what happens to the acceleration (of a ball going around in a circle tethered to a string) if:
1) radis of the circle halves and the period of rotation stays the same
2) the peiod of rotation halves and the radius stays the same
3) the period doubles and the speed doubles (the radius changes as necessary)
this is my reasoning:
1) v=2(pi)r divided by T, therefore, if radius is halved, then the velocity will be halved, and thus the acceleration would be halved.
2)velocity is twiced, therefore, the accerleration is quadrupled based on a=v^2/r
3)i'm not sure at all.
the answers are:
-quadruples, quarters, stays the same, halves, doubles.
any help is appreciated!
acceleration= 4PI^2 r/T
a) acceleration is halved, right?
2) if T is halved, then acceleration is doubled
3) If speed doubles, and period doubles, what happened to r?
v=2PI r/T
TV=2PI r so r quadrupled
a= v^2/r so a stays the same.
Let's go through each scenario step by step and analyze the changes in acceleration.
1) When the radius of the circle halves and the period of rotation stays the same:
In circular motion, the centripetal acceleration is given by the equation a = v^2/r, where a is the acceleration, v is the velocity, and r is the radius. If we consider the initial conditions as r1 and v1, and the final conditions as r2 and v2, we can write the equation as:
a1 = v1^2/r1 and a2 = v2^2/r2.
Given that the period of rotation stays the same (T1 = T2), we know that the velocity (v) is directly proportional to the radius (r) according to the equation v = 2πr/T. Therefore, we can express v2 and r2 as v2 = (v1/2) and r2 = (r1/2).
Now, let's substitute these values into the acceleration equations:
a1 = v1^2/r1 and a2 = v2^2/r2
a1 = v1^2/r1 and a2 = (v1/2)^2/(r1/2)
a1 = v1^2/r1 and a2 = (v1^2/4)/(r1/2)
a1 = v1^2/r1 and a2 = (v1^2/4)*(2/r1)
Simplifying a2, we get:
a2 = (v1^2/4)*(2/r1)
a2 = (v1^2/2r1)
Comparing a2 to a1, we can see that a2 is halved compared to a1. So, when the radius is halved but the period stays the same, the acceleration is halved as well.
2) When the period of rotation halves and the radius stays the same:
In this scenario, the period of rotation is inversely proportional to the velocity (T = 2πr/v). Therefore, if the period halves, the velocity doubles (v2 = 2v1), while the radius remains the same (r2 = r1).
Let's substitute these values into the acceleration equation:
a1 = v1^2/r1 and a2 = v2^2/r2
a1 = v1^2/r1 and a2 = (2v1)^2/r1
a1 = v1^2/r1 and a2 = (4v1^2)/r1
a2 = 4(v1^2/r1)
Comparing a2 to a1, we can see that a2 is quadrupled compared to a1. So, when the period halves but the radius stays the same, the acceleration is quadrupled.
3) When the period doubles and the speed doubles (the radius changes as necessary):
In this scenario, the speed doubles, so we can write v2 = 2v1. The period doubles, which implies T2 = 2T1. The radius changes to accommodate these changes.
Since the speed (v) is directly proportional to the radius (r) according to the equation v = 2πr/T, when the speed doubles, the radius doubles as well (r2 = 2r1). Therefore, T2 = 2(2πr1)/(2v1) = 2πr1/v1.
Now, let's substitute these values into the acceleration equation:
a1 = v1^2/r1 and a2 = v2^2/r2
a1 = v1^2/r1 and a2 = (2v1)^2/(2r1)
a1 = v1^2/r1 and a2 = (4v1^2)/(2r1)
Simplifying a2, we get:
a2 = (4v1^2)/(2r1)
a2 = 2(v1^2/r1)
Comparing a2 to a1, we can see that a2 is the same as a1. So, when the period doubles and the speed doubles, the acceleration stays the same.
Therefore, the correct answers are:
1) The acceleration halves.
2) The acceleration quadruples.
3) The acceleration stays the same.