Question

A child exerts a tangential 42.2 N force on the rim of a disk shaped merry go round with a radius of 2.40 m. If the merry-go-round starts at rest and acquires an angular speed of 0.0860 rev/s in 3.50s, what is its mass?

Answer 1

w1= 0

w2 = 0.086*2*pi = 0.54 rad/s

alfa = (w2-w1)/t = 0.54/3.5 = 0.154 rad/s^2

Tnet = I*alfa

r*F = 0.5*m*r^2*alfa

==> m = 2*F/(r*alfa)

= 2*42.2/(2.4*0.154)

= 229 kg

Answer 2

Angular acceleration is, t 0.0860 (2)rad/s - 0 3.5 s 27(0.0860 rad/s) 3.5 s =0.154 rad/s Expression for torque is, Expression for moment of inertia is, -mr2 I Relation between torque and angular acceleration is rF (2.4 m)42.2 Nm(2.4 m) (0.154 rad/s 2(2.4 m)42.2 N (2.4 m)(0.154 rad's?) m= =228.35 kg