The 1-slug mass m rotates around the vertical pole in a...

The 1-slug mass $m$ rotates around the vertical pole in a horizontal circular path. The angle $\theta = 30^\circ$ and the length of the string is 4 ft. Which equation of motion can be used to obtain the tension in the string directly (without using the other equations)? $\sum F_t = ma_t$ $\sum F_\theta = 0$ $\sum F_n = ma_n$ None of the above

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00:01 Here in this question the two string and their point of attachment to the pole form an equilateral triangle therefore the angle theta will be equal to 60 degree each and the diagram is as follows so here this is the diagram here there is tension t 1 t and here the theta will be equal to 60 degree since it is an equilateral triangle so here given that length l is equal to 1 .5 meter theta is given which is equal to 60 degree and mass m is also given which is 0 .5 kilogram let omega represents the angular velocity of the pole in this case then mass m which is this one makes an angular path a with the tangential velocity.

01:28 That is here, ay is equal to 0.

01:35 Therefore, the force in the y direction is equal to mass into acceleration ay.

01:41 Since ay is equal to zero, the force also becomes zero.

01:44 And here from the diagram we can say that t1 cos 60 degree minus t2 goes 60 degree minus m g is equal to 0 or we can say that here t1 minus t 2 is equal to 2 times m g for force in x direction we have mass into acceleration in x direction for this is equal to mass into acceleration in x direction is given by r omega square and sigma f x is equal to zero therefore t1 sine 60 degree plus t2 sine 60 degree is equal to m r omega square here radius r is equal to length l.

02:43 Therefore, we can say that root 3 by 2 times t1 plus root 3 by 2 times t2 is equal to m l omega square.

02:56 Or we can say that let this be equation number 2 and this one be equation number 1.

03:05 So from equation number 1 and 2 we can say that we can say that 2 root 3 m g plus 2 root 3 t 2 is equal to 2 m r omega square let this be equation number 3 now let us draw the triangle representing them this is the triangle here we have one of the sides here represents the radius r so this will be l by 2 and hyper continuous will be l.

03:45 So from the triangle we can say that r square is equal to l square minus l by 2 all square for solving we will get r is equal to root 3 by 2 times l now we have to substitute for r in equation number 3 upon substituting we will get 2mg plus 2 t2 is equal to m l omega square or we can say that here t 2 will be equal to half into m into l omega square minus m g now in the first part we need to find out the minimum speed which is required to ensure that the lower string is taught so in order to ensure that the lower string is out we need to find out the minimum speed.

04:49 So here the minimum speed will be square root of 2 times t2 plus m g divided by m l is equal to omega...

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