Asked by Justin
Suppose you exert a force of 180 N tangential to a 0.280-m-radius 75.0-kg grindstone (a solid disk). (a)What torque is exerted? (b) What is the angular acceleration assuming negligible opposing friction? (c) What is the angular acceleration if there is an opposing frictional force of 20.0 N exerted 1.50 cm from the axis?
Answers
Answered by
Elena
τ=FR= 180•0.28 = 50.4 N•m
I=mR²/2 = 75•0.28²/2=2.94 N•m²
τ=Iε =>
ε= τ/I = 50.4/2.94 = 17.14 rad/s²
τ₁=FR-F(fr)r =180•0.28 - 20•0.015 = 50.4 – 0.3 = 50.1 N•m
ε₁= τ₁/I = 50.1/2.94 = 17.04 rad/s²
I=mR²/2 = 75•0.28²/2=2.94 N•m²
τ=Iε =>
ε= τ/I = 50.4/2.94 = 17.14 rad/s²
τ₁=FR-F(fr)r =180•0.28 - 20•0.015 = 50.4 – 0.3 = 50.1 N•m
ε₁= τ₁/I = 50.1/2.94 = 17.04 rad/s²
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