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Section 12.6 Rotational DynamicsSection 12.7 Rotation
Chapter 12, Problem 24E(choose chapter or problem)
An object whose moment of inertia is \(4.0 \mathrm{~kg} \mathrm{~m}^{2}\) experiences the torque shown in Figure EX12.24. What is the object’s angular velocity at \(t = 3.0 s\)? Assume it starts from rest.
Equation Transcription:
Text Transcription:
4.0 kg m^2
t = 3.0 s
Questions & Answers
QUESTION:
An object whose moment of inertia is \(4.0 \mathrm{~kg} \mathrm{~m}^{2}\) experiences the torque shown in Figure EX12.24. What is the object’s angular velocity at \(t = 3.0 s\)? Assume it starts from rest.
Equation Transcription:
Text Transcription:
4.0 kg m^2
t = 3.0 s
ANSWER:
Step 1 of 2
We have to find the object's angular velocity at \(t=3.0 \mathrm{~s}\)
The angular acceleration of the object can be found using the equation
\(\omega_{f}=\omega_{i}+\) area under the angular acceleration curve between \(t_{i}\) and \(t_{f}\)
The graph of angular acceleration versus time is as shown in the below figure, the graph is similar to that of torque versus time except the values for angular acceleration is \(\alpha=\frac{\tau}{I}\).
\(I=4.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\)