Question

A wheel starts from rest and has an angular acceleration that is given by α(t) = (4.00 rad/s^4)t^2 where t is in seconds. After it has turned through 6.00 rev, its angular velocity is:

a) 48 rad/s
b) 72 rad/s
c) 96 rad/s
d) 120 rad/s

Answer 1

The question requires using kinematic equations for rotational motion.Therefore, the angular velocity of the wheel after it has turned through 6.00 rev is approximately a. 48 rad/s.

Step-by-step explanation:

The student's question explores the concept of angular velocity in the context of rotational motion in physics. For a wheel starting from rest and undergoing angular acceleration described by the function α(t) = (4.00 rad/s4)t2, after turning through 6.00 revolutions (which is equivalent to 6.00 rev × 2π rad/rev = 37.70 rad), we aim to find its angular velocity. To calculate this, we use the relationship between angular displacement (θ), initial angular velocity (ω0), angular acceleration (α), and time (t), given by θ = ω0t + (1/2)αt2. Since the wheel starts from rest, the term involving ω0 vanishes. By rewriting the equation and taking the integral of the angular acceleration, we can obtain the angular velocity. This example uses kinematic equations to solve for the final angular velocity.

In this case, the initial angular velocity is 0 rad/s, the angular acceleration is given by α(t) = (4.00 rad/s^4)t^2, and the wheel has turned through 6.00 rev, which is equivalent to 12π rad.

Plugging in the values, we have:

ω = 0 + (4.00 rad/s^4)(6π)^2

ω ≈ 48 rad/s

Therefore, the angular velocity of the wheel after it has turned through 6.00 rev is approximately 48 rad/s.