Frisbee you throw a frisbee. Ignoring the translations motion of the frisbee ent

Question

Frisbee you throw a frisbee. Ignoring the translations motion of the frisbee entirely, and concentrating only on the spin you give the frisbee by snapping your wrist just before you release it, estimate the frisbee's angular acceleration during the throw the torque you exert on the frisbee during the throw the frisbee's rotational kinetic energy at release As in prior estimates, you're to make no assumptions about the time involved. Suggested quantities to make assumptions about arc: the initial and final angular velocities of the frisbee, the angular displacement of the frisbee during the wrist snap, and the mass of the frisbee. Assumed and calculated quantities should have no more than 2 SF, and all should be justified. Explain your work as you go.

Explanation / Answer

let mass of the frisbee is  =m let initial angular velocity of the frisbee is = 1 and final angular velocity of the frisbee is =2 when getting the angular velocity 2 the angular displacement covered by frisbee is = __________________________________________________________________________ from the equation of motion 2^2-1^2=2....................(1) where is angular acceleration of the frisbee from equation (1) angular acceleration of the frisbee is   =(2^2-1^2)/2 __________________________________________________________________________ let the length of the wrist is =L moment of inertia of the frisbee when it move in a circle of radius L is I =mL^2 the torque exerted on the frisbee is    = I*                                                           =(mL^2)*(2^2-1^2)/2 since = (2^2-1^2)/2 __________________________________________________________________________ rotational kinetic energy of the frisbee at released is K=1/2 I*1^2                                                                                  =1/2(mL^2)(1^2) __________________________________________________________________________ the relation between linear acceleration and angular acceleration is linear acceleration a =L*                                =L*(2^2-1^2)/2 __________________________________________________________________________ torque =force * perpendicular distance force excerted on the frisbee is F =/L                                                    =(mL^2)*(2^2-1^2)/2L                                                    =(mL)*(2^2-1^2)/2 __________________________________________________________________________ total kinetic energy of the frisbee is K =1/2 I (2^2+1^2)

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