Relation between Torque and Angular Acceleration

Suppose that the angular speed of a body changes from ω to (ω+Δω) in a very small time interval Δt. If the angular displacement of the body during this time interval is Δθ = (ωΔt) then

From  Work Energy Theorem,

Work done = Change in K.E

τ Δθ = ΔK.E. (of rotation) ; where τ = average torque during this time interval.

$\displaystyle \tau \Delta\theta = \frac{1}{2}I [(\omega + \Delta \omega)^2 – \omega^2] $

$\displaystyle \tau \Delta\theta = \frac{1}{2}I(2\omega \Delta \omega) $

⇒ τ ω Δt = I ω Δω  (as Δθ = ωΔt)

⇒ $\large \tau = I \frac{\Delta \omega}{\Delta t} $

⇒ $\large \tau = I \alpha $

Illustration : A uniform disc of radius R and mass M is free to rotate about a fixed horizontal axis perpendicular to its plane and passing through its centre. A string is wrapped over its rim and a block of mass m is attached to the free end of the string. The block is released from rest. If string does not slip on the rim then find the acceleration of the block. Neglect the mass of the string.

Also Read :

→ Motion of Rigid body & Rotational Kinematics
→ K.E of Rotation & Moment of Inertia
→ Moment of inertia of a thin ring & Radius of Gyration
→ Parallel & Perpendicular axes Theorem
→ Torque & Pseudo Torque
→ Angular momentum & its Conservation
→ Combined Rotation & Translation

Next Page → 

 ←Back Page

Leave a Reply