A small, solid sphere of mass 0.8 kg and radius 35 cm rolls without slipping alo

Question

A small, solid sphere of mass 0.8 kg and radius 35 cm rolls without slipping along the track consisting of slope and loop-the-loop with radius 3.25 m at the end of the slope. It starts from rest near the top of the track at a height h, where h is large compared to 35 cm. What is the minimum value of h (in terms of the radius of the loop R) such that the sphere completes the loop? The acceleration due to gravity is 9.8 m/s2. The moment of inertia for a solid sphere is 2/5 mr2. Answer in units of m What axe the force component in the horizontal direction on the sphere at the point P. which has coordinates (-R, 0) if we take the center of the loop as origin, and if h = 3 R ? Answer in units of N

Explanation / Answer

For completing the loop, balancing forces at the top,

mv^2/(R-r)=mg

v=sqrt(g(R-r))=sqrt(9.81*3.25-0.35)=5.62m/s

w=v/r=5.62/0.35=16.06rad/s

I=2/5mr^2=0.4*0.8*0.35^2=0.0392kgm^2

Conserving energy between initial point and top of the loop,

mgh=mg(R-r)+0.5mv^2+0.5Iw^2

0.8*9.81*h=0.8*9.81*(3.25-0.35)+0.5*0.8*5.62^2+0.5*0.0392*16.06^2

Solving for h,

h=5.154m

b)Balancing energy between starting point and P,

0.8*9.81*3.25*3=0.8*9.81*3.25+0.5*0.8*v^2+0.5*0.0392*v^2/0.35^2

v=9.54m/s

Fx=mv^2/(R-r)=0.8*9.54^2/(3.25-0.35)=25.11N

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