to increase mass to stay by adding a second mass to and see what happens, so I would have two times mass to then I would have this higher velocity here. Now, how is that going to work? Well, when I add the second mass. Okay, so let's come back to your original question. We have to m two G. It's equal to M one. BT squared over our When I add this second mass, what do we expect to happen? Well, um, m b squared over r was a was related to f T f t got bigger. Which means over here something has to get bigger. And it could have been our and it could have been the We'll get investigate this relationship more on later when we talk about torque and moment of inertia. But we could have had our get smaller or we could have had V get bigger. Generally speaking, what's gonna happen is a little bit of both. It will get pulled in a little bit, and as it gets pulled in, it will start to go faster and faster. And this is related to a concept called conservation of angular momentum, which we will investigate later. But for now, we can see at least that there is this relationship between these two objects. Um, and I could ask if I gave you enough of the variables I could ask you for a result such as What's the Radius? Or how fast is it rotating or how much mass is hanging underneath the table? Um, any of those would be fair game because we were able to come up with this equation here. Remember, the two came after I changed the Mass, so it's M two over em. One R G Square to the whole thing is equal to the tangential velocity of the puck.

Work

Kinetic Energy

Potential Energy

Equilibrium and Elasticity

Energy Conservation

04:39