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University of Sheffield
University of Winnipeg
I) How much tension must a rope withstand if it is used to accelerate a 1210-kg car horizontally along a frictionless surface at 1.20 m/s$^2$ ?
(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?
(II) A person has a reasonable chance of surviving an automobile crash if the deceleration is no more than 30 $g$'s. Calculate the force on a 65-kg person accelerating at this rate.What distance is traveled if brought to rest at this rate from 95 km/h?
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welcome to our second video in two dimensional dynamics. In this video, we're going to consider a car going around a curve. And we're going to say that this car isn't just traveling with some constant VT. They also have some tangential acceleration. And the question that we want to resolve is at what speed? At what speed does the car slide off? And not just at what Speed, but when does that happen? Okay, so looking at this, then we say, All right, I know if I think about the free body diagram here, I'm gonna have a centripetal force. And then I'm also gonna have a tangential force that I have to deal with. Um, my centripetal force is actually coming from friction. Okay, that's the force that cause me this way. And then my tangential force comes from me pushing down on the accelerator. So I know that I have tangential force is equal to tangential. Acceleration is equal to tangential force divided by mass. And we could go ahead and just plug in a number here will say that we're accelerating tangentially at 2 m per second squared and then for centripetal we say. Okay, well, I know that my centripetal force has to be equal to the friction force. So that's going to be in this case, an F M U S. Because we don't want to slide in the our direction. Can the radio direction, Um and so this is going to be an FM us between the tire and the ground. Um, and that's got to be less than or equal to m us times fn, which is the weight of the car. So that means that we need a centripetal force has to be less than or equal to m us times. Mg centripetal force is equal to am the squared over r so we can cancel out on em here. And I find that I need tangential velocity to be less than or equal to the square root of them us R. G. In order for me to continue traveling around this circle. Okay, so this is my condition for when I start to slide. Here's my tangential acceleration so I can ask myself, When will I start sliding? Well, I know that velocity is a function of time is equal to V not plus one hat R v not plus a times time on DSO I say. Okay, well, I know my saying my initial velocity is zero. I start from rest going around this thing or we could even give ourselves something Say it's ah, 10 m per second plus acceleration of 2 m per second squared times time then I can plug in over here. I say I need this to be less than or equal to the square root of U. S. Times are g the time at which we will violate that will end up being U S R G minus 10 m per second, divided by two meters per second. Okay, so at the time at which we are equal to this the very next moment so 0.1 seconds after that is when we will start to slide s So this is kind of Ah, simple little problem. You could also do this by converting into omega. Remember that Omega is equal to V over r and we do Havenaar here. It's the radius of curvature off the turn. Assuming we have a constant one, we could have plugged this all into omega and then solve for it. Using that way, and that would have been just fine. To instead of an 80 were to remember that we have an 80 is equal to our Alfa and solve for Alfa and plug that in using omega is a function of time is equal to Omega, not plus Alfa Times time and we would have come up with the exact same result as we had before.
Equilibrium and Elasticity