Question

a wheel rotation at an angular speed of 20 rad s is brought to rest by a constant torque in 4s if the M.I os 0.2 kg m square the work done in first 2sec is
can any one answer this?​

Answer 1

Explanation:

◆ Answer -

Work done = 30 J

● Explanation -

# Given -

w0 = 20 rad/s

w = 0 rad/s

t = 4 s

I = 0.2 kgm^2

# Solution -

Angular acceleration is -

α = (w-w0)/t

α = (0-20)/4

α = -5 rad/s^2

Angular displacement is given by -

θ = w0t + 1/2 αt^2

θ = 20×2 + 1/2 × (-5)×2^2

θ = 30 rad

Torque is calculated as -

τ = Iα

τ = 0.2 × -5

τ = 1 Nm

Work done is given by -

W = τθ

W = 1 × 30

W = 30 J