Find the centripetal acceleration of a dust particle on the edge of a vinyl record, 0.3m in diameter

Find the centripetal acceleration of a dust particle on the edge of a vinyl record, 0.3m in diameter, rotation at 33.3 rev/min.

Please show your working clearly.

THANKYOU!!!! <3

Okey dokey... lets do this:

A few symbols for you:
a = centripetal acceleration (in metres per second squared, m/s/s)
r = radius of the motion (in metres, m. In this case, that's the radius of the record)
T = time taken for one rotation (in seconds)

Now, we know the record is spinning at 33.3 revolutions per minute. That means it is doing 33.3 revolutions every 60 seconds. So to find T, we divide 33.3 by 60, which is 0.555 rev/sec.

So we have a value of 0.555 rev/sec, but we want a value for T, which is seconds only. So we divide 1 second by 0.555 revs and get T = 1.8 seconds

This means the values we now have are:
r = 0.3 m
T = 1.8 s

With these, we can use the equation: v = 2πr divided by T (π is a button on your calculator, and v is the speed of the dust particle).

v = 2 x π x 0.3 / 1.8
v = 1.05 m/s

And finally, we put this into the equation for centripetal acceleration:

a = (v x v) divided by r
a = (1.05 x 1.05) / 0.3

a = 3.675 m/s/s

hope you understand the equations in this form :)

Haha XD oops, diameter and radius mix-up there.

That's what a weeks worth of holidays does to you :P

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