A gyroscope is essentially a spinning wheel attached to an axle that is free to rotate in any direction about a pivot. In the figure below, a gyro wheel of mass \(M\) and radius \(R\) spins with angular velocity \(\omega\) in a counterclockwise direction when viewed from the right. The distance between the wheel and the pivot \(O\) is \(D\). Assume that the axle is of negligible mass and that the wheel's entire mass is concentrated around the rim. a) Is there an axis of symmetry in the problem? If so what is it? b) What is the magnitude of the angular momentum of the gyroscope? Draw the direction the angular momentum vector is pointing on the diagram. c) Are there any external forces acting on the gyro wheel? If so, draw them on the diagram. d) What is the magnitude of any torque acting on the wheel? Draw the direction of the torque on the diagram. e) Does the gyroscope move in any direction? If so which?

Step by step solution

01

a) Axis of symmetry

A gyroscope has an axis of symmetry which is the axle itself, as it is the axis about which the wheel spins. To visualize this, imagine looking at the gyroscope from the right side where the wheel spins counterclockwise. The axle would appear as a straight line, making it an axis of symmetry.

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b) Magnitude of angular momentum

The magnitude of the angular momentum \(\textbf{L}\) of the gyroscope can be calculated using the formula: $$ L = I \omega $$ Where \(I\) is the moment of inertia of the spinning wheel, which can be calculated as the mass of the wheel (\(M\)) multiplied by its radius \(R\) squared and \(\omega\) is the angular velocity. $$ I = M R^2 $$ Thus, the magnitude of the angular momentum is: $$ L = M R^2 \omega $$ The direction of the angular momentum vector \(\textbf{L}\) is along the axle pointing to the left following the right-hand rule.

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c) External forces acting on the gyro wheel

There are two external forces acting on the gyro wheel. One is the gravitational force acting downward on the center of mass of the wheel, which can be represented as: $$ \textbf{F}_{g} = M \textbf{g} $$ Where \(\textbf{g}\) is the acceleration due to gravity vector. The other external force comes from the pivot itself, providing an upward force balancing the wheel's weight. We can denote the force by the pivot as \(\textbf{F}_{p}\). Both \(\textbf{F}_{g}\) and \(\textbf{F}_{p}\) should be drawn on the diagram.

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d) Magnitude of torque and its direction

The torque \(\textbf{T}\) acting on the wheel can be calculated using the cross product of the position vector of the wheel's center of mass \(\textbf{r}_{cm}\) (with respect to the pivot) and the gravitational force acting on the wheel \(\textbf{F}_{g}\): $$ \textbf{T} = \textbf{r}_{cm} \times \textbf{F}_{g} $$ The position vector of the center of mass \(\textbf{r}_{cm} = D\hat{\textbf{i}}\). So, the torque becomes: $$ \textbf{T} = D\hat{\textbf{i}} \times M\textbf{g} $$ The magnitude of the torque is given by: $$ T = D M g $$ Using the right-hand rule, the direction of the torque vector is perpendicular to the plane formed by \(\textbf{r}_{cm}\) and \(\textbf{F}_{g}\), which is into the plane of the wheel.

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e) Gyroscope movement direction

The gyroscope moves as a result of the torque applied on it. According to the conservation of angular momentum, the change in angular momentum equals the applied torque. As the torque is in the plane of the wheel (pointing into the plane), the gyroscope will rotate in a direction perpendicular to both the torque and the angular momentum. This motion is called precession. The gyroscope moves in a clockwise direction when viewed from above, which can be determined using the right-hand rule with the direction of the torque and angular momentum vector.

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