A small sphere, with radius 1.5 cm and mass 6.4 kg, rolls without slipping on th
ID: 2265420 • Letter: A
Question
A small sphere, with radius 1.5 cm and mass 6.4 kg, rolls without slipping on the inside of a large fixed hemispherical bowl with radius 1.24 m and a vertical axis of symmetry. It starts at the top from rest. What is the kinetic energy of the sphere at the bottom? What fraction of its kinetic energy at the bottom (in percent) is associated with rotation about an axis through its center of mass?
Explanation / Answer
Here
Loss in Potential Energy = Gain in Linear + Rotational Kinetic Energy
Therefore
Total Kinetic Energy = mg*(R-r)
= 6.4*9.8*(1.24 - 0.015)
= 76.832 J
Total Kinetic Energy = 0.5*I*w^2 + 0.5*m*v^2
76.832 = 0.5*(2/5)*mv^2 + 0.5*m*v^2
76.832 = 0.7*m*v^2
v = 4.141 m/sec
Rotational Kinetic Energy = 0.5*I*w^2
= 0.5*(2/5)*mr^2*w^2
= 0.5*(2/5)*m*v^2
= 0.5*(2/5)*6.4*4.141^2
= 21.95 J
Therefore
required Fraction = 21.95/76.832
= 0.28569 = 28.569 %
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.