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

A fly wheel of moment of inertia 3 xx 10^(2) kg mạis rotating with u

1.	A fly wheel of moment of inertia 3 xx 10^(2) kg mạis rotating with uniform angular speed of 4.6 rad s^(-1). If a torque of 6.9 xx 10^2 N m retards he wheel, then the time in which the wheel comes to rest is
Answer» 1.5 s 2 s 0.5 s 1 s Solution :Here , moment of inertia , `I = 3 xx 10^(2) KG m^(2)` TORQUE `tau = 6.9 xx 10^(2)` N m Initial angular SPEED `omega_(0) = 4.6 rad s^(-1)` Final angular speed , `omega = 0 rad s^(-1)` As `omega = omega_(0) + alpha t` `therefore alpha = (omega - omega_(0))/(t) = (0 - 4.6)/(t) = - (4.6)/(t) rad s^(-2)` Negative sign is for DECELERATION . Torque , `tau = I alpha` `6.9 xx 10^(2) = 3 xx 10^(2) xx (4.6)/(t) or t = (3 xx 10^(2) xx 4.6)/(6.9 xx 10^(2)) = 2 s`