The angular velocity of a spinning gyroscope is measured every 0.5 s. The results and the best-fit line from a spreadsheet are shown in FIGURE P4.63. What is the gyroscope's initial angular velocity at t = 0 s?

Question

The angular velocity of a spinning gyroscope is measured every 0.5 s. The results and the best-fit line from a spreadsheet are shown in FIGURE P4.63. What is the gyroscope's initial angular velocity at t = 0 s?

Accepted Answer

Hi, everyone. In this practice problem, we're being asked to find a wheel's initial angular velocity. When T equals zero seconds, we're being given a graph showing an electronically generated line of the best fit using measured angular velocities for a wheel. The velocity is measured every 0.4 seconds. And we're being asked to find the wheel's initial angular velocity. When T equals zero seconds, the graph or the figure given is for an omega or angular velocity in radius per seconds versus times squared in seconds squared. The line of best have an equation of Y equals 20 X plus 10. And the options given for the wheel's initial angular velocity R A 20 radiant per seconds, B 10 radiant per seconds, C 30 radiant per second and D 3. radiant per seconds. So we will assume that the wheel is a rigid body. And we want to remember that in the figure given the horizontal axis is N T squared and not T. So in the case of this particular practice problem, we will not be able to use the rotational kinematic equation. We wanna actually remember that in the case of the T or in the case of T, the line will actually show a constant angular acceleration. But in this case, because it is in T squared, that is not the case. So in this case, X equals T squared will describe the X axis and at T equals zero seconds, then X equals T squared. And in this case, then X will equals zero seconds squared. So we are being asked to find the wheel's initial angular velocity at P equal zero seconds. And we want to realize the equation line for the best fit given in our figure in order to do that. So the Y in this case is going to be the angular velocity and the X is going to be RT squared, which is why we are finding the P squared value for RT equals zero seconds, which in this case is also zero seconds squared. So we want to utilize the equation Y equals 20 X plus 10, which in this case, the Y is going to be omega equals 20. X X is going to be the T squared plus 10. So our omega value at T equals zero seconds will equals 20 T squared at D equals zero seconds plus 10. And in this case, our omega at T equals zero seconds. Well, then equals to B 20 multiplied by the T squared value which is going to be zero seconds squared, zero seconds squared plus 10, 20 multiplied by zero seconds. Squared is just going to be equals to zero. So our omega at D equals zero seconds is going to equals to 10 radiance per seconds just like. So, so the wheels initial angular velocity when B equals zero seconds is going to equals to 10 gradients per seconds, which will correspond to option B in our answer choices. So option B is going to be the answer to this particular practice problem and that'll be it for this one. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and they'll be it for this one. Thank you.

The angular velocity of a spinning gyroscope is measured every 0.5 s. The results and the best-fit line from a spreadsheet are shown in FIGURE P4.63. What is the gyroscope's initial angular velocity at t = 0 s?

Verified video answer for a similar problem:

Key Concepts

Angular Velocity

Graph Interpretation

Linear Equation