Understanding Radial Acceleration in Circular Motion

When a particle moves in a circular path, it experiences an acceleration directed towards the center of the circle. This acceleration is called radial acceleration or centripetal acceleration. It is responsible for changing the direction of the velocity vector, keeping the particle on the circular path. The magnitude of the radial acceleration depends on the linear speed of the particle and the radius of the circle.

The formula for radial acceleration ($\text{a}_\text{r}$) is given by:

$$ \text{a}_\text{r} = \frac{\text{v}^2}{\text{R}} $$

where $\text{v}$ is the linear speed of the particle and $\text{R}$ is the radius of the circular path.

Given Information and Calculating Linear Speed

We are given the following information about the particle's motion:

• Radius of the circle, $\text{R} = 30 \, \text{cm}$.
• Linear speed of the particle as a function of time, $\text{v} = 3\text{t}$, where $\text{v}$ is in meters/second ($\text{m/s}$) and $\text{t}$ is in seconds ($\text{s}$).
• We need to find the radial acceleration at time $\text{t} = 5 \, \text{s}$.

First, we need to ensure all units are consistent. The radius is given in centimeters, so we convert it to meters:

$$ \text{R} = 30 \, \text{cm} \times \frac{1 \, \text{m}}{100 \, \text{cm}} = 0.3 \, \text{m} $$

Next, we calculate the linear speed ($\text{v}$) of the particle at $\text{t} = 5 \, \text{s}$ using the given formula $\text{v} = 3\text{t}$:

$$ \text{v} = 3 \times 5 \, \text{m/s} = 15 \, \text{m/s} $$

So, at $\text{t} = 5 \, \text{s}$, the linear speed of the particle is $15 \, \text{m/s}$.

Calculating Radial Acceleration at t = 5s

Now we can calculate the radial acceleration using the formula $\text{a}_\text{r} = \frac{\text{v}^2}{\text{R}}$. We have the linear speed $\text{v} = 15 \, \text{m/s}$ and the radius $\text{R} = 0.3 \, \text{m}$.

Substitute these values into the formula for radial acceleration:

$$ \text{a}_\text{r} = \frac{(15 \, \text{m/s})^2}{0.3 \, \text{m}} $$

$$ \text{a}_\text{r} = \frac{225 \, \text{m}^2/\text{s}^2}{0.3 \, \text{m}} $$

$$ \text{a}_\text{r} = \frac{225}{0.3} \, \text{m/s}^2 $$

To simplify the division:

$$ \text{a}_\text{r} = \frac{2250}{3} \, \text{m/s}^2 $$

$$ \text{a}_\text{r} = 750 \, \text{m/s}^2 $$

The radial acceleration of the particle at $\text{t} = 5 \, \text{s}$ is $750 \, \text{m/s}^2$. This high value of radial acceleration is necessary to keep the particle undergoing this particular particle motion in a relatively small circle while having a high instantaneous linear speed.

Thus, the calculated radial acceleration at $\text{t} = 5 \, \text{s}$ is $750 \, \text{m/s}^2$. This type of problem helps understand the relationship between radial acceleration, linear speed, and the radius of circular motion.