System of Particles & Rotational Motion Test - 7

Hence the torque is zero .Because the line of action of the tension passes through the centre .So lever arm of tension is zero So the torque acting is also zero .

The power delivered by the torque τ exerted on rotating body is given by
P=τω or τ=P/ω
Here P=360KW=360000 Watt
ω=30 x 2π rad/sec,
ω=60π rad/sec
now,
τ=360000 /60×3.14Nm
τ= 1910.8 Nm

Torque of a bar is proportional to force applied tangential to bar, and hence torque can be treated as direct proportional with angular displacement.

Torque is an axial vector. Axial vectors are those vectors which represent rotational effect and act along the axis of rotation. E.g. angular velocity, torque, angular momentum etc. are axial vectors.

Torque = moment of inertia × angular acceleration
3.5 × 25 =87.5 Nm

For a system to be in equilibrium torques on the system must be equal to zero
This is true only if the torques are taken about the centre of mass as the whole mass of the object seems to be situated at centre of mass.

Force (in linear) = torque (in rotational)
Mass (in linear motion) = moment of inertia ( l ) (in rotational motion)
Acceleration(a) (in linear) = angular acceleration (α) (in rotational)
F = ma (in linear) and torque = Iα (in rotational)

Rotational analogue of force in linear motion is torque. Torque is the turning effect of force.

τ = dI/dt
If τ = 0 then L is conserved.