00:01
So in this question we are given that there is a mass m which is attached to two strings of length l equal lengths and both these strings are attached to a rotating rod and the distance between the points where they are attached to the rod is d and in this question specifically we are given that d is equal to l.
00:20
So basically this triangle between these two points and the mass m is an equilateral triangle which we can then say that this theta is equal to this theta is equal to this theta and theta.
00:31
Is 60 degrees that's the angle for an equal little triangle and these strings are taught so they can be considered as fixed length bodies so this is the situation we have and we are first asked to find draw all the forces on the mass m so let's say this is the mass m so there will be a force draw a little bit here so there will be force m g the weight force there will be a since it is rotating there will be a centrifugal force fc which will try to throw it away from the radius or the axis of rotation and then the two forces countering that motion is one is tension t1 in the other is tension t2 and both these have angle theta from the vertical.
01:30
Let me mark these here as well.
01:32
So there's tension t1 in the top string and tension t2 in the bottom string.
01:37
So these are the forces that act on this body of mass.
01:42
The next question then asks us what is the radius of the rotation which means this distance from the axis directly all the way to m.
01:52
Let's call this r.
01:53
So if you see the triangle, any half triangle which has one side r in one side l, you will see that r is equal to l cos of theta.
02:05
So if you see this triangle over here, this triangle, you will see that it is r is l cost theta.
02:13
I'm sorry, r is l sine theta.
02:19
Sorry about that.
02:21
So r is l sine theta.
02:24
And sine of 60 is point, roughly point 87.
02:27
So you can say in this case that sine of 60 degrees and that makes r equal to 0 .87 times l.
02:36
So this is the answer for part b.
02:39
Now we are asked to find the expression for the velocity of this velocity of the mass in terms of as a function of the tension t1, the mass, the gravity constant or the acceleration to gravity, the theta, and the as well as the uh i think that's it so in these four uh we need to find the velocity so basically we have tension t1 and t2 we need to find a solution such that we eliminate t2 and make it in terms of t1 only so for that let's use this free body diagram over here and write the equations for this body let me draw line here so this a new part so let's write down the equations uh so in the vertical direction, we can see that tension t1, cost theta component.
03:45
So basically this direction.
03:46
Let me mark down.
03:48
Let's say this is y and this is x.
03:52
So t1 cost theta is in the positive y direction.
03:56
And negative y direction, you will have two components, which will be t2 cost theta and mg, both facing downwards.
04:06
So we can write this or rewrite this equation.
04:09
As t1 minus t2 is equal to m g by cost theta it's called this equation one so we take the t2 term on the left and divide the whole equation by cost theta similarly for the vertical direction we see that the positive extraction has the centrifugal force and in this case that would be mv square by r and for the negative x direction we have two forces t1 sine theta component and t2 sine theta component both add up in the negative side of the x direction we can simplify this equation again and we can say that t1 plus t2 is equal to mv square by r we can write from this equation on top l sine theta because we want to avoid any parameters which are not tension t1 mass acceleration root to gravity or theta so l sine theta and one more sine theta comes by dividing the whole equation by sine theta so this becomes l sine square theta this is equation two so now we see that why have done this because we can now add equation one and two and just eliminate t2 which we want to do so i will say that equation one plus equation two gives me that two t1 is equal to m g by cost theta minus uh sorry plus mv square by l sine square theta or we can write now we need the expression for velocity so if you rearrange this you get the expression of velocity as square root of 2 t1 minus m g by cost theta l sine square theta by m so this is the expression for the velocity in terms of t1 m g and theta as well as hell sorry i have forgot this see over here i was thinking i've forgotten one more parameter and that is l and we know that for this situation given in this diagram on the right side theta it will be 60 because if the three sides of this triangle have to be at the same length of l the thetas are fixed so these these bodies are rotating the tensions are changing in the strings but the lengths are not extending so everything is taught and we assume that none of these strings slack out so theta will remain 60 for this whole problem so that we already know in this case.
07:05
Alright so next part then gives us the following problem.
07:10
Let me draw all i know here.
07:14
It says that if m is 1 .34 kg's tension t1 is 35 newton's and length l is 1 .7 meters and we will take acceleration to gravity is 9 .81 meter per second square.
07:35
If this is the case what is the velocity? that's the question.
07:40
So now that we have the equation from here, we can just take it to the right side and add the values in.
07:45
So velocity becomes equal to square root of 2 into 35 minus 1 .34 times 9 .81.
07:57
Cost of 60 is 0 .5 times 1 .7.
08:02
Sine square 60 is roughly 0 .75 divided by 1 .34.
08:08
And you take the square root of this whole thing.
08:11
And if you solve this, you get this velocity to be equal to 6.
08:14
4 4 meters per second.
08:17
This is the answer for part i think this is part d.
08:22
This is part c.
08:24
This is part b and this was part a...