The spring is uncompressed when the uniform slender bar AB...

The spring is uncompressed when the uniform slender bar AB is in the vertical position shown. Bar AB with a mass of 50 kg and a length of 1 m is released from rest in the position where the bar has been rotated 30 clockwise from the position shown. This initial position of the bar AB is illustrated using the dashed line in the figure. Neglect any sag of the spring as well as the spring mass. (a) Determine the initial angular acceleration of the bar when it is just released from rest. (b) Find the angular velocity of the bar when it reaches the vertical position. Answers: (a) ? = 198.951 rad/s² (b) ?? = 10 rad/s

Transcript

00:02 I so we are given that the rod of mass m is equals to 50 kt length l is equals to 1 meters string stiffness k 12 kilo newtons per meter so in the first part for uncompressed spring length as you can see from you can see which we have in material let's say in triangle c o t by using the theorem which says that cd square is equals to c square plus or d square we can find it out this length so cd should become equals to the under root over c it will have the point of one by four plus one by four that whole square plus one square so this will comes to 1 .118 meters so therefore cd we have been found 1 .118 meter which is the answer for first part of this question now let's have a look to second part so weight of the rod can be given by the formula w is equals to m into g now mass is given 50 multiply value of g will be 9 .8 meter per second square so on solving weight will comes out 490 .5 newtons so basically the rod ab is rotated by 30 degree clockwise that we haven't given according to question so in this case the triangle o c c dash that means to say the new figure will be formed something like in this way so by perpendicular this making an angle of 30 degree as you can see so this is what o c c dash triangle o c c dash means to say we will draw the perpendicular to this so you will get a new point and that is going to be the required c point so in this triangle o c c dash means this complete triangle we can say that by using the trigonometry property sin of 30 this should be equals to x c dash by o c dash this is equals to x by 1 by 4 plus 1 by 4 so you will get the value of x is equals to 1 by 4 which will be equals to 0 .25 a meter so therefore the answer for the second part of this question we have been found the length x is equals to 0 .25 meter now next according to this for the a part we have to find the initial angular acceleration which is denoted by alpha so you remember from newton's second law summation of torque t naught this is equals to moment of inertia times of angular acceleration alpha of vector so according to the given figure summation of t naught vector this should be equals to plus of f of s vector along the o c dash minus of m g sin of 30 along o g dash of the radial vector so this can be equals to c force is…

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