# How things work college course/Conceptual physics wikiquizzes/Uniform circular motion

The force required to sustain uniform circular motion is $F=ma=mv^{2}/r$ , where $a$ is acceleration and $r$ is the radius of the circle. The period of orbit, $T$ , is related to velocity $v$ by the fact that the distance traveled in one period of orbit is the circumference of the circle, $2\pi r$ .

You will be given the aforementioned equations as you are asked to do the following problems:

## Plug in the numbers

• What is the acceleration (in m/s2) of a particle that is traveling on a circle with a radius of 2 meters at a speed of 3 meters per second?

$a={\frac {v^{2}}{r}}={\frac {3^{2}}{2}}={\frac {9}{2}}={\frac {8}{2}}+{\frac {1}{2}}=4.5$

• What is the acceleration (in m/s2)of a particle that is traveling on a circle with a radius of 2 meters at a speed of 2 meters per second?

$a={\frac {v^{2}}{r}}={\frac {2^{2}}{2}}=2$

• What is the acceleration (in m/s2)of a particle that is traveling on a circle with a radius of 3 meters at a speed of 3 meters per second?

$a={\frac {v^{2}}{r}}={\frac {3^{2}}{3}}={\frac {9}{3}}=3$

## Proportional reasoning

The force required to sustain uniform circular motion is $F=ma=mv^{2}/r$ , where $a$  is acceleration and $r$  is the radius of the circle. The period of orbit, $T$ , is related to velocity $v$  by the fact that the distance traveled in one period of orbit is the circumference of the circle, $2\pi r$ .

Mr. Smith is using a string to swing a rock so fast that gravity may be neglected. What happens to the tension in the string when the velocity doubles?

$F={\frac {mv^{2}}{r}}\rightarrow F=kv^{2}$

where k is a constant that depends on units. Adopting units such that F=v=1 initially, we see that k=1. In these units the force after changing v equals 2. The final force in these units is

$F=v^{2}=2^{2}=4$  or 4 times the initial force.

Answer: The force increases by a factor of 4 when the speed is doubled.

Mr. Smith is using a string to swing a rock so fast that gravity may be neglected. What happens to the speed of the rock if the period is cut in half while the radius is tripled?

$v={\frac {distance}{time}}={\frac {2\pi r}{T}}\rightarrow v=k{\frac {r}{T}}$
$v={\frac {r}{T}}={\frac {3}{\frac {1}{2}}}={\frac {3\cdot 2}{{\frac {1}{2}}\cdot 2}}=6$