Answer:
τ = 0.26 N.m
Step-by-step explanation:
First we find the moment of inertia of the wheel, by using the following formula:
I= mr²
where,
I = Moment of Inertia = ?
m = mass of wheel = 2.3 kg
r = radius of wheel = 50 cm/2 = 25 cm = 0.25 m
Therefore,
I = (2.3 kg)(0.25 m)²
I = 0.115 kg.m²
Now, we find the angular acceleration of the wheel:
α = (ωf - ωi)/t
where,
α = angular acceleration = ?
ωf = final angular velocity = (120 rpm)(2π rad/1 rev)(1 m/60 s) = 12.56 rad/s
ωi = Initial Angular Velocity = 0 rad/s
t = time = 5.5 s
Therefore,
α = (12.56 rad/s - 0 rad/s)/(5.5 s)
α = 2.28 rad/s²
Now, the torque is given as:
Torque = τ = Iα
τ = (0.115 kg.m²)(2.28 rad/s²)
τ = 0.26 N.m