# Why is there angular acceleration in a non-spinning gyroscope?

• JC2000
In summary, the author derives that the angular acceleration is g/r which is non zero. This is due to the disc being constrained to move in a circle by the force at the hinge and the radial acceleration is perpendicular to the circle.f

#### JC2000

I was watching the above video which is part of a series explaining the mechanics behind a gyroscope. In the video the author explains the mechanics of the gyroscope when stationary (the disc is not rotating). Here he derives a result that the angular acceleration is g/r which is non zero.

1.Does the angular acceleration refer to the rotation of the disc and the rod about the hinge (due to the torque)?
2.If so, how is the relationship between the angular acceleration and tangential acceleration derived?
3. Is the tangential acceleration tangential to the curve charted by the falling disc?
4. If I understand correctly, the angular acceleration is a function of r, which is the perpendicular distance between the line of action of the force (m*g) and the axis about which rotation occurs. Thus as the disc falls at an angle, r changes, and hence angular acceleration changes (?). If so why is the tangential acceleration equal to acceleration due to gravity only initially?

Summary:: In the video linked below the gyroscope disc is not rotating. How can it have an angular acceleration in that case?

I was watching the above video which is part of a series explaining the mechanics behind a gyroscope. In the video the author explains the mechanics of the gyroscope when stationary (the disc is not rotating). Here he derives a result that the angular acceleration is g/r which is non zero.

1.Does the angular acceleration refer to the rotation of the disc and the rod about the hinge (due to the torque)?
2.If so, how is the relationship between the angular acceleration and tangential acceleration derived?
3. Is the tangential acceleration tangential to the curve charted by the falling disc?
4. If I understand correctly, the angular acceleration is a function of r, which is the perpendicular distance between the line of action of the force (m*g) and the axis about which rotation occurs. Thus as the disc falls at an angle, r changes, and hence angular acceleration changes (?). If so why is the tangential acceleration equal to acceleration due to gravity only initially?

1) Yes.

2) For circular motion, if the angle is measured in radians, then we have ##a = r\alpha##. This follows from the definition of the radian.

3) The disc is constrained to move in a circle: the tangential acceleration is defined as the component acceleration tangential to the circle; the radial acceleration is defined as the component perpendicular to circle.

4) Initially (only instantaneously) there is negligible force at the hinge. As the disc falls a force at the hinge and along the arm is required to constrain the disc to its circular path: gravity alone would accelerate it vertically. The acceleraion after the first instant is, therefore, due to the sum of these forces.

• • vanhees71, Dale and JC2000
1) Yes.

2) For circular motion, if the angle is measured in radians, then we have ##a = r\alpha##. This follows from the definition of the radian.

3) The disc is constrained to move in a circle: the tangential acceleration is defined as the component acceleration tangential to the circle; the radial acceleration is defined as the component perpendicular to circle.

4) Initially (only instantaneously) there is negligible force at the hinge. As the disc falls a force at the hinge and along the arm is required to constrain the disc to its circular path: gravity alone would accelerate it vertically. The acceleraion after the first instant is, therefore, due to the sum of these forces.
Thank you for the clear explanation!