A wheel has a moment of inertia 5×10^-3 kg-m^2 and is making 20 revolution per second. the torque needed to stop it in 10 second is ......10^-2
Concepts: Moment of Inertia: Moment of Inertia is a property of an object related to mass distribution. It plays the same role in rotational dynamics that mass plays in linear dynamics. The moment of inertia r of a body with mass m at a distance from the axis center is given by Angular Acceleration (α): The rate of change of angular velocity is called angular acceleration. Torque: The cross product of the force from the point of action and the force path is called the torque. Torque is also defined as the product of moment of inertia and angular acceleration, and force is defined as the product of mass and linear acceleration. τ = Iα -- (1) The first equation of motion for rotary motion with constant torque is given as follows. angular acceleration, t is time Calculation: Given Moment of inertia I = 5 x 10-3 kg m2 torque τ = Angular acceleration α time given t = 10 sec Initial angular speed ωi = 20 rev sec -1 = 40 π sec -1 (1 revolution = 2 π ) Final angular speed ω f = 0 (it will be at rest) 0 = 20 + α (10) ⇒ α = - 2 rev sec -2 Now using eqution (1) Torque τ = 5 x 10-3 kg m2 × 2 rev sec -2 = 5 x 10-3 kg m2 × 40 π sec -2 ⇒ τ = 2 π x 10-2 Nm So, the correct option is 2 π x 10-2 Nm
