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Explain work done by torque. |
Answer» Solution :Suppose a rigid body rotating about a fixed axis, which is taken as the Z-axis. This axis is perpendicular to plane X.Y..  Let force `vec(F_(1))` acting on a particle of the body at point `P_(1)` and it rotates on a circle of radius `r_(1)` with centre C on the axis `CP_(1)=r_(1)`. In TIME `Deltat`, the point MOVES to the position `P._(1)` from `P_(1)`. The displacement of the particle is `DeltaS_(1)=r_(1)Deltatheta` and it is in the direction tangential at `P_(1)therefore Deltatheta=angleP_(1)OP._(1)` is the angular displacement of the particle. The WORK done by the force `vec(F_(1))` on the particle is `dW_(1)=vec(F_(1)).dvec(S_(1))` `=F_(1)DeltaS_(1)cosphi_(1)` `=F_(1)r_(1)Deltathetacos(90^(@)-alpha_(1))` where `DeltaS_(1)=r_(1)Delta thetaand phi_(1)+alpha_(1)=90^(@)` where `phi_(1)` is the angle between `vec(F_(1))` and the tangent at `P_(1) and alpha_(1)` is the angle between `vec(F_(1))` and the radius vector `vec(OP)=vec(r_(1)), phi` The torque due to `vec(F_(1))` about the origin `vectau=vec(OP_(1))xxvec(F_(1))""....(1)` but ACCORDING to figure (b) `vec(OP)_(1)=vec(OC)+vec(CP)_(1)` Since `vec(OP)` is along the axis, `therefore TAU=OC(F_(1))sin0^(@)=0`, hence `vec(OP)_(1)=vec(CP)_(1)` From eqn. (1), `vec(tau_(1))=vec(CP)_(1)xxvec(F_(1))` `therefore vec(tau_(1))=vec(r_(1))xxvec(F_(1))[because vec(CP_(1))=vec(r_(1))]` `therefore tau_(1)=r_(1)F_(1)sinalpha_(1)` and work `DeltaW_(1)=tau_(1)Deltatheta` If there are more than one forces acting on the body the work done by all of them can be added to give the total work done on the body `DeltaW=(tau_(1)+tau_(2)+tau_(3)+....tau_(n))Deltatheta` `therefore DeltaW=tauDeltatheta""....(2)` where `tau_(1)+tau_(2)+tau_(3)+....tau_(n)=tau` total torque If `Deltathetararr0` total work done, `dW=taud theta` is similar to eqn. `dW=Fds` in linear motion Diving (2) on both in eqn. (2) `(DeltaW)/(Deltat)=tau.(Deltatheta)/(Deltat)` If `Deltatrarr0=(DeltaW)/(Deltat)=(dW)/(dt)and(Deltatheta)/(Deltat)=(d theta)/(dt)` `therefore` Instantaneous power, `P=tau.(d theta)/(dt)` `therefore P=tauomega` This is the instantaneous power. It is similar to equation `P=Fv` in linear motion. |
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