A particle starts moving at t = 0 in a circle of radius R = 2 m with constant angular acceleration of 3 rad/sec^2. Initial angular speed of the particle is 1 rad/sec . At the instant when the angle between the acceleration vector and the velocity vector of the particle is 37º, calculate the value of ‘t’ at this moment

To solve this problem, we need to analyze the motion of a particle moving in a circular path with constant angular acceleration. We have the following parameters: the radius \( R = 2 \, \text{m} \), the initial angular speed \( \omega_0 = 1 \, \text{rad/sec} \), and the angular acceleration \( \alpha = 3 \, \text{rad/sec}^2 \). We need to find the time \( t \) when the angle between the acceleration vector and the velocity vector is \( 37^\circ \).

Understanding Angular Motion

In circular motion, two types of acceleration are present: tangential acceleration and centripetal (or radial) acceleration. The tangential acceleration \( a_t \) is due to the change in the speed of the particle along the circular path, while the centripetal acceleration \( a_c \) is directed towards the center of the circle and is responsible for keeping the particle in circular motion.

Calculating Tangential and Centripetal Acceleration

The tangential acceleration can be calculated using the formula:

• Tangential Acceleration: \( a_t = \alpha R \)

Substituting the values:

• \( a_t = 3 \, \text{rad/sec}^2 \times 2 \, \text{m} = 6 \, \text{m/sec}^2 \)

The centripetal acceleration is given by:

• Centripetal Acceleration: \( a_c = \frac{v^2}{R} \)

To find \( v \) (the linear speed), we first need to determine the angular speed \( \omega \) at time \( t \). The angular speed can be calculated using:

• Angular Speed: \( \omega = \omega_0 + \alpha t \)

Substituting the known values:

• \( \omega = 1 + 3t \)

The linear speed \( v \) is related to angular speed by the equation:

• Linear Speed: \( v = \omega R \)

Thus, we have:

• \( v = (1 + 3t) \times 2 = 2 + 6t \)

Now, substituting \( v \) into the centripetal acceleration formula:

• \( a_c = \frac{(2 + 6t)^2}{2} \)

Finding the Angle Between Acceleration Vectors

The angle \( \theta \) between the acceleration vector and the velocity vector can be found using the tangent of the angle:

• tan(θ) = a_t / a_c

Given that \( \theta = 37^\circ \), we can find \( \tan(37^\circ) \) which is approximately \( 0.7536 \). Therefore, we have:

• \( 0.7536 = \frac{6}{\frac{(2 + 6t)^2}{2}} \)

Rearranging gives:

• \( 0.7536 \cdot \frac{(2 + 6t)^2}{2} = 6 \)

Multiplying both sides by 2:

• \( 0.7536 (2 + 6t)^2 = 12 \)

Now, dividing both sides by \( 0.7536 \):

• \( (2 + 6t)^2 = \frac{12}{0.7536} \approx 15.92 \)

Taking the square root of both sides:

• \( 2 + 6t = \sqrt{15.92} \approx 3.98 \)

Now, solving for \( t \):

• \( 6t = 3.98 - 2 \)
• \( 6t = 1.98 \)
• \( t \approx \frac{1.98}{6} \approx 0.33 \, \text{seconds} \)

Final Result

Thus, the time \( t \) at which the angle between the acceleration vector and the velocity vector is \( 37^\circ \) is approximately \( 0.33 \) seconds. This analysis illustrates the relationship between angular motion and the components of acceleration in circular motion.