Question

A solid 0.595-kg ball rolls without slipping down a track toward a loop-the-loop of radius R = 0.7350 m. What minimum translational speed vmin must the ball have when it is a height H = 1.091 m above the bottom of the loop, in order to complete the loop without falling off the track?

Answer #1

**let v be the speed at the top point,, then from gravity
relation we have**

**v^2/r = g**

**v = sqrt(g*r)**

**velocity v = sqrt(9.8*0.7350) = 2.68 m/s**

**let now u is the speed at height H.**

**Work done by gravity = change in kinetic
enrgy**

**W = 0.5*m*u^2 + 0.5*I*w^2 - (0.5*m*v^2 +
0.5*I*w^2)**

**W = (0.5*m*u^2 + 0.5*(2/5)*m*r^2*w^2 - (0.5*m*v^2 +
0.5*(2/5)*m*r^2w^2))**

**W = 0.7*m*u^2 - 0.7*m*v^2 = 0**

**m*g*(2*R - H) = 0.7*m*(u^2-v^2)**

**g*(2*R - H) = 0.7*(u^2-v^2)**

**u^2 - v^2 = g*(2*R - H)/0.7**

**u = sqrt(v^2 + g*(2*R-H)/0.7)**

**intial speed u = sqrt( 2.68^2 + 9.8*(2*0.7350 -
1.091)/0.7)**

**initlal speed u = 3.533 m/s**

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