1 Rotational Equilibrium If the angular velocity of a rigid body is not to change, no net external torque (τ) can act on the object, since τi = dL dt =Idω dt , (61) where L is the angular momentum and I is the moment of inertia. The complete conditions for the equilibrium of a rigid body are as follows: The resultant external force is zero: F=0. The resultant external torque is zero about any origin: τ =0. For statics, the object is at rest and not translating or rotating. When dealing with an object whose mass is distributed, treat all of the mass as if it is concentrated at the centre of mass (CM). The gravity force exerts no torque about the centre of mass point. Example: Abe and Mary carry a uniform log of length 6 cm and weight 150 N. Abe is 1m from one end and Mary is 2 m from the other end. What weight does each person support? Solution: If we imagine a pivot at Abe’s position, and take a counter-clockwise torque as positive, then G×3=150×2 G= 150×2 3=100N Now, F +G=150N F =150N −G=150N −100N =50N
1 Rotational Equilibrium If the angular velocity of a rigid body is not to change, no net external torque (τ) can act on the object, since τi = dL dt =Idω dt , (61) where L is the angular momentum and I is the moment of inertia. The complete conditions for the equilibrium of a rigid body are as follows: The resultant external force is zero: F=0. The resultant external torque is zero about any origin: τ =0. For statics, the object is at rest and not translating or rotating. When dealing with an object whose mass is distributed, treat all of the mass as if it is concentrated at the centre of mass (CM). The gravity force exerts no torque about the centre of mass point. Example: Abe and Mary carry a uniform log of length 6 cm and weight 150 N. Abe is 1m from one end and Mary is 2 m from the other end. What weight does each person support? Solution: If we imagine a pivot at Abe’s position, and take a counter-clockwise torque as positive, then G×3=150×2 G= 150×2 3=100N Now, F +G=150N F =150N −G=150N −100N =50N
