A 2kg mass swings at the end of a light string (length 3m). Its speed at the lowest point on its circular path is 6m/s. What is the tension in the string when it makes a angle of 40 degrees with the vertical.
I get v = sqrt[36-2gr(1-cos40)] = 4.76m/s
then T = m[v^2/r+gr(1-cos40)] = 29.84N
The answer is supposed to be 22.6
If I use sin instead of cos I get the right answer. I have this hole handout describing how mgcos(thea) opposes the centripetal force of a mass on a string in circular motion (between pi and 2pi, excluding 3pi/2). What the heck is going on???
I get v = sqrt[36-2gr(1-cos40)] = 4.76m/s
then T = m[v^2/r+gr(1-cos40)] = 29.84N
The answer is supposed to be 22.6
If I use sin instead of cos I get the right answer. I have this hole handout describing how mgcos(thea) opposes the centripetal force of a mass on a string in circular motion (between pi and 2pi, excluding 3pi/2). What the heck is going on???