Вопрос
d)A fixed solid disc pulley is acted upon by a torque of 12 Nm. If the mass of the disc is 0.6 Kg, calculate the acceleration of the pulley if the radius of the pulley is 0.2 m (5 Marks) e) A rod of diameter 4 cm, length 48 cm and mass 1.2 kg rotates through its centre with axis parallel to its length Find the Moment of inertia (5 Marks) f) Define a couple of force (2 Marks) g) For a solid cylinder of radius 24 cm and mass 3.2 kg find the radius of gyration (5 Marks)
Решения
4.2
(320 Голоса)
Чавдар
элита · Репетитор 8 лет
Ответ
d) To calculate the acceleration of the pulley, we can use the formula:Torque = Moment of Inertia * Angular AccelerationThe moment of inertia for a solid disc is given by:Moment of Inertia = (1/2) * Mass * Radius^2Given:Torque = 12 NmMass = 0.6 kgRadius = 0.2 mSubstituting the values in the formula:12 Nm = (1/2) * 0.6 kg * (0.2 m)^2 * Angular AccelerationSolving for Angular Acceleration:Angular Acceleration = 12 Nm / ((1/2) * 0.6 kg * (0.2 m)^2)Angular Acceleration = 100 rad/s^2Therefore, the acceleration of the pulley is 100 rad/s^2.e) The moment of inertia for a rod rotating about its center is given by:Moment of Inertia = (1/12) * Mass * Length^2Given:Diameter = 4 cmLength = 48 cmMass = 1.2 kgRadius = Diameter / 2 = 2 cmLength = 48 cmSubstituting the values in the formula:Moment of Inertia = (1/12) * 1.2 kg * (0.48 m)^2Moment of Inertia = 0.0288 kg*m^2Therefore, the moment of inertia of the rod is 0.0288 kg*m^2.f) A couple of force refers to a pair of forces that are equal in magnitude but opposite in direction, resulting in a net force of zero. These forces cause a rotation or torque without causing a linear acceleration.g) The radius of gyration for a solid cylinder is given by:Radius of Gyration = sqrt((2/3) * Radius^2)Given:Radius = 24 cmMass = 3.2 kgSubstituting the values in the formula:Radius of Gyration = sqrt((2/3) * (0.24 m)^2)Radius of Gyration = 0.24 mTherefore, the radius of gyration for the solid cylinder is 0.24 m.