A 0.75-kg ball is attached to a 1.0-m rope and whirled in a vertical circle. The rope will break when the tension exceeds 450 N.What is the maxuimum speed the ball can have at the bottom of the circle without breaking the rope?

Question

anonymous · Answer

$$v=\sqrt{T*r/m}$$
=(450*/0.75)^1/2=24.94

anonymous · Answer

mv^2/r + mg =f
0.75*v^2/i + 0.75*10 = 450
v^2=450-7.5/0.75
v^2=590
v=24.28 m/sec
 
at lowest postion
mv^2/r+mg=f

at top postion
mv^2/r-mg=f

at horizontal
mv^2/r = f