Calculating Centripetal Acceleration of Stone

[ffrpg]

The problem reads, A boy whirls a stone in a horizontal circle, 1.6m above the ground, by means of a string. The radius of the circular motion is 1.0m. The string breaks and the stone hits the ground 5.0m from the point directly below where the string broke. What was the magnitude of the centripetal acceleration of the stone while it was experiencing circular motion, before the string broke?


I used the formula a=v^2/r. I started off by trying to find v by using the formula v = squareroot of rg. So v=sqrt(1.0)(9.81). After finding v I used the formula T= 2 pi r/ v. I didn't get the right answer. I've looked through my book, and can't find an example similar to this. I don't know how to use the other info given to me, like knowing the circle was 1.6m above the ground and knowing that it landed 5.0m away.

 

[quantumdude]

Originally posted by ffrpg
I used the formula a=v^2/r.

Yes, that's right.

I started off by trying to find v by using the formula v = squareroot of rg. So v=sqrt(1.0)(9.81).

That is not right.

You have to solve a projectile motion problem to find v. That is how you use the information given about the stone landing 5.0m away. If you call the breaking point of the string the origin (0m,0m), then the stone lands at the point (-1.6m,5.0m). The only unknown in the projectile motion problem is v_i, the speed with which the stone left its circular path. That is also the speed it had in its orbit.

 

[M98Ranger]

It seems like you are on the right track with using the formula a=v^2/r to calculate the centripetal acceleration. However, the formula you used to find v is incorrect. The correct formula to find the velocity of an object in circular motion is v = sqrt(r * a), where r is the radius and a is the centripetal acceleration. In this case, r = 1.0m and a is what we are trying to find. So, v = sqrt(1.0 * a).

Now, we can use the information given about the stone hitting the ground to find the velocity of the stone right before the string broke. Since the stone travels 5.0m horizontally before hitting the ground, we can use the formula x = vt to find the time it took for the stone to reach the ground. x is the distance (5.0m), v is the horizontal velocity (which is the same as the velocity of the stone in circular motion), and t is the time. So, t = x/v.

Now that we have the time, we can use the formula T=2*pi*r/v to find the period of the circular motion. T is the period, r is the radius, and v is the velocity. We already know r, and we can use the value of v we found earlier (sqrt(1.0 * a)). So, T = 2*pi*1.0/sqrt(a).

Finally, we can use the formula v=2*pi*r/T to find the velocity of the stone in circular motion before the string broke. v is the velocity, r is the radius, and T is the period. We already know r and T, so we can substitute those values and solve for v.

Now that we have the velocity of the stone in circular motion before the string broke, we can use the formula a=v^2/r to find the centripetal acceleration. Substitute the value of v we found earlier and the value of r (1.0m) into the formula and solve for a. This will give us the magnitude of the centripetal acceleration of the stone while it was in circular motion before the string broke.

I hope this helps clarify the steps you need to take to solve this problem. It's important to carefully read and understand the given information and use the correct formulas to solve the problem. Good luck!