A wheel with rotational inertia 0.04 kg•m2 and radius 0.02 m is turning at the rate of 10 revolutions per second when a frictional torque is applied to stop it. How much work is done by the torque in stopping the wheel?

-78.96 J

Explanation:

The workdone by the torque in stopping the wheel = rotational kinetic energy change of wheel.

So W = 1/2I(ω₁² - ω₀²) where I = rotational inertia of wheel = 0.04 kgm², r = radius of wheel = 0.02 m, ω₀ = initial rotational speed = 10 rev/s × 2π = 62.83 rad/s, ω₁ = final rotational speed = 0 rad/s (since the wheel stops)

W = 1/2I(ω₁² - ω₀²) = 1/2 0.04 kgm² (0² - (62.83 rad/s)²) = -78.96 J  

t.                 w               e                 n                   t                 y                 c             h               a           r           a               c                 t             e             r             s        

the components of a vector are the parts of the vectors that are to each other.

b perpendicular