A block of mass m slides down a frictionless track, then around t...

Question

A block of mass m slides down a frictionless track, then around the inside of a circular loop-the-loop of radius R . From what minimum height h must the block start to make it around without falling off? Give your answer as a multiple of R.

Accepted Answer

Hey, everyone. So this problem is going to use the conservation of energy. and Newton 2nd law, let's see what they're asking us. A toy designer creates a roller coaster type track using a smooth ramp and a frictionless vertical loop of diameter D. The car starting at rest slides down the incline and should go around the vertical loop by always being in contact with the loop. The designer asks a physicist to find the minimum height the car must be, must start at and still successfully negotiate the loop. What should be the physicist answer? We are asked to write our answer as a function of D, our multiple choice answers here are a 0. DB DC, 1.25 D or D, 1.5 D. So the first thing I'm going to do here is kind of draw a diagram of what we have going on. So we have a ramp, it's gonna make that cool loop de loop and that ramp is gonna start from a height of H and this minimum height age is what we are going to be solving for. Now, we'll look at the car at the top of this loop as a particle. And we can recognize that there are two forces acting on the car when it is at the top of the loop that has the normal force because it is in contact with the surface. And then it has the weight working in those, both of those forces are working in the negative y direction. And then we're also told that the diameter of this circle is deep. OK. So from Newton's second law, we can recall that the sum of the forces is equal to mass multiplied by acceleration. So we have our normal force and our weight is equal to mass times acceleration. We are asked to find H though and neither of these or none of these terms have that age component. So let's look at conservation of energy. Once we drop the car from the top of this ramp, there are no other energy forces acting, there are no other forces acting on this um scenario, this system. So the energy of the system will be conserved. So our conservation of energy equation we can recall is given by U I plus K I for initial potential energy plus initial kinetic energy is equal to U F plus K F for final potential energy plus final kinetic energy. And so if we recall that potential energy is given as M G H, we can write that as M G H I. And on the other side of the equation M G H F and then recall kinetic energy is given as one half M V squared. So initially, that will be one half M V I squared and one half M B F squared for the final position. And so this equation is our conservation of energy equation. And now here we do have the height component. So we can look at the givens that we know and see if we have enough to solve for height. So we know that initially the I is zero because the car starts at rest. We don't know V F, we don't know that final speed. We do know the initial height. That's that age that we're solving for. And our final height when the car reaches just at the top of the loop, de loop, that height is deep. OK. So we need to solve for age. We need to figure out what that final speed is that V F Going back to our, going back to our Newton 2nd law Problem. We can use that um to solve four VF. So we can recall that cental centripetal acceleration is given by A equals V squared over R. And we know the definition of radius in diameter two r equals d. And then for our normal force because we're looking for the minimum height, that normal force is going to be zero, it's going to just stay attached to or just stay in contact with the loop. And so then plugging that in we have zero plus can recall weight is given by mass multiplied by gravity equals mass multiplied bye V squared over R. And then in turn, that R is equal to D over two. So the masses will cancel, we can isolate velocity, which is what we are looking for. It's gonna be our V F is equal to the square root of D multiplied by G over two. Coming back down to our con uh conservation of energy equation. We can now say that V F is equal to The Square Root of DG Over two. So from here, we can solve this conservation of energy equation for H V I is zero. So that whole term goes to zero. And then we will plug in all of our other known values and see what we can come up with and isolate for that age. So each of these terms has a mass term so that all cancels. So we have G H equals G D plus one half V squared or one half square root of D G over two squared, the square root and the squared cancel or we can simplify that G H equals G D plus one half Multiplied by DG over two. Each of these terms has uh gravity G. So those can all cancel. And we are left with H equals D plus D over four or 40.25 D, All right. And we add that together and we get h equals 1.25 d. So that is the final answer to this problem and that aligns with answer choice C so that's all we have for this one. We'll see you in the next video.

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A block of mass m slides down a frictionless track, then around the inside of a circular loop-the-loop of radius R . From what minimum height h must the block start to make it around without falling off? Give your answer as a multiple of R.

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