When a heavy object is suspended from a spring and the spring is rotated, the dynamics involve both centripetal and gravitational forces. The centripetal force, calculated as mv²/r, acts horizontally, while the vertical component of the spring force counteracts the weight of the object. To determine the resultant spring force, one must apply Hooke's Law to find the extension of the spring. This analysis combines principles of circular motion and elasticity in physics.
PREREQUISITESPhysics students, mechanical engineers, and anyone interested in the dynamics of rotating systems and spring mechanics.
It's the same as rotating the object hung on a rope: if swung in a horizontal circle at constant speed, v, the centripetal force is mv^2/r, which is the horizontal component of the spring force, and the verticlal force component of the spring force is the weight of the object. Once you calculate the resultant spring force, use Hooke's law to determimne its extension.ambroa said:hi,
what happens, when I hand on a spring a heavy object and then we'll rotate the spring? I've tried to write down some maths for it but I don't know anyway...