Question

The moment of inertia of a wheel is 1000 kg m2
. Its rotation is uniformly
accelerated. At some instant of time, its angular speed is 10 rad s–1. After the
wheel has rotated through an angle of 100 radians, the angular velocity of
the wheel becomes 100 rad s–1. Calculate the torque applied to the wheel and
the change in its kinetic energy.

Answer 1

Dear Student,

◆ Answer -

τ = 49500 J

∆KE = 4950000 J

● Explanation -

# Given -

I = 1000 kgm^2

ω1 = 10 rad/s

ω2 = 100 rad/s

θ = 100 rad

# Solution -

Using 2nd law of kinematics -

ω2² = ω1² + 2αθ

100² = 10² + 2 × α × 100

10000 = 100 + 200α

100 = 1 + 2α

α = 99/2

α = 49.5 rad/s²

Torque applied to the wheel is given by -

τ = Iα

τ = 1000 × 49.5

τ = 49500 J

Change in kinetic energy of the wheel is -

∆KE = I/2 (ω2² - ω1²)

∆KE = 1000/2 (100² - 10²)

∆KE = 500 (10000 - 100)

∆KE = 500 × 9900

∆KE = 4950000 J

Hope this helps you...