A grinding wheel of mass 5.0 kg and diameter 0.20 m is rotating with an angular speed of 100 rad s-1. Calculate its kinetic energy. Through what distance would it have to be dropped in free fall to acquire this kinetic energy? ( Take g= 10.0 m s-2)
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(1/2)Iw^2
= (1/2)(mr^2/2)(w^2) ....... w = omega
= (5×0.2×0.2×100×100)/4
= 500 J
now 1/2 mv^2 = 500
1/2×5×v^2 = 500
v = 10√2m/s
now by Newton's third law
v^2 - u^2 = 2as
200 - 0 = 2 × 10 × s
s = 10m
so this should be dropped upto 10m to get that much of kinetic energy.
plz mark that as BRAINLIEST
and hope it would be a BRAINLIEST answer!!
:)
= (1/2)(mr^2/2)(w^2) ....... w = omega
= (5×0.2×0.2×100×100)/4
= 500 J
now 1/2 mv^2 = 500
1/2×5×v^2 = 500
v = 10√2m/s
now by Newton's third law
v^2 - u^2 = 2as
200 - 0 = 2 × 10 × s
s = 10m
so this should be dropped upto 10m to get that much of kinetic energy.
plz mark that as BRAINLIEST
and hope it would be a BRAINLIEST answer!!
:)
naaznagma9634:
But answer is 1.25J and 2.5 m
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