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 with radius \( R \) and constant angular acceleration \( \alpha \) equals its tangential acceleration.

### Step 1: Understand the definitions of radial and tangential accelerations.

- **Radial Acceleration (\( a_r \))**: This is given by the formula:
\[ a_r = R \omega^2 \]
where \( \omega \) is the angular velocity.

- **Tangential Acceleration (\( a_t \))**: This is given by the formula:
\[ a_t = R \alpha \]
where \( \alpha \) is the angular acceleration.

### Step 2: Determine the expression for angular velocity (\( \omega \)).

Since the point mass starts from rest and has a constant angular acceleration \( \alpha \), the angular velocity at time \( t \) can be expressed as:
\[ \omega = \alpha t \]

### Step 3: Substitute \( \omega \) into the radial acceleration formula.

Substituting \( \omega \) into the radial acceleration formula, we have:
\[ a_r = R (\alpha t)^2 = R \alpha^2 t^2 \]

### Step 4: Set the radial acceleration equal to the tangential acceleration.

Now, we set the radial acceleration equal to the tangential acceleration:
\[ R \alpha^2 t^2 = R \alpha \]

### Step 5: Simplify the equation.

We can cancel \( R \) from both sides (assuming \( R \neq 0 \)):
\[ \alpha^2 t^2 = \alpha \]

### Step 6: Solve for \( t \).

Dividing both sides by \( \alpha \) (assuming \( \alpha \neq 0 \)):
\[ \alpha t^2 = 1 \]
\[ t^2 = \frac{1}{\alpha} \]
\[ t = \sqrt{\frac{1}{\alpha}} \]

### Final Answer:

The time needed after motion begins for the radial acceleration to be equal to the tangential acceleration is:
\[ t = \frac{1}{\sqrt{\alpha}} \]

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