Two horizontal turntables are in contact with each other at their outer edges, and the static friction between them allows them to turn together (in opposite directions) without slipping. Thelarger turntable has a radius of 0.36m, and the smaller turntable a radius of 0.12m. The larger turntable accelerates from rest at a rate of 1.5rad/s2, and just when it completes its third revolution, a ball of clay that was (until now) stuck to the outer edge of the smaller turntable flies off. Find the speed of the ball of clay as it leaves the turntable.

Question

Two horizontal turntables are in contact with each other at their outer edges, and the static friction between them allows them to turn together (in opposite directions) without slipping. Thelarger turntable has a radius of 0.36m, and the smaller turntable a radius of 0.12m. The larger turntable accelerates from rest at a rate of 1.5rad/s2, and just when it completes its third revolution, a ball of clay that was (until now) stuck to the outer edge of the smaller turntable flies off. Find the speed of the ball of clay as it leaves the turntable.

Answer 1

consider the gear ratio:

w_final/w_first=(.36/.12)/1.5

or wfinal=wfirst*3
and the same holds for angular acceleration

ok, when the larger turns once, the second turns three, so when the first rotates 3, the smaller rotates 9. The angular acceleration of the smaller is 3 times the larger.

at the end...
wfinal^2=2*acceleration*displacement
wfinal^2=2*1.5*3 * 3*3*2PI
solve for wfinal, then
linear speed is wfinal*radius

Answer 2

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 * 2π radians. Therefore, the angular displacement is 3 * 2π = 6π 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.