Hi, I am trying to solve a problem but I can't figure how to continue.. There's an object moving in a circular path with radius 0.5m in an horizontal surface. The rope will break if the tension exceeds 16N. What is the maximum kinect energy? So r = 0.5m T = 16N K=[itex]\frac{1}{2}[/itex]m[itex]v^{2}[/itex] How can I find m and v? I tried with T = mg <=> 16 = m(9.8) <=> m = 1.63kg but I don't know if it is correct. But besides that I can't figure how to find v
Show what you did with it. You mean the value of the mass you calculated in post #1? No, that's not correct. For some reason, you set the tension equal to the weight of the mass. Why would you do that? Hint: You don't need the mass to answer the question. It asks for kinetic energy, not mass.
a = [itex]\frac{v^{2}}{r}[/itex] and then F = m*a <=> F = m*[itex]\frac{v^{2}}{r}[/itex] <=> 16 = 1.63 * [itex]\frac{v^{2}}{0.5}[/itex] But since 1.63 isn't right it will give wrong results I did that because I couldn't figure any other way to find the mass.. trial and error But K=[itex]\frac{1}{2}[/itex]m[itex]v^{2}[/itex] so I need the mass
Perfect so far. Yes, your value for mass is wrong. (Hint: There's not enough information to determine the mass. But you don't need it!) You do not need the mass, you need the kinetic energy. So you just need to solve for 1/2mv^{2}. Go back and stare at the first equation in this post.
Ok, finally I figured that :P So, i did this: F = ma <=> a = [itex]\frac{F}{m}[/itex] a = [itex]\frac{v^{2}}{r}[/itex] <=> [itex]v^{2}=a*r[/itex] <=> [itex]v^{2}=\frac{F}{m}*r[/itex] K = [itex]\frac{1}{2}*m*\frac{F*r}{m} = \frac{1}{2}*F*r[/itex] Thanks!
Cool. A short cut would be to recognize mv^{2} when you see it: [tex]F = \frac{m v^2}{r}[/tex] Thus: [tex]m v^2 = F r[/tex] [tex]\frac{1}{2}m v^2 = \frac{1}{2}F r[/tex]