QuestionDetails: A rope of negligible mass is wrapped around a 225-kg solid cyli

Question

QuestionDetails: A rope of negligible mass is wrapped around a 225-kg solid cylinderof radius 0.400 m. The cylinder is suspended several meters off theground with its axis oriented horizontally, and turns on that axiswithout friction. (a) If a 75.0-kg man takes hold of the free endof the rope and falls under the force of gravity, what is hisacceleration?
1 m/s2

(b) What is the angular acceleration of the cylinder?
2 rad/s2

(c) If the mass of the rope were not neglected, what would happento the angular acceleration of the cylinder as the manfalls? QuestionDetails:

Explanation / Answer

F = mg - /R The torque is a product of moment of inertia I andangular acceleration = I F = mg - I/R since =a/R we have F = mg - Ia/R also F=ma we have ma= mg - Ia/R2 a= mg /( m + I/ R2 ) I= 0.5mcR2 ( moment of inertia fora cylinder of mass mc and radius R ) a= mg /( m + (0.5mcR2 )/ R2 ) a= mg /( m + (0.5mc ) ) a= 75.0 x 9.81 /( 75.0 + 0.5 x 225) a= 3.92 m/s2 - b) = a /R = 3.92 /0.400 = = 9.81 rad /s2 - - c) As the rope would unroll the load woud increase while themoment of inertia of the cyliner would decrease m'= m + mrope and mc'= mc -mrope Same equation/relationships would hold

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