The linear and angular velocities of a body in rotatory motion are `3 ms^(-1)` and `6 rad//s` respectively. If the linear acceleration is `6 m//s^2` then its angular acceleration in `rad s^(-2)` is

Options

A. `6`
B. `10`
C. `12`
D. `2`

Explanation

To find the angular acceleration of the body in rotatory motion, we can use the relationship between linear and angular quantities. Here are the steps to solve the problem:

### Step 1: Identify the given values
- Linear velocity (v) = 3 m/s
- Angular velocity (ω) = 6 rad/s
- Linear acceleration (a) = 6 m/s²

### Step 2: Relate linear velocity to angular velocity
The relationship between linear velocity (v) and angular velocity (ω) is given by the formula:
\[ v = ω \cdot r \]
where r is the radius of the circular path.

### Step 3: Solve for the radius (r)
From the equation \( v = ω \cdot r \), we can rearrange it to find r:
\[ r = \frac{v}{ω} \]
Substituting the known values:
\[ r = \frac{3 \, \text{m/s}}{6 \, \text{rad/s}} = \frac{1}{2} \, \text{m} \]

### Step 4: Relate linear acceleration to angular acceleration
The linear acceleration (a) is related to angular acceleration (α) by the formula:
\[ a = α \cdot r \]
We can rearrange this to find α:
\[ α = \frac{a}{r} \]

### Step 5: Substitute the known values to find angular acceleration (α)
Now substitute the known values of a and r:
\[ α = \frac{6 \, \text{m/s}^2}{\frac{1}{2} \, \text{m}} \]
\[ α = 6 \cdot 2 = 12 \, \text{rad/s}^2 \]

### Conclusion
The angular acceleration of the body is \( 12 \, \text{rad/s}^2 \).

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