A meterstick l 1 m has a mass of m 0134 kg initially it...

A meterstick (L = 1 m) has a mass of m = 0.134 kg. Initially, it hangs from two short strings: one at the 25 cm mark and one at the 75 cm mark. 1) What is the tension in the left string? N 2) Now the right string is cut! What is the initial angular acceleration of the meterstick about its pivot point? (You may assume the rod pivots about the left string, and the string remains vertical) rad/s^2 3) What is the tension in the left string right after the right string is cut? N 4) After the right string is cut, the meterstick swings down to where it is vertical for an instant before it swings back up in the other direction. What is the angular speed when the meterstick is vertical? rad/s 5) What is the acceleration of the center of mass of the meterstick when it is vertical? m/s^2 6) What is the tension in the string when the meterstick is vertical? N

Transcript

00:01 The length of the meter stick is 1 meter.

00:02 The mass of the meter stick is 0 .134 kg.

00:09 So it is hanging from two strings.

00:15 So the first string is at 25 centimetre mark and the second string is at 75 centimetre mark.

00:24 So this is how it is hanging.

00:27 So if we are talking about the forces here.

00:30 At 50 centimeter mark we have mg acting and there is going to be tension reduced in the.

00:37 String so we know that this is 25 centimeter this is 75 centimeter and this is 50 centimeter mark so firstly we are supposed to calculate what is the tension in the left string so to calculate the tension in the string in string we know that by translational equilibrium the vertically upward forces should balance out the vertically downward force so 2 t is going to be be equal to m g or t is simply going to be m g divided by two so we are going to get the tension in each string for example let's string particularly is going to be 0 .134 times 9 .8 divided by 2 hence tension in the string equal to 6 .56 newton now further the right string is cut so we have to calculate the initial angular acceleration of the meter stick about the the pivoted point.

01:48 So if right string is cut, so the situation is going to be the left string is still acting at 25 centimeter mark.

02:07 So the right string is cut and at 50 centimeter mark we have mg acting downward.

02:15 So this mg with this acting downward will tend to rotate this given meter stick about this point so the perpendicular distance curve is going to be 25 centimeter and the pivot point is here so this is one axis the standard axis passing through the center this is another axis so we can calculate what is the initial angular acceleration so for initial angular acceleration we know that tau is equal to i alpha so alpha is going to be equal to tau by i so what is the torque by torque is provided by weight so m g times the perpendicular distance which is 25 centimeter this equal to moment of inertia so moment of inertia about this left string so let's call this point b so moment of inertia about p times angular acceleration so mass we already know with the 0 .134 times 9 .8 let's keep this in centimeter moment of inertia about this point can be found out using the parallel axis theorem we know we know standard axis is the axis which passes through the center and this is a parallel axis so that is going to be nothing but ml squared divided by 12 plus md squared so this distance is nothing but 25 times alpha l is nothing but one meter so let's keep that in 100 and continue it in centimeter squared so we can eliminate mass so this is going to be 9 .8 times 25 is equal to hundred squared divided by 12 plus 25 squared multiplied with alpha so we get alpha value as we get it as 16 .8 radiant per second square so that is the initial angular acceleration.

04:13 We find out the tension in the left string right after the string is cut so we know that there is put translation and rotational motion so we can say that m g minus p is equal to ma but to find out a value, we can write it as in terms of a is equal to r alpha.

04:33 So mg minus t is equal to m times r.

04:37 R is a perpendicular distance from the pivot which is nothing but l by 4 times alpha which is 16 .8.

04:45 So tension in the left string when the string is cut is going to be mg minus m times 16 .8 times l divided by 4.

04:56 So that is going to give us the tension in the left string as 0 .75 newton.

05:05 Now we have to note here that we have used the condition a is equal to our alpha.

05:11 Further, due to the cutting of the right string, the left string would swing down along with a meter stick and it would take a vertical position.

05:21 So as it takes vertical position at vertical position by conservation of energy, we can say that whatever potential energy was possessed by the given meter stick is converted into kinetic energy.

05:38 So we can say that half mv squared is going to be equal to mgh.

05:44 Therefore velocity at the vertical position is going to be root 2gh.

05:49 So the height at which it was initially from the pivot to the center is 75 centimeter.

05:54 So v is going to be equal to 2 times 9 .8 times...

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