A fly wheel of `M.I. 6xx10^(-2) kgm^(2)` is rotating with an angular velocity of `20 rad s^(-1)`. The torque required to bring it to rest in `4s` is

A flywheel of M.I. 6xx10^(-2)kgm^(2) is rotating with an angular velocity of 20rads^(-1) . The torque required to bring it to rest in 4s is

A wheel of moment of inertia 2.0 xx 10^3 kgm^2 is rotating at uniform angular speed of 4 rads^-1 . What is the torque required to stop it in one second.

A wheel of moment of inertia 2 xx 10^3 kg m^2 is rotating at uniform angular speed of 4 rad/s,.then . the torque required to stop it in one second is,

A flywheel of moment of inertia 3 xx 10^(2) kg m^(2) is rotating with uniform angular speed of 4.6 rad s^(-1) . If a torque of 6.9 xx 10^(2)Nm retards the wheel, then the time in which the wheel comes to rest is

A whel of M.I. = 1 kg-m^(2) is rotating at an angular speed of 40 rad/s. Due to friction on the axis, the wheel comes to rest in 10 minute. What is the angular momentum of the wheel in kg m^(2)//s two minutes before it comes to rest ?