2) A very lightweight spring of spring constant k is attached to a wheel in the form of a disk of mass M and radius R at a point that is a distance r from the center. The entire setup is in a horizontal plane (the view in the picture is from above) so gravity does not affect the system. The spring is attached such that it is initially unstretched until the wheel is turned slightly. a) (2pts) Show how you would find the moment of inertia of this system. b) (2pts) For small angles, what would the torque on the wheel due to the spring be

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Xema8 Xema8 · Answer 1 · 2020-05-11T04:45:38+02:00

Answer:

Explanation:

a )

Moment of inertia will be of the wheel which is rotating about its centre .

So

I = 1/2 m R² ; m is mass and R is radius of the wheel

b )

If wheel is turned by small angle dθ , spring is stretched by length r dθ which is almost linear due to small value of dθ.

force developed in the spring = k x r dθ

torque of this force about the centre

= force x distance from the centre

= k x r dθ x r

Torque = k r² dθ