Question

a disk of radius 2.09 m rotates about its axis. points on the disk's rim undergo tangential acceleration of magnitude 1.71 m/s^2. at a particular time the rim has a tangential speed of 1.15m/s. at a time 0.911 s later, what is the tangential speed, v, of a point of the rim, the magnitude of the point's radial acceleration,ar,adn the magnitude of it total acceleration, atot?
please help, i'm confused of how to work this...been stuck on this for an hour..help is much appreciated.

Answer 1

The tangential acceleration of a point on the rim is 1.71m/s^2
Given the tangential speed at one time you can use v=u+at to determine the tangential speed at a later time t, using the tangential accerleration for a.
I get v = 2.71m/s for the tangential speed at t=0.911s.

Then you know that for circular motion at speed v, there is a radial acceleration given by v^2/r, where v is the tangential speed.
So at time t=0.911, with a tangential speed of 2.71 and radius 2.09 the magnitude of radial acceleration is 3.51m/s^2.

Now since the radial and tangential accelerations are perpendicular, pythagoras will give you the magnitude of the total acceleration \[a _{tot}=\sqrt{a _{r}^2 +a _{t}^2}\]
I got the magnitude of atot = 3.9m/s^2

Answer 2

@armi "by radial acceleration you mean
α=dω/dt
?"
yes that is what i mean....

Answer 3

@ProfBrainstorm thanks so much, you helped out alot

Answer 4

you're welcome