Physics Friday 143

Consider a bead sliding down a straight wire inclined by an angle θ, with a coefficient of kinetic friction μ. One can see that the component of the bead’s weight parallel to the wire is , while the component perpendicular is . The normal force exerted on the bead by the wire is thus , and so the frictional force, acting parallel to the wire in the opposite direction to the parallel component of weight, is .
For the net force to be zero, allowing the bead to slide at constant velocity, we thus need , which means . If , then the friction force is less than the component of the weight, and the bead will accelerate down the wire. If , then the friction force is greater than the component of the weight, and the sliding bead will decelerate, coming to a stop somewhere along the wire.

Now, let us take the bead on a helix from last week, and add a friction force, with a coefficient of kinetic friction μ. Here, our constraint force (normal force) exerted by the wire on the bead has, in the Frenet frame, both normal and binormal components, and has magnitude dependent on velocity of
.
We note that this has a minimum value at v=0 of
, meaning the minimum value of the frictional force is .
Now, the tangential component of the weight is . Thus, we see that if , then the friction will exceed the tangential component of the weight, and the sliding bead will decelerate, coming to a stop somewhere along the helical wire. This condition is

which simplifies to
.
Note, however, that for , the frictional force increases with velocity, meaning that there is a speed where the forces balance, and acceleration is zero; the bead can slide along at that constant speed, and if it starts at a different speed, it will accelerate (or decelerate) toward that speed. Setting friction and tangential component of the weight equal, and solving for the zero-net-force velocity vC, we have
.
We note that this requires that , which is equivalent to .

Advertisements

Tags: , , , ,

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s


%d bloggers like this: