# Circular motion

1. Aug 1, 2008

### steven10137

1. The problem statement, all variables and given/known data
A particle starts from rest at t=0s.
It moves along a circular path of radius 18m and has an acceleration component along its path of 6.7m/s^2.
What is the magnitude of the acceleration when t=2s

2. Relevant equations
$$\begin{array}{l} \left| a \right| = \sqrt {a_n ^2 + a_t ^2 } \\ a_n = \frac{{v^2 }}{p} \\ v = \frac{{2\pi r}}{T} \\ \end{array}$$

3. The attempt at a solution
The particle is acting in circular motion, hence:
$$v = \frac{{2\pi r}}{T} = \frac{{2\pi \left( {18} \right)}}{2} = 18\pi \;ms^{ - 1}$$

The normal component of acceleration is then given by:
$$a_n = \frac{{v^2 }}{p} = \frac{{\left( {18\pi } \right)^2 }}{{18}} = \frac{{18^2 .\pi ^2 }}{{18}} = 18\pi ^2 \;ms^{ - 2}$$

Then I can find the magnitude of the total acceleration:
$$\left| a \right| = \sqrt {a_n ^2 + a_t ^2 } = \sqrt {\left[ {18\pi ^2 } \right]^2 + 6.7^2 } \approx 178ms^{ - 2}$$

Look OK? My first response would be that the figure seems quite high...

2. Aug 1, 2008

### Staff: Mentor

That's an equation for uniform circular motion where T is the period of the motion. Not relevant here.

Instead, use kinematics to calculate the tangential speed of the particle at the end of 2 seconds.

3. Aug 1, 2008

### steven10137

OK thanks, I thought this might have been where I went wrong.

$$\begin{array}{l} a_t = \frac{{dv}}{{dt}} \Rightarrow dv = a_t .dt \\ \int {dv} = \int {a_t .dt} \Rightarrow v = \int\limits_0^2 {a_t .dt} = \int\limits_0^2 {6.7.dt} = \left[ {6.7t} \right]_0^2 = 13.4\,ms^{ - 1} \\ a_n = \frac{{v^2 }}{r} = \frac{{13.4^2 }}{{18}} = 9.98\;ms^{ - 2} \\ \left| a \right| = \sqrt {a_t ^2 + a_n ^2 } = \sqrt {6.7^2 + 9.98^2 } = 12.0167\;ms^{ - 2} \\ \end{array}$$

Cheers.

4. Aug 1, 2008

Much better.

5. Aug 1, 2008

### steven10137

ok thanks :)

Just to confirm my understanding of this, the tangential acceleration is defined as the rate of change of velocity, in the 't' direction, yeah?
There is no other way to calculate this than to use kinematics?

6. Aug 1, 2008

### Staff: Mentor

Right.
Not sure what you mean.

7. Aug 1, 2008

### steven10137

Well I was just curious as to whether or not there was another way to get the same result.

8. Aug 1, 2008

### Staff: Mentor

Not that I can see.