About a screw (circular motion)

[fara0815]

Hello folks! I am about to finish the chapter "Circular motion" in my script but there is the following problem that is keeping me busy and I would very appreciate every help I can get!

A M10-Screw with a pitch of 1.5 mm (means, that the screw goes 1.5 mm per revolution into the thread) is being screwed into a thread with a constant angular speed of 360 deg/s. The screw's head has a diameter of 16mm.What is the veloctiy of a point on the outside diameter of the head and what is its acceleration?
According to my professor, the velocity is 50,29 mm/s and the acceleration is 0,3158 m/s^2.

For the veloctiy I got with :

f= 1/s
w=2\pif= 6,28 \frac{rad}{s}
v=wr= 6,28 \frac{rad}{s} 8 mm = 50,24 \frac{mm}{s}
which is close but I assume that it is a three-dimensional problem ant I do not how to solve it in 3d.

 

[Kurdt]

Seems fine. Perhaps your lecturer made a mistake or remembered the wrong thing, they are human too.

 

[berkeman]

In addition to the angular v = wr term, you need to add in the velocity component of the 1.5mm/s in the axis of the screw. If you've used vectors before, just add the vectors. If not, draw a triangle with the radial velocity and the axial velocity as the sides, and calculate the hypotenuse.

 

[fara0815]

Thank you very much guys! It makes a lot of sense if I think about it!
I actually get as absolute value of v 50,26 mm/s which is closer to my prof's answer. But what about the acceleration? There is not time I can work with !?

 

[berkeman]

Do you know the formula for the centripetal acceleration in uniform circular motion? That is the only component of acceleration, since the linear motion in the axis of the screw does not contribute to the acceleration.

 

[fara0815]

oh man, you are totally right! I have tried the whole time to get the acceleration with the tangential velocity but I forgot that I can use the angular velocity. Thus
\alpha=\omega^2*r