Question

A 1-lb collar is attached to a spring and slides without friction along a circular rod in a vertical plane. The spring has an undeformed length of 5 in. and a constant k =25 lb/ft. Knowing that the collar is released from being held at A determine, the speed of the collar and the normal force between the collar and the rod as the collar passes through B.

Answer 1

0.4167 ft/s

Step-by-step explanation:

The law of conservation is applied to point A and B.

This gives:

`T_(A) + V_(A) = T_(B) + V_(B)`

Hence,
`T_(A)` is the kinetic energy at the position A

`V_(A)` is the velocity at point A

Considering point A, the kinetic energy at the point will be:

`T_(A) = 0`

The potential energy will be:

`V_(A) = (V_(A))_(g) + (V_(A))_(c)`

Hence,
`V_(Ag)` is the potential energy and
`V_(Ac)` is the kinetic energy

The potential energy is given by the following:

`V_(Ag) = mgr`

substituting 1 lb for mg gives 0.4167 ft for r

Then the velocity,
`V_(Ag) = 1lb * 0.4167ft\\ = 0.4167 ft`