To find the speed of the ball of clay as it leaves the turntable, we need to use the concepts of rotational kinematics.
First, let's determine the angular displacement of the larger turntable. We know that the larger turntable completes 3 revolutions, which is equivalent to 3 * 2Ï€ radians. Therefore, the angular displacement is 3 * 2Ï€ = 6Ï€ radians.
Next, we can use the formula of rotational kinematics, which relates angular displacement (θ), angular velocity (ω), initial angular velocity (ω₀), and angular acceleration (α) as follows:
θ = ω₀t + 0.5αt²
Since the larger turntable starts from rest (ω₀ = 0), the equation simplifies to:
θ = 0.5αt²
Plugging in the values, we have:
6π = 0.5 * 1.5 * t²
Simplifying the equation:
6π = 0.75t²
Dividing both sides by 0.75:
8π = t²
Taking the square root of both sides:
t = √(8π)
Now, the smaller turntable will make the same number of revolutions as the larger turntable, given that they are in contact and not slipping. We can find the angular displacement of the smaller turntable using its radius:
θ = 2π * (r / R) = 2π * (0.12 / 0.36) = 2π / 3 radians
Finally, we can find the speed of the ball of clay as it leaves the turntable using the formula:
v = ω * r
where ω is the angular velocity and r is the smaller turntable's radius.
Since the angular displacement is 2π / 3 radians and the time is √(8π), we can find the angular velocity as:
ω = θ / t = (2π / 3) / √(8π)
Simplifying the equation:
ω = (2 / 3√2) rad/s
Now, we can calculate the speed of the ball of clay as it leaves the turntable:
v = ω * r = (2 / 3√2) * 0.12
Simplifying the equation:
v ≈ 0.0604 m/s
Therefore, the speed of the ball of clay as it leaves the turntable is approximately 0.0604 m/s.