A solid steel ball starts from rest at a height H, it rolls around two loop to l

Question

A solid steel ball starts from rest at a height H, it rolls around two loop to loops. The first loop has a radius of 0.5 m and the second loop has a radius of 0.25 m What is the minimum velocity that the ball travel at to make it around the larger loop (the one with a radius of 0.5 m) (marked as point A) (show this don , just state it) At what height H must the bail start from to have the velocity that you found in a? (marked as point B) Assuming that the ball is traveling at the minimum velocity that you found in pan A; what is the ball's velocity at the top of the second (0.25 m) loop (marked point C)

Explanation / Answer

A) net force on the ball at A is mg-T = m*v^2/R

if tension T = 0 for the minimum velocity ,then

mg = m*v^2/R

v = sqrt(R*g) = sqrt(0.5*9.81) = 2.21 m/s

B) using law of conservation of energy

Energy at the B = energy at A

(m*g*H) = (0.5*m*v^2) + (m*g*2*R)

m cancels

H = ((0.5*2.21*2.21)+(9.81*2*0.5))/9.81 =1.24 m

C) using law of conservation of energy

(0.5*m*v^2) + (m*g*2R) = (0.5*vc^2*m)+(m*g*2R1)

m cancels

(0.5*2.21*2.21)+(9.81*2*0.5) = (0.5*vc^2) +(9.81*2*0.25) = 3.83 m/s

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