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A student sits on a rotating stool holding two 3.0 kg objects. When his arms are

ID: 1746513 • Letter: A

Question

A student sits on a rotating stool holding two 3.0 kg objects. When his arms are extendedhorizontally, the objects are 1.0 m from the axis of rotation, andhe rotates with an angular speed of 0.75 rad/s. The moment ofinertia of the student plus stool is 3.0 kg·m2 and isassumed to be constant. The student then pulls the objectshorizontally to 0.20 m from therotation axis. (a) Findthe new angular speed of the student.
rad/s

(b) Find the kinetic energy of the student system before and afterthe objects are pulled in. before J after J (a) Findthe new angular speed of the student.
rad/s

(b) Find the kinetic energy of the student system before and afterthe objects are pulled in. before J after J before J after J

Explanation / Answer

             Given that the mass fo the object is m = 3.0 kg              The distance between the object and the axis is d = 1.0 m               The initial angular speed is 1 = 0.75 rad/s              The moment of inertia of the student plus stool is I = 3.0kg·m2               The new distance between the each mass and axis is x = 0.20 m -------------------------------------------------------------------------------------------------       (a) since there isno external force on the system then the angular momentum isconserved .                         Initial angular momentum = final angular momentum                                                     I1*1 = I2*2                                         ( I + 2md2)1 = ( I +2m*x2)2                                                           2 = ( I +2md2)1 / ( I + 2m*x2)                                                                  =----------- rad/s           This isthe new angular speed of the student                (b) The initial kinetic energy is K1= (1/2)I1*12                                                         = (1/2)(I + 2m*d2 )*12                                                         = ------------- J                                                                = (1/2)(I + 2m*d2 )*12                                                         = ------------- J                    Theinitial kinetic energy is K2 = (1/2)I2*22                                                         = (1/2)(I + 2m*x2 )*22                                                         = ------------- J                                                                = (1/2)(I + 2m*x2 )*22                                                         = ------------- J