A 50 kg child sits on the edge of a merry-go-round at a radius of 10m. The ride

Question

A 50 kg child sits on the edge of a merry-go-round at a radius of 10m. The ride has a rotational inertia of 1000 kgm2. The child pushes with a constant force of 100N at the edge and parallel to the edge. A frictional force of 20Nm acts at the axle of the merry-go-round.
a. What is the net torque acting on the merry-go-round about its axle?
b. What is the rotational acceleration of the merry-go-round?
c. At this rate, what will the rotational velocity of the merry-go-round be after 3 sec if it starts from rest?

Explanation / Answer

Given Data: Mass of the child, m = 50 kg Radius of the merry-go-round, r = 10 m Rotational inertiaof the ride, I = 1000 kg-m2 Force , F = 100 N Frictional force, f = 20 Nm Solution: (a) Torque , = F * r - f                       = 100 N (10 m) - 20 Nm                       = 980 Nm ---------------------------------------------------------- (b) Rotational acceleration is calculated as,                      = I            980 Nm = (1000 kg-m2)                         = 0.98 rad/s2 ------------------------------------------------------------- (c) Initial velocity, i =0 rad /s      Time interval, t = 3.0s      The rotational velocity after the given time interval is,           = i + t               = 0 + (0.98 rad/s2) (3.0 s)               = 2.94 rad/s ------------------------------------------------------------- (c) Initial velocity, i =0 rad /s      Time interval, t = 3.0s      The rotational velocity after the given time interval is,           = i + t               = 0 + (0.98 rad/s2) (3.0 s)               = 2.94 rad/s

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