A mass m = 0.5 kg is hung from a pulley with moment of inertia I = 0.20kg⋅m2 and radius R = 10cm. There is friction in the pulley. The mass is dropped from rest, and after 2s it traveled 0.4m. What is the torque due to friction?
s=at²/2 => a=2s/t²=2•0.4/2²=0.2 m/s²
The equation of the load motion
ma=mg-T => T=m(g-a)
The equation of the pulley motion
Iε=M-M(fr)
M(fr) = M-Iε = TR-Ia/R=
=m(g-a)R – Ia/R = ...
To calculate the torque due to friction, we need to determine the angular acceleration of the pulley and then use the equation τ = Iα, where τ represents torque, I represents moment of inertia, and α represents angular acceleration.
First, let's find the angular acceleration α of the pulley. We can use the kinematic equation:
θ = ω₀t + (1/2)αt²
Where:
θ = angular displacement of the pulley
ω₀ = initial angular velocity of the pulley (which is 0 since it starts from rest)
t = time (2s in this case)
α = angular acceleration (what we want to find)
Rearranging the equation, we have:
0.4m = (1/2)α(2s)²
Now, let's solve for α:
0.4m = α(0.5s²)
α = (0.4m) / (0.5s²)
α = 0.8 m/s²
Now that we have the angular acceleration, we can calculate the torque due to friction:
τ = Iα
τ = (0.20 kg⋅m²)(0.8 m/s²)
τ = 0.16 N⋅m
Therefore, the torque due to friction in the pulley is 0.16 N⋅m.