A 1 Kg bead is confined to move on a vertical circle of...

A 1 Kg bead is confined to move on a vertical circle of radius 2 meters. There is no friction between the bead and the circular track. The bead has a speed of 10m/s when at the top of the circle at point "a" and the total mechanical energy of the bead is assumed to be constant as it moves around the circle. 10m/s "a" 2m 30° "b" y x When the bead reaches point "b", the radius from the center of the circle to the bead makes a 30 degree angle to the vertical as shown. Calculate the following: 1) the magnitude and direction of the velocity of the bead at point "b" 2) the magnitude and direction of the acceleration of the bead at point "b"

Transcript

00:01 So in this problem we have our circular track with an object in it.

00:05 Now the a position, we have a speed of 10 meters per second, the radius is 2 meters, and at some point here we're 30 degrees off a vertical below.

00:19 So of course we have a tangential velocity vb, which we want to find, and so we're told the energy is constant through the loop so there's no friction energy, friction or anything else to consider.

00:33 Now what that tells us is the energy at a should equal the energy at b, and this is going to be made of the kinetic and potential energy at each of these states.

00:52 So kinetic energy is of motion, so that's where we're going to get vb, and we have the velocity here, but for the height we want to pick a axis.

01:00 Now we could pick this point, that's probably a good way to do it.

01:04 I'll pick the bottom of the circle here to be h equals zero.

01:08 We can pick our point there to be arbitrarily.

01:14 So now we can plug in what these values are.

01:18 Ke is one half m, one half mv squared, and potential energy mg times h.

01:29 And so we're expanding this.

01:34 Firstly we notice that the mass cancels out from everything, so we don't need to worry about what the mass is yet, and so we can use this to find vb.

01:44 So we're going to subtract this term over and multiply by two and take the square root, so we end up with va squared plus 2gha minus 2ghb.

02:00 So we have va, it is 10, acceleration due to gravity 9 .81, and i'm going to factor out this 2g to save some writing.

02:14 So ha, well this radius is two meters, which means the diameter, the whole height here is going to be four meters.

02:23 And what about the height of b? well the height of b is here.

02:29 Now we know this is two meters, and we can find the length of this section here as 2cos30, because we see we have a right triangle there, and we're adjacent to the angle, so it's a cosine term, because you know this is of course two meters as hypotenuse.

02:51 And so we see the space here is going to be 2 minus 2cos30.

02:57 So minus 2 minus 2cos30.

03:06 Take the square root, gives us vb equal to 13 .16 meters per second, so faster, which makes sense because we've taken some of the height and turned it to kinetic energy.

03:19 Now on our axis system x and y, our velocity is somewhere up here.

03:25 Now we're told that this angle is 30 degrees from the string, so rotating this picture 90 degrees in our heads counterclockwise, we'd see that this angle is 30 degrees.

03:38 So that's the direction, you can report it however you typically would in your class, but it says 30 degrees above the horizontal x as the direction.

03:50 Now in part two we can find the acceleration.

03:53 So there's a couple terms to consider.

03:54 We have the centripetal acceleration, or the radial acceleration, which is keeping the object on a circular path...

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