The minimum torque required for rotating a flywheel of moment of inertia 2.1 kgm² from rest to a speed of 1200 rpm in 6 s? (π = 22/7)

CONCEPT:

• The torque of a body undergoing circular motion with angular acceleration α is related to its moment of inertia as follows:
• τ = Iα 
• Where, τ = torque, I = moment of inertia, α = angular acceleration

Equation of motion:

• The mathematical equations used to find the final velocity, displacements, time, etc of a moving object without considering force acting on it are called equations of motion.
• These equations are only valid when the acceleration of the body is constant and they move on a straight line.

There are three equations of motion:

• For a constant acceleration the equations of motion for translational and rotational motion are as follows:

Where, v, ω = final velocity, u, ω0 = initial velocity, s, θ = distance traveled by the body under motion, a, α = acceleration of body under motion, and t = time taken by the body under motion.​

​SOLUTION:

Given: Moment of Inertia, I = 2.1 kg m2, N = 1200 rpm, t = 6 sec

As the body is in rest, initial angular velocity of the body, ω0 = 0 rad/sec

As the body is in motion, the angular velocity of the body, \(\omega =\frac{2 \pi N}{60}=\frac{2 \times \pi \times 1200}{60}=40\pi ~rad/sec\) 

Using the first equation of Rotational Motion:

ω = ω0 + αt

\(\Rightarrow 40\pi=0+α \times 6 \Rightarrow α = \frac{20\ \pi}{3}~ rad/sec^2\)

Now, using \(τ=I α\):

\(τ=2.1\times\frac{20 \ \pi}{3}=44~ Nm\)