Final answer:
The rotational speed at which the accelerometer experiences full scale acceleration is approximately 655.57 rpm. The output voltage produced by the accelerometer for an acceleration of +40 g is +40000 mV, and for an acceleration of -40 g is -40000 mV.
Step-by-step explanation:
To find the rotational speed at which the accelerometer experiences full scale acceleration, we need to calculate the centripetal acceleration at the distance of 6 inches from the axis of rotation. The centripetal acceleration is given by the formula a = rw^2, where a is the acceleration, r is the radius, and w is the angular velocity. Since the accelerometer experiences full scale acceleration at 40 g, we can convert this to meters per second squared by multiplying by the acceleration due to gravity (9.8 m/s^2). Setting the centripetal acceleration equal to this value, we can rearrange the equation to solve for angular velocity: w^2 = a/r.
Using the given distance of 6 inches, which is equal to 0.1524 meters, and a full scale acceleration of 40 g (392 m/s^2), we can plug these values into the equation to solve for w^2. Taking the square root of both sides, we find the angular velocity w to be approximately 68.72 rad/s.
To convert this angular velocity to revolutions per minute (rpm), we can use the conversion factor of 1 rpm = 2π rad/min. Therefore, the rotational speed at which the accelerometer experiences full scale acceleration is approximately 655.57 rpm.
For the output voltages produced by the accelerometer, we can use the formula V = k * a, where V is the output voltage, k is the sensitivity of the accelerometer, and a is the acceleration. The sensitivity of the MMA2201D accelerometer is 1000 mV/g. So, for an acceleration of +40 g, the output voltage would be 40 * 1000 mV = +40000 mV. For an acceleration of -40 g, the output voltage would be -40000 mV.