Just looking for confirmation for a torque calculation

Thread starter pete94857
Start date Nov 11, 2024
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Energy Engineering Torque calculations

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A user seeks confirmation on their torque calculation for rotating a 10 mm cube weighing 7.5 grams by 180 degrees in 0.001 seconds, arriving at a value of 0.0000617 Joules, which they feel is low. The discussion highlights the importance of torque in rotation and questions the assumptions made in the calculation, particularly regarding the moment of inertia and the method of applying torque. Participants suggest that the energy required depends on the acceleration profile used during the rotation, with some arguing that energy can be recouped in a specific strategy. The conversation emphasizes the need for clarity in calculations and the relationship between angular speed and energy expenditure. Ultimately, the complexity of the problem requires careful consideration of various factors to achieve an accurate torque calculation.

Nov 14, 2024

pete94857

jbriggs444 said:

The correct formula is ##KE = \frac{1}{2} \times I \times {\omega_f}^2##

Note that the ##\omega_f## is squared.

If you do not apply the factor of ##\frac{1}{2}## and do not square ##\omega_f## then the result will be angular momentum. We use the symbol "L" for angular momentum: ##L = I \times \omega_f##.

In the car problem, the same applies. The correct formula is ##KE = \frac{1}{2} \times m \times v^2##

If you do not apply the factor of ##\frac{1}{2}## and do not square ##v## then the result will be momentum. We use the symbol "p" for momentum: ##p = m \times v##

Notes taken thank you