A ring of radius R is rotating about axis of ring such that angular velocity is given as = 5t. Find acceleration of a point P on rim after 5 sec

Options

A. 5R
B. 25R
C. `sqrt(650)`R
D. None of these

Explanation

To solve the problem of finding the acceleration of a point P on the rim of a rotating ring after 5 seconds, we can break down the solution into several steps:

### Step 1: Understand the given information
We know that the angular velocity \( \omega \) of the ring is given by:
\[ \omega = 5t \]
where \( t \) is the time in seconds. The radius of the ring is \( R \).

### Step 2: Calculate the angular velocity at \( t = 5 \) seconds
To find the angular velocity after 5 seconds, we substitute \( t = 5 \) into the equation for \( \omega \):
\[ \omega = 5 \times 5 = 25 \, \text{rad/s} \]

### Step 3: Find the tangential acceleration
The tangential acceleration \( a_t \) can be calculated using the formula:
\[ a_t = \alpha R \]
where \( \alpha \) is the angular acceleration. Angular acceleration \( \alpha \) is the derivative of angular velocity with respect to time:
\[ \alpha = \frac{d\omega}{dt} = \frac{d(5t)}{dt} = 5 \, \text{rad/s}^2 \]
Now substituting \( \alpha \) into the equation for \( a_t \):
\[ a_t = 5 \times R = 5R \]

### Step 4: Calculate the centripetal acceleration
Centripetal acceleration \( a_c \) is given by:
\[ a_c = \omega^2 R \]
Substituting the value of \( \omega \) we found earlier:
\[ a_c = (25)^2 R = 625 R \]

### Step 5: Calculate the resultant acceleration
The resultant acceleration \( a \) is the vector sum of the tangential and centripetal accelerations. Since these two accelerations are perpendicular to each other, we can use the Pythagorean theorem:
\[ a = \sqrt{a_t^2 + a_c^2} \]
Substituting the values we calculated:
\[ a = \sqrt{(5R)^2 + (625R)^2} \]
\[ = \sqrt{25R^2 + 390625R^2} \]
\[ = \sqrt{390650R^2} \]
\[ = R \sqrt{390650} \]

### Step 6: Final result
The acceleration of point P on the rim after 5 seconds is:
\[ a \approx 625.02 R \]