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Consider a solid sphere of mass m and radius r being released from a height h...

Consider a solid sphere of mass m and radius r being released from a height h (i.e., its center of mass is initially a height h above the ground). It rolls without slipping and passes through a vertical loop of radius R.

Consider a solid sphere of mass m and radius r bei

a. Use energy conservation to determine the tangential and angular velocities of the sphere when it reaches the top of the loop.

b. Draw a force diagram for the sphere at the top of the loop and write down Newton´s second law in the radial direction.

c. What is the minimum height from which the sphere must be released so that it doesn’t fall off the track at the top of the loop?

d. Neglecting its moment of inertia, would the minimum height be more or less than the one you calculated? Explain.

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Answer #1

a) mgh=mg2R+0.5mv^2+0.5 (Moment Of Inertia) Angular Velocity^2

mgh=2Rmg+0.5mv^2+0.5(2/5mr^2) (v/r)^2

mgh=2Rmg+0.5mv^2+0.2mv^2

mgh=2Rmg+0.7mv^2

gh-2Rg/0.7=v^2

v= root(gh-2Rg)

Angular velocity=v/r

b) On the top most pont Normal reaction and weight both will be point down. So N+mg=mv^2/R

c) To complete the circle minimum speed required is root(5gr) at the bottom so mgh=0.5m5gr (considering centre of mass only)

h=2.5g

d) If we neglect the MOI that rotaion the minimum heoght will be greater as we need to give more energy so must fall from greater height

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