Finding a in circular motion problem

[infraray]

The problem is this: sling with r=1.5m spinning 2m above ground. Stone is released and lands 10m away. What is acceleration while in circular motion? The answer is supposed to be 160 m/s. I keep coming up with approx. 145 m/s. I assume I am to figure t first to then find v, where t= sqrt(2y/g), which yields approx. 0.64s. Where am I going wrong? I have tried this many different ways and am pulling my hair out. Any help or leads will be greatly appreciated. Thanks.

 

[quantumdude]

Originally posted by infraray
I assume I am to figure t first to then find v, where t= sqrt(2y/g), which yields approx. 0.64s.

That equation is for free fall, not for circular motion.

The acceleration of a particle undergoing uniform circular motion is
a_C=v²/r. You need to know how fast the mass is moving.

 

[infraray]

Maybe I am missing something, but the problem isn't doesn't contain velocity as a given, the only givens are: r=1.5m, y=2m, and x=10m. I assume I am to figure out the initial v as the stone leaves the sling and work backward to find a. Once again I preemtively thank anyone who can offer some advice.

 

[HallsofIvy]

You know the height of the stone when it is "released" from the sling and you know it was moving horizontally at that time. The motion of the stone from that point is y= (-g/2)t^2+ 2 and x= v t.
You know that when y= 0, x= 10 m. You should be able to solve the equations (-g/2)t^2+ 2= 0 for t (yes, t= sqrt(2y/g) which, since y= 2, is t= 2/sqrt(g)) and then v t= 10 for v. THAT gives you the speed of the stone. Knowing that you can use the formulas for circular motion (particularly a= v^2/R) to find the acceleration, a.

By the way, you say that "The answer is supposed to be 160 m/s."

Of course, that's impossible: I'm sure you mean 160 m/s^2.

Actually, I get 163 1/3 meters per second squared.

 

[infraray]

Thanks HallsofIvy! What was messing me up was trying to calculate the velocity by using the formula v=c/t where c is the circumference. I was assuming that because I was looking for the acceleration while in circular motion I should have been using that formula. It all works out fine now. Oh, yeah I realize the 160 m/s should have contained the square I apparently did not proof read well enough. Thanks again for your help.