Verify that the units of the rotational form of Newton's second law [Eq. (8-9)] are consistent. In other words, show that the product of a rotational inertia expressed in \(\mathrm{kg} \cdot \mathrm{m}^{2}\) and an angular acceleration expressed in \(\mathrm{rad} / \mathrm{s}^{2}\) is a torque expressed in \(\mathrm{N} \cdot \mathrm{m}\).

Short Answer

Expert verified
Answer: Yes, the units of the rotational form of Newton's second law are consistent. The product of rotational inertia (\(\mathrm{kg} \cdot \mathrm{m}^{2}\)) and angular acceleration (\(\mathrm{rad}/\mathrm{s}^{2}\)) results in the same units as torque (\(\mathrm{N}\cdot\mathrm{m}\)), which confirms the consistency.

Step by step solution

01

Identify the units of rotational inertia, angular acceleration, and torque

Rotational Inertia (\(I\)): \(\mathrm{kg} \cdot \mathrm{m}^{2}\) Angular Acceleration (\(\alpha\)): \(\mathrm{rad} / \mathrm{s}^{2}\) Torque (\(\tau\)): \(\mathrm{N} \cdot \mathrm{m}\)
02

Express Newton's second law in rotational form

Newton's second law for rotation states that the torque is equal to the rotational inertia multiplied by the angular acceleration: \(\tau = I * \alpha\)
03

Check the units by multiplying the units of rotational inertia and angular acceleration

Now we perform the multiplication \(\mathrm{kg} \cdot \mathrm{m}^{2} * \dfrac{\mathrm{rad}}{\mathrm{s}^{2}}\). It's also important to know that \(\mathrm{rad}\) is a dimensionless unit, so it can be removed in the multiplication. \(\mathrm{kg} \cdot \mathrm{m}^{2} * \dfrac{\mathrm{1}}{\mathrm{s}^{2}} = \dfrac{\mathrm{kg} \cdot \mathrm{m}^{2}}{\mathrm{s}^{2}}\)
04

Convert the units of torque to the standard unit

We have to convert the unit of torque from \(\mathrm{N} \cdot \mathrm{m}\) to the standard units of mass, length, and time to compare it with the result obtained in Step 3. We know that \(\mathrm{N}\) (newton) is equal to \(\mathrm{kg} \cdot \mathrm{m}/\mathrm{s}^{2}\), so we can substitute it back: \(\mathrm{N} \cdot \mathrm{m} = \mathrm{(kg} \cdot \mathrm{m}/\mathrm{s}^{2}) \cdot \mathrm{m} = \dfrac{\mathrm{kg} \cdot \mathrm{m}^{2}}{\mathrm{s}^{2}}\)
05

Compare the results

Now we compare our final result from Step 3 and Step 4: \(\dfrac{\mathrm{kg} \cdot \mathrm{m}^{2}}{\mathrm{s}^{2}} = \dfrac{\mathrm{kg} \cdot \mathrm{m}^{2}}{\mathrm{s}^{2}}\) Since both units are the same, we can conclude that the units of the rotational form of Newton's second law are consistent.

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Most popular questions from this chapter

An experimental flywheel, used to store energy and replace an automobile engine, is a solid disk of mass \(200.0 \mathrm{kg}\) and radius $0.40 \mathrm{m} .\( (a) What is its rotational inertia? (b) When driving at \)22.4 \mathrm{m} / \mathrm{s}(50 \mathrm{mph}),$ the fully energized flywheel is rotating at an angular speed of \(3160 \mathrm{rad} / \mathrm{s} .\) What is the initial rotational kinetic energy of the flywheel? (c) If the total mass of the car is \(1000.0 \mathrm{kg},\) find the ratio of the initial rotational kinetic energy of the flywheel to the translational kinetic energy of the car. (d) If the force of air resistance on the car is \(670.0 \mathrm{N},\) how far can the car travel at a speed of $22.4 \mathrm{m} / \mathrm{s}(50 \mathrm{mph})$ with the initial stored energy? Ignore losses of mechanical energy due to means other than air resistance.
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