In the previous problem, how much energy is dissipated in the form of frictiorml
ID: 1508655 • Letter: I
Question
In the previous problem, how much energy is dissipated in the form of frictiorml heat losses to slow the centrifuge to a complete stop? A sphere of radius 20 cm and mass 2 kg begins rolling down a 16 m high incline with an angle of 16degree after starting from rest. a) Calculate the potential energy of the sphere at the top of the incline. b) Calculate the translational and rotational kinetic energy of the sphere at the bottom? c) What is the ratio of the translational kinetic energy to tin; rotational kinetic energy? d) What quantities in the problem does this ratio depend on?Explanation / Answer
10)
given,
r = 20 cm = 0.2 m
m = 2 kg
h = 16 m
a) Initial potentail energy = m*g*h
= 2*9.8*16
= 313.6 J
b) let v is the speed of the sphere at the bottom
apply conservation of energy
0.5*m*v^2 + 0.5*I*w^2 = m*g*h
0.5*m*v^2 + 0.5*(2/5)*m*r^2*w^2 = m*g*h
0.5*m*v^2 + 0.2*m*r^2*w^2 = m*g*h
0.5*m*v^2 + 0.2*m*v^2 = m*g*h
0.7*m*v^2 = m*g*h
v = sqrt(h*g/0.7)
= sqrt(16*9.8/0.7)
= 14.97 m/s
translational kinetic energy = 0.5*m*v^2
= 0.5*2*14.97^2
= 224 J
Rotational kinetic energy = 0.2*m*v^2
= 0.2*2*14.97^2
= 89.6 J
c) KE_translational/KE_rotational = 0.5*m*v^2/(0.2*m*v^2)
= 5/2
= 2.5
d) It does not depend on any thing.
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