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

Answer:

0.4167 ft/s

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, $V_{Ag} = 1lb * 0.4167ft\\ = 0.4167 ft$