Since the wheel start from rest. angular acceleration, θ=1/2αt² 14=1/2α x 8.7² α= 0.3699 rad/s² moment of inertia of loop= mr²= 4.1x0.37=1.517 kgm² torque=T= lα T= 0.5611Nm= 0.56Nm to significant figure Disc moment of inertia of disc= 1/2mr² Required torque value= 0.28Nm So, I= 1/2X 4.1X 0.37²= 0.280 Kgm² T= Iα = 0.280 X 0.3699= 0.10 to two significant figure