Torque on a Merry-Go-Round
ID: 1754013 • Letter: T
Question
A day-care worker pushes tangentially on a smallhand-driven merry-go-round and is able to accelerate it from restto a spinning rate of 19.4 rpm in 10.3 s. Assume the merry-go-roundis a disk of radius 2.60 m and has a mass of 890 kg, and twochildren (each with a mass of 25.2 kg) sit opposite each other onthe edge.Calculate the torque required to produce the acceleration,neglecting frictional torque?
What force isrequired? A day-care worker pushes tangentially on a smallhand-driven merry-go-round and is able to accelerate it from restto a spinning rate of 19.4 rpm in 10.3 s. Assume the merry-go-roundis a disk of radius 2.60 m and has a mass of 890 kg, and twochildren (each with a mass of 25.2 kg) sit opposite each other onthe edge.
Calculate the torque required to produce the acceleration,neglecting frictional torque?
What force isrequired?
Explanation / Answer
wo= 0 w= (19.4Revolutions/Minute)(2p radians/revolution)(1 minute/60 sec) = 2.03156rad/s t = 10.3 s The moments of inertia and a uniform cylinder through its centeris: I = 1/2mr2 so I = 1/2(890 kg)(2.6 m)2= 3008.2 kgm2 But this is just the Merry-go-round, we need the totalmoment of inertia. Treating the children as point 25 kgmasses at the edge (r = 2.6 m) their moment of inertia would beeach: I = mr2 = (25.2kg)(2.6 m)2 = 170.35 kgm2 The total moment of inertia is I = 3008.2 kgm2 + 170.35 kgm2 + 170.35kgm2 = 3348.9 kgm2 Now apply: w = wo + at : wo= 0; w = 2.03156 rad/s; t = 10.3 s a= 0.197239rad/s/s And finally the angular equivalent of F = ma: t = Ia t = (3348.9 kgm2)(0.1972391rad/s/s) = 660.534Nm of torque Since force is applied at a 90o angle to theradius, the torque is: t = rF So F = t/r = (660.534 Nm)/(2.6m) = 254.05 N