1. ## conical pendulum,, could somebody help please,,

calculate the tension and angular speed in a 2m string attached to a 2kg bob that is moving in horizontal circles of 0.5m radius.

i have traweld the net and the same formula keeps poping up Tcos(angle)=mg.
after doing a sketch of the problem the first thing i did was to find the upper angle,

i used SOH(OPP/HYP) 0.5M/2M=0.25 tan-1 0.25= 14 degrees.so far so good.its now i get confused, i understand that the T refers to tension, cos is cosine and mg is mass x gravity.

the book says T=mg/cos(angle),,,,,T=2x9.8/0.25= 78.4N where does the 0.25 come from i thought a cosine needed adj/hyp. the above 0.25 i got was from the SOH value, sorry if this is a bit long winded for you people but this is the problem with home study i just need it simplifying and explaining thanks,

2. Originally Posted by prs
calculate the tension and angular speed in a 2m string attached to a 2kg bob that is moving in horizontal circles of 0.5m radius.

i have traweld the net and the same formula keeps poping up Tcos(angle)=mg.
after doing a sketch of the problem the first thing i did was to find the upper angle,

i used SOH(OPP/HYP) 0.5M/2M=0.25 tan-1 0.25= 14 degrees.so far so good.its now i get confused, i understand that the T refers to tension, cos is cosine and mg is mass x gravity.

the book says T=mg/cos(angle),,,,,T=2x9.8/0.25= 78.4N where does the 0.25 come from i thought a cosine needed adj/hyp. the above 0.25 i got was from the SOH value, sorry if this is a bit long winded for you people but this is the problem with home study i just need it simplifying and explaining thanks,
Dear prs,

If $\theta$ is the angle between the string and the horizontal $\cos\theta=\frac{0.5}{2}$. I am sure this is what is meant in the book. Even if you take the angle between the string and the verticle axis (hope this is what you mean by the upper angle) you should end up with, $\sin\theta=\frac{0.5}{2}$. Then you should find the angle between the string and the horizontal by $(90-\theta)^o$ , find the cosine of this angle; eventually you would end up with the same answer. Hope you understood.

3. thanks for that sud. its helped a lot