Question

A solid cast iron disk, 1 m in diameter and 0.2 m thick, is used as a flywheel. It is rotating at 350 rpm. It is brought to rest in 1.5 s by means of a brake. If the mass density of cast iron is 7200 kg/m³, the torque exerted by the brake will be nearly
1. 3.5 kNm
2. 4.5 kNm
3. 5.3 kNm
4. 6.3 kNm

Answer 1

Correct Answer - Option 1 : 3.5 kNm

Concept:

Torque exerted is given by:

T = Iα

where I = moment of inertia

α = angular acceleration

Calculation:

Given:

diameter of disc (d) = 1 m, thickness (t) = 0.2 m, mass density (ρ) = 7200 kg/m³,

Angular speed (N) = 350 rpm, ω = 0, t = 1.5 s

Now,

Mass of the disc (m) = density × volume of disc

\(m = \rho \times \frac{\pi }{4}{d^2} \times t = 7200 \times \frac{\pi }{4} \times {1^2} \times 0.2 = 1130.973\;kg\)

Moment of inertia of disc (I)

\(I = \frac{{m{r^2}}}{2} = \frac{{1130.973 \times {{\left( {0.5} \right)}^2}}}{2} = 141.371\;kg.{m^2}\)

Initial angular velocity (ω_o) is given by:

\({\omega _o} = \frac{{2 \times \pi \times N}}{{60}} = \frac{{2 \times \pi \times 350}}{{60}} = 36.65\;rad/s\)

Now, from newton’s equation of motion

ω = ω_o + αt

0 = 36.65 + α(1.5)

\( \Rightarrow \alpha = \; - \frac{{36.65}}{{1.5}} = 24.43\;rad/{s^2}\)

Torque required is:

T = Iα

T = 141.371 × 24.43 = 3454.34 Nm = 3.45 kNm ≃ 3.5 kNm