part a centripetal = gravitational at the top v^2/R = g
v^2 /1.2 = 9.81
v^2 = 11.77
v = 3.43
at the bottom your force up is T
your force down is m g
your acceleration up is v^2/R
so
T - m g = m v^2/R
T = m (g + v^2/R)
= 0.225 (9.81 + 3.43^2/R ) { but we already know v^2/R = g :) }
T = 0.225 (2*9.81)
You are playing with a yo-yo with a mass of 225 g. The full length
of the string is 1.2 m.
(a) Calculate the minimum speed at which you can swing the yo-yo
while keeping it on a circular path. (Hint: At the top of the swing FT = 0)
(b) At the speed just determined, what is the tension in the string at
the bottom of the swing. (Hint: F: Fc = FT + Fg)
I've already solved a) and I got the velocity of 3.4 m/s but I'm really confused on how to solve b)? I tried switching out Fc with mv^2/r and Fg with mg making the equation mv^2/r = Ft + mg but I'm not sure what else to do since when I filled it out, I got a different answer than my teacher
2 answers
what happened to the m on the left? how did you get rid of only 1 m and not both?