A coin is rolling in a plane surface .what fraction of its total kinetic energy is rotational
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Answered by
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In case of rolling without slipping the angular velocity (ω) and the velocity of centre of mass (v) are related as: v = Rω.
Moment of inertia of the coin (disc), I = 1/2mr2
Total kinetic energy, K = Translational kinetic energy + Rotational kinetic energy
or, K = 1/2 mv2 + 1/2 Iω2
We can write rotational kinetic energy as:
1/2 Iω2 = (1/2 )×(1/2 mr2) (ω2) = 1/4 m (rω)2 = 1/4 mv2
Thus,
K total = Kt + Kr
2Kr = Kt
Ktotal = 3 Kr
Fraction of rotational kinetic energy
Kr/3 Kr = 1/3
Moment of inertia of the coin (disc), I = 1/2mr2
Total kinetic energy, K = Translational kinetic energy + Rotational kinetic energy
or, K = 1/2 mv2 + 1/2 Iω2
We can write rotational kinetic energy as:
1/2 Iω2 = (1/2 )×(1/2 mr2) (ω2) = 1/4 m (rω)2 = 1/4 mv2
Thus,
K total = Kt + Kr
2Kr = Kt
Ktotal = 3 Kr
Fraction of rotational kinetic energy
Kr/3 Kr = 1/3
riya558:
thanks ann
Answered by
0
Explanation:
KE rotational=1/4mv^2
total KE= 3/4mv^2
fraction of total kE is rotational
KE rotational ÷total KE = 1/3
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