A mass of .375kg hangs from a string that is wrappedaround disk with a radius of .26m. When the mass isreleased, the mass accelerates downward at 5.67m/s^2 and thepulley rotates about its axis as the string unwinds. What is themoment of inertia. What I did was use the equation = I. So my torque was the mass* acceleration and my angularacceleration was acceleration divided by the radius. I divided my by but that was incorrect. Where am I goingwrong? A mass of .375kg hangs from a string that is wrappedaround disk with a radius of .26m. When the mass isreleased, the mass accelerates downward at 5.67m/s^2 and thepulley rotates about its axis as the string unwinds. What is themoment of inertia. What I did was use the equation = I. So my torque was the mass* acceleration and my angularacceleration was acceleration divided by the radius. I divided my by but that was incorrect. Where am I goingwrong? What I did was use the equation = I. So my torque was the mass* acceleration and my angularacceleration was acceleration divided by the radius. I divided my by but that was incorrect. Where am I goingwrong?
Explanation / Answer
Given that the mass is M = 0.375kg radiusis R = 0.26m accelarationis a = 5.67 m/s2---------------------------------------------------------------------------------- Apply Newtons laws forthe mass and the pully then Mg - T =Ma (for mass M) -----------(1) T*R=I (for pully) T*R = Ia/R T = Ia/R2 ---------(2) From the equations (1) and (2) we get Mg - Ia/R2 = Ma where I is the moment of inertia of the solidcylinder I = (Mg- Ma)R2/a =--------- kg.m2 Apply Newtons laws forthe mass and the pully then Mg - T =Ma (for mass M) -----------(1) T*R=I (for pully) T*R = Ia/R T = Ia/R2 ---------(2) From the equations (1) and (2) we get Mg - Ia/R2 = Ma where I is the moment of inertia of the solidcylinder I = (Mg- Ma)R2/a =--------- kg.m2