(II) A pendulum consists of a mass M hanging at the bottom end...

Question

(II) A pendulum consists of a mass M hanging at the bottom end of a massless rod of length ℓ, which has a frictionless pivot at its top end. A mass m, moving horizontally as shown in Fig. 9–44 with velocity v, impacts M and becomes embedded. What is the smallest value of v sufficient to cause the pendulum (with embedded mass m) to swing clear over the top of its arc?

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with collisions. So in this problem, a lantern with a mass of capital M is hanging from the end of a massless chain of length. L, the chain is fixed to a pivot point at the top allowing the lantern to swing freely. A bird with a mass of little M flies directly towards the lantern with a velocity V and crashes into it becoming embedded, determine the minimum velocity that the bird bird must have in order to ensure that the lantern. But now with the bird inside of it swings completely over the top of its arc. Below the question, we're given a diagram of what was described in the problem. We're given four possible choices as our answers. A is the square root of GL, multiplying the quantity of one plus capital M divided by little MB is the square root of two gl, multiplying the quantity of one plus capital M divided by little M choice C, we have two multiplied by the square root of GL which is multiplying the quantity of one plus capital M divided by low M. And choice D is four multiplying the square root of GL which is multiplying in the quantity of one plus capital M divided by little M. Now, in this problem, we have the bird crashing into the lantern and crashing indicates a collision. And in collisions, we have momentum being conserved. That's the, where we're going to start. We recall that for momentum to be conserved, we have to have our total initial momentum P I total is equal to our total final momentum PF total. And here momentum is a vector quantity. So this is a vector equation to recall your definition for momentum, we'll have P is equal to MV M is the mass and V is the velocity and both momentum and velocity are vector quantities. Now, here in this case, I have the bird finding in the horizontal direction towards the lantern. And so um I'm just dealing with motion in the X direction. So I'm only since I'm only dealing with one component, I'm gonna drop vector notation and only work in the X action. So my total initial momentum is just gonna be the momentum from the bird. So that's gonna be a little MV. And since the velocity of the lantern is zero, it has no initial momentum. So my total initial momentum is just gonna be that of the bird. And so that's MV, and that's gonna be equal to the total final momentum. Since the bird gets embedded into the lantern, they're gonna move together as a single object. So its total mass is going to be little M plus capital M and that quantity gets multiplied by their common speed, which we'll call V prime. So the question asks for V, so I'll solve for V. So V is going to be equal to and I'm just going to divide through by the little M. So I'll have little M plus capital M divided by little M and that fraction gets multiplied by V prime. Now, it was given the capital M and little M as parameters in our problem and they will show up in our final answer, but I don't have any information about the prime. So V prime is going to be the speed at which the lantern and bird begin to move around this arc. And if we're looking for the minimum speed that the bird has to hit the lantern in order for it to reach the top of the arc, that means that the top of the arc, its final speed is going to be zero. And so as I move up the arc, it's gonna be going against gravity and that's what's going to cause it to slow down. So initially, I'll have all kinetic energy at the bottom of our arc and all gravitational potential energy at the very top of the arc. And since I'm dealing with energies here, I want to use our conservation of energy equation. Recall that for energy to be conserved we have K I plus UISB equal to KF plus UF PK is our connect energy and U is our potential energy. Now, we're free to set the 0.4 our gravitational potential energy. And the easiest thing to do is set our initial gravitational potential energy to zero. So that will eliminate that term all of our equation. And we already discussed at the very top if we're looking for the minimum velocity that the lantern needs to reach the top, that means its final speed should be zero. And so that means our final kinetic energy gonna be zero. So this equation ends up being K I is equal to uf our initial connect energy is equal to our final gravitational potential energy. To recall our formulas for um connect energy in gravitational potential energy. For K connect energy is gonna be one half MV squared. We'll have one half the mass. In this case is the combined mass of the bird and lantern uh will be little M plus capital M. And the speed here is just gonna be the V prime and that gets squared and that's gonna have to be equal to our gravitational potential energy that it's gonna be mGy. And since we set the initial um gravitational potential energy to zero, that means our initial why value is gonna be zero? And so we can label that in our diagram and at the very top of this arc, the final Y value is gonna be to L. And that's because we have a radius of basically L and we've gone from the one end of the semicircle to the other. So we basically have the diameter. So twice the radius that's gonna be two L, that's gonna be our wide value that we plug into our gravitational potential energy formula. So the right hand side of our um energy conservation equation going to be little M plus capital M. That quantity is multiplied by G and that is multiplied by two L. We need to solve this for V prime. And so first thing I can do is cancel out the masses. So I'll get B prime is equal to the square root of four GL, just multiplied over the one half uh multiply or divide over by the one half. And then took the square root, I can simplify this a little bit by pulling out the um four outside the square root sign. So have the prime is equal to two multiply the square root of GL. So I can plug that back into our equation. For V. We have V is equal to, I'm gonna simplify this fraction a little bit. I have little M divided by little M which is one, don't have plus capital M divided by little M. And then that gets, that quantity gets multiplied by the prime, which is two multiplied by the square root of GL. So that is gonna be my final answer. And if I look at what answer choices corresponds to this corresponds to answer. C So just a quick little recap here, we initially started off with the conservation of momentum since we had a collision between the bird and the lantern to find out the speed at which the bird collided into the lantern. But in order to solve that, we need to figure out the speed at which the lantern and bird moved away after the collision. And for that, we had to use our conservation of energy. In that case, our initial kinetic energy at the bottom of the arc that we're looking for to be equal to the gravitational potential energy at the top of the loop. So I hope that this has been helpful and I'll see you in the next video.

Key Concepts

Conservation of Momentum

Energy Conservation

Pendulum Motion

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