A point mass moves along a circle of radius `R` with a constant angular acceleration `alpha`. How much time is needed after motion begins for the radial acceleration of the point mass to be equal to its tangential acceleration ?

Options

A. `sqrt((1)/(alpha))`
B. `sqrt((R)/(alpha))`
C. `sqrt((2)/(alpha))`
D. `sqrt((1)/(2alpha))`

Answer

To solve the problem, we need to find the time \( t \) when the radial acceleration of a point mass moving in a circle equals its tangential acceleration.

### Step-by-step Solution:

1. **Understanding Radial and Tangential Acceleration:**
- The radial acceleration \( a_r \) is given by the formula:
\[ a_r = \omega^2 R \]
- The tangential acceleration \( a_t \) is given by:
\[ a_t = \alpha R \]
where \( \omega \) is the angular velocity and \( R \) is the radius of the circle.

2. **Relating Angular Velocity to Time:**
- Since the point mass starts from rest and has a constant angular acceleration \( \alpha \), the angular velocity \( \omega \) at time \( t \) can be expressed as:
\[ \omega = \alpha t \]

3. **Substituting Angular Velocity into Radial Acceleration:**
- We can substitute \( \omega \) into the equation for radial acceleration:
\[ a_r = (\alpha t)^2 R = \alpha^2 t^2 R \]

4. **Setting Radial Acceleration Equal to Tangential Acceleration:**
- We want to find the time \( t \) when the radial acceleration equals the tangential acceleration:
\[ a_r = a_t \]
- Substituting the expressions we have:
\[ \alpha^2 t^2 R = \alpha R \]

5. **Simplifying the Equation:**
- We can cancel \( R \) (assuming \( R \neq 0 \)) and \( \alpha \) (assuming \( \alpha \neq 0 \)):
\[ \alpha t^2 = 1 \]

6. **Solving for Time \( t \):**
- Rearranging gives:
\[ t^2 = \frac{1}{\alpha} \]
- Taking the square root of both sides:
\[ t = \sqrt{\frac{1}{\alpha}} = \frac{1}{\sqrt{\alpha}} \]

### Final Answer:
The time needed after motion begins for the radial acceleration of the point mass to be equal to its tangential acceleration is:
\[ t = \frac{1}{\sqrt{\alpha}} \]