The disk weights 40 lb and has a radius of gyration is 0.6 ft. A 15 lb/ft moment is applied and the spring has a spring constant of 10 lb/ft. The system was initially at rest and the disk is rolling without slipping. The spring is initially unstretched. Find the angular velocity of the wheel when disk moves to the right 0.5 ft.

angular velocity = 2.6543 rad/s

Explanation:

To find the angular velocity of the wheel when the disk moves to the right 0.5 ft, we need to be aware that the spring will stretch twice the value of gyration with any slight change in the position or movement of gyration since the top of the wheel is holding the spring.

The work done here:

= ((distance moved by the wheel) X spring constant X (Final displacement^2 - Initial displacement^2)) + Mass (q2 – q1)

Where q2 = 0.5ft

            q1 = 0.8 lb      

Note that linear velocity = radius X angular velocity

= -0.5(10)(1^2 – 0) + 15(0.5/0.8) = 4.375 ft·lb    

Then, the kinetic energy :

 Since the spring is initially unstretched, the initial tension in the spring = 0

So the final tension = ((distance moved by the wheel) X (linear velocity)^2 X (angular velocity) ^2 + (distance moved by the wheel) X (linear velocity) X ( radius of gyration) ^2 X (angular velocity) ^2

= 0.5(40/32.2)(0.8w) ^2 + 0.5(40/32.2)(0.6)^2 X w2

final tension = 0.621 w2

So the angular velocity of the wheel when disk moves to the right 0.5 ft = The Initial workdone + Initial kinetic energy will be equal to the final workdone + the final Kinetic energy

0 + 4.375 ft·lb = 0.621 w2

 angular velocity = 2.6543 rad/s