Determine the minimum force needed to keep an object of...

Determine the minimum force needed to keep an object of 7.88kg mass in a vertical circle with a radius of 2.42m, if it is travelling at a velocity of 13.8(m)/(s). a. 546N b. 620N c. 68.9N d. 44.9N e. Not shown here. f. 543N g. 679N h. 353N Determine the minimum force needed to keep an object of 7.88 kg mass in a vertical circle with a radius of 2.42 m,if it is travelling at a velocity of 13.8 m/s. a.546N b.620N 68.9N d.44.9N e.Not shown here f.543N g.679 N h.353N

Transcript

00:01 In this question we have given two wires that is ac and bc and are tied at point c to a small sphere of a mass that is 5 kilogram now it is told that it rotates revolves at a constant speed v in the horizontal circle of radius that is 1 .6 meter now we have to find the minimum value of this speed v.

00:21 First of all i am drawing the diagram for this given qion so we have given two wires one is this that is ac and one wire is this is a c and one wire is a given to us that is bc so this is the bc and we have given this angle that is 45 degree and this is 30 degree now there is a small point and this radius is given to us that is 1 .6 meter and this is revolving in this horizontal direction so this is the circular path and the speed of this is given to us that is v now we have to find this speed so i'm going to show the forces that is acting on this system so one is the tension force that is let this tension force in this string is t1 and tension force in this string is t2 so i'm going to draw the free body diagram so we can say that for this point mass that is see there is a tension force in this section this is given by t1 and tension force in this direction that is t2 so this is t2 this is t2 one is the weight acting vertically downward that is mg this is our horizontal action and this is our vertical direction now as we can say that this angle is 30 degrees so this angle is 30 degree and this angle is 45 degree so we can say the component of this tension force in this direction is given by that is t1 cost 30 degree and component of this tension force in this direction is t2 cost 45 degree similarly in this direction this is t1 sine 30 and this and this t2 sign sorry this is this is c2 sign 45 and there is a one force acting in this direction that is centipater force due to with there is a centipater acceleration so let the centipater acceleration is ac so finally we can say that in vertical direction there is no net force net force is zero so we can say that t1 cost 30 sorry this is t1 cost 30 plus t2 cost 45 will balance the weight that is mg and in horizontal direction we can say that net force is equal to that is mass into centimeter acceleration so net force in inward direction so this is t1 sine 30 plus this is t2 sine 45 is equal to this is mv square by r this is our equation second and this is our equation first now after solub equation 1 and equation 2 we get the value of t 1 and this will comes out to be mg minus mv square by r upon this is root 3 minus 1 upon 2 as we know that t1 is positive so we can say that t1 is greater than 0 so finally this is m g minus mv square by r upon this is root 3 minus 1 by 2 and this is greater than 0 so from here we get the value that is m g is greater than mv square by r or we can say that v is less than under root of g r now put the value of g and r so this will comes out to be that is under root of this is 9 .8 and radius is 1 .6 so finally we will get so this is the so we have to find the maximum value so maximum value is equal to this is undrote of g r so this comes out to be 3 .96 meter per second...

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