A wheel of moment of inertia 2.5 kg m^2 has an initial angular velocity of 40 rad s^(-1). A constant torque of 10 N m acts on the wheel. The time during which the wheel is accelerated to 60 rad s^(-1) is

A wheel of moment of inertia 2.5 kg m^2 has an initial angular velocit

1.	A wheel of moment of inertia 2.5 kg m^2 has an initial angular velocity of 40 rad s^(-1). A constant torque of 10 N m acts on the wheel. The time during which the wheel is accelerated to 60 rad s^(-1) is
Answer» 4.5 s 6 s 5 s 2.5s Solution :Here , I = `2.5 kg m^(2) , omega_(0) = 40 rad s^(-1)` `tau = 10 N m , omega = 60 rad s^(-1)` As `tau = I alpha or alpha = (tau)/(I) = (10)/(2.5) = 4 rad s^(-2)` USING `omega = omega_(0) + alpha t, t = (omega - omega_(0))/(alpha)` Substituting the values , we GET `t = (60 - 40)/(4 ) = 5`s