Question

The flywheel of an engine has moment of inertia 1.9 kg m2 about its rotation axis. What constant torque is required to bring it up to an angular speed of 537 rev/min in 6.78 s, starting from rest, in N-m

Answer 1

τ = 15.76 N-m

Step-by-step explanation:

• For rigid bodies, the net external torque acting on a rotating body, is equal to the product of the moment of inertia about its rotation axis, times the angular acceleration of the body, as follows:

`\tau = I * \alpha (1)`

• Since I is a given of the question, we need to find out the angular acceleration.
• By definition, the angular acceleration is the rate of change of the angular velocity with respect to time:
• α = Δω/Δt = (ωf-ω₀) / (tfi-t₀)
• Since ω₀ = 0, and choosing t₀ =0, as tfi is a given, we need just to find out ωf.
• We have this value in rev/min, so we need first to convert it to rad/sec, as follows:

`537 rev/min * (1min/60 sec)*(2*\pi *rad/rev) = 17.9 * \pi rad/sec (2)`

• Replacing in the definition of α, we have:
• α = 17.9*π rad/sec / 6.78 s = 8.29 rad/sec²
• Replacing I and α in (1), we finally get:

τ = 15.76 N-m.