9. A point particle of mass M is attached to the end of a...

9. A point particle of mass M is attached to the end of a massless, ideal string of length L, while the other end of the string is fixed, and the mass is launched into a circular trajectory in a vertical plane near the surface of the earth, such that the string always remains taught. Use standard polar coordinates centered on the fixed point of the string, where ? = 0 corresponds to the string oriented horizontally and the mass moving upwards with a total speed v0. (a) What is the minimum value of v0 required so that the string remains taught at all times, and the particle actually travels in a circle? (b) At a given angle ?, what is the tangential (or circumferential) acceleration a? of the point mass?

Transcript

00:01 All right, in this problem, we have a mass tied on the end of a string, and it has length l, which is going to be the radius of the circle, it's moving in a circle of radius l.

00:13 The initial velocity is v0, which points upward when the string is horizontal.

00:21 I also drew on this picture a line where the potential energy due to gravity is zero.

00:28 We have to use energy conservation as part of the problem.

00:32 So i'm drawing that in to begin with.

00:35 Here's a free body diagram for the mass.

00:38 So it's got gravity pointing downward.

00:41 It's got tension pointing inward along the string.

00:44 Our centrifugal acceleration points inward.

00:48 Here's our angle theta.

00:51 So the radial part of the radial equation tells me t plus mgc sine theta is mv squared over r because this is the component of gravity that points along that direction okay the tangential velocity or the tangential part of the force law tells us m g coast theta is minus um m alpha r so alpha r would be the tangential part of the acceleration so alpha is the acceleration.

01:43 And if you think about this originally, when theta is equal to zero, cosine theta is equal to one.

01:51 And so my alpha has to be negative because in fact it's slowing down.

01:58 As it moves upward here, gravity is going to be slowing it down as we expect our angular acceleration at that point to be negative.

02:08 So that's why we get that minus sign.

02:13 So in order for the string to remain taught, the tension has to be positive.

02:21 As soon as the tension is negative, the string becomes loose.

02:26 And the equals zero, one could argue about equals zero, whether it makes sense or not.

02:32 But when it's equal to zero, there's no tension.

02:36 So strictly speaking, it could start to loosen up as soon as the tension is zero.

02:43 So i'm making it strictly greater than.

02:48 And i can solve my basic equation for t is mv squared over l minus mg sine theta.

03:00 Now in order to analyze this further, i need information about the velocity.

03:07 I can't get information about the velocity from the tangential equation...

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