A pendulum is made from a bob of mass m and an ideal string...

A pendulum is made from a bob of mass m and an ideal string of length L. Suppose the bob is pulled back and released from rest at a position with the string at an angle of ?? relative to the vertical where ?? has some particular but arbitrary value less than 90 degrees. (part a) Draw a careful sketch of what is going on here. (part b) What is the maximum speed of the bob during the swing of the pendulum? (part c) What is the tension on the string immediately after the bob is released? (part d) What is the magnitude of the acceleration of the bob immediately after the bob is realeased? (part e) What is the work done by the force of Weight as the bob falls from its initial position to the position corresponding to the bottom of the arc? (part f) What is the work done by the Tension force as the bob falls from its initial position to the position corresponding to the bottom of the arc? (part g) What is the magnitude of the force of tension on the string when the bob is at the position corresponding to the bottom of the arc? General Hint for Problem 5: Here are some potentially useful diagrams that might be applicable to solving this problem in an efficient way.

Transcript

00:01 All right, hello.

00:01 In this question, we're going to investigate a bob being released from an initial angle theta naught.

00:07 And so i've drawn here on the left the initial position.

00:10 We have the bob.

00:11 It's got a string length l, and there's a mass m on the end, and it's elevated to some altitude of theta naught, and that is going to be some distance h above where it's going to end up at equilibrium.

00:25 So that's part a.

00:26 Part b is what is the maximum speed of the bob during the swing? well, in order to do this, we can just use conservation of energy.

00:33 We know initially we have some gravitational potential energy, and at the end we have some kinetic energy, the end being here when it's at the bottom.

00:41 It's going to move at speed v here.

00:45 And so our masses cancel out, and we're going to have v is the square root of 2 times g times h.

00:50 Well, what is h here? in order to answer that, i'm going to use this diagram over here.

00:56 I know that this distance here is l, and this is going to be what my h is here, but this entire distance here is l.

01:03 So i know if i call this some distance d here, and this is going to be this side length here, i know that l equals h plus d.

01:12 I also know that the cosine of my initial angle equals my adjacent component d over my hypotenuse, which is l.

01:21 Well, i know that d is going to be l minus h over l, so my h is going to be l minus l times the cosine of theta naught.

01:31 So my maximum speed is going to be the square root of 2 times g times l times 1 minus cosine of my initial angle there.

01:40 That's part b.

01:41 Part c says, what is the tension in the string immediately after the bob is released? so over here, if i draw a free body diagram, i have the weight pulling down, and i'm going to have the tension pulling up.

01:57 Well, if i go ahead and write the sum of the forces in the y direction here, the only component in the y direction, this is going to be my theta naught here, i'm going to have tension times the cosine of theta naught pulling up, and then in the other direction, i'm going to have my weight pulling down.

02:14 And that's going to equal m a y, so our acceleration in the y direction.

02:18 Well, what does that really mean? i don't know what my acceleration in the y direction is, right? so using this equation, i can't actually solve for t because i don't know what t is, and i don't know what my acceleration is.

02:29 I do know, however, that because this is a fixed length string, the tension in the bob is going to cause it to accelerate in a circle.

02:40 So i'm going to have some circular motion here.

02:43 So in order to do this, i need to look at this component of my weight, right? so my, what i'll call my parallel component to my tension here.

02:56 So some of the forces in that direction, i'm going to have my tension pulling in, and then pulling out, i'm going to have my gravity, but just the cosine component of that gravity, and that's going to equal m times my centripetal acceleration, which is m v squared over r.

03:12 When i initially release it, v is going to be equal to zero.

03:16 So my tension is going to be just the x component of my weight.

03:21 That is how we get tension out of that.

03:25 So now that we have that, we can use that to figure out what our acceleration in the y direction is, because we know what the tension is.

03:32 So our acceleration in the y direction is going to be tension, which is mg cosine theta naught, times cosine theta naught, minus the weight.

03:41 And if i divide all that by m, i'm going to end up with my acceleration in the y direction.

03:46 So that will be my acceleration in the y.

03:47 My acceleration in the x will come from a sum of the forces in the x direction.

03:52 And in that, i only have the tension, but i just have the sine of theta naught component of the tension.

03:57 And that's going to equal m, excuse me there, m times ax.

04:02 Well, i know that tension is mg cosine theta naught...

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