Question

A wheel having a moment of inertia of 1.0 kg m2

, is rotating about an axis through its

centre and perpendicular to its plane, at a speed of 30 rad s−1

. Due to friction on its axis,

it comes to rest in 10 minutes. Calculate i) the total work done by friction, ii) average

torque exerted by the force of friction and iii) the angular momentum of the wheel 2.0

minutes before it stops rotating.

Answer 1

Welcome student,

● Answer -
i) W = 450 J
ii) τ = 0.05 Nm
iii) L = 6 kgm/s^2

● Explanation -
# Given -
I = 1 kgm^2
ω1 = 30 rad/s
ω2 = 0 rad/s
t = 10 min = 600 s

# Solution -
Work done by fiction is given by -
W = 1/2 I(ω1^2 - ω2^2)
W = 1/2 × 1 × (30^2-0)
W = 450 J

Torque exerted is given by formula -
τ = Iα
τ = 1 × (30-0)/600
τ = 0.05 Nm

Angular speed of wheel 2 minutes before stopping,
i.e. t = 10-2 = 8 min = 480 s
ω2 = ω1 + αt
ω2 = 30 + (-0.05×480)
ω2 = 6 rad/s

Angular momentum 2 min before stopping -
L = Iω2
L = 1 × 6
L = 6 kgm/s^2

Hope this helped you...