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Chapter 8: Problem 47

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

A tower outside the Houses of Parliament in London has a famous clock commonly referred to as Big Ben, the name of its 13 -ton chiming bell. The hour hand of each clock face is \(2.7 \mathrm{m}\) long and has a mass of \(60.0 \mathrm{kg}\) Assume the hour hand to be a uniform rod attached at one end. (a) What is the torque on the clock mechanism due to the weight of one of the four hour hands when the clock strikes noon? The axis of rotation is perpendicular to a clock face and through the center of the clock. (b) What is the torque due to the weight of one hour hand about the same axis when the clock tolls 9: 00 A.M.?
A spinning flywheel has rotational inertia $I=400.0 \mathrm{kg} \cdot \mathrm{m}^{2} .\( Its angular velocity decreases from \)20.0 \mathrm{rad} / \mathrm{s}\( to zero in \)300.0 \mathrm{s}$ due to friction. What is the frictional torque acting?
Verify that \(\frac{1}{2} I \omega^{2}\) has dimensions of energy.
A sign is supported by a uniform horizontal boom of length \(3.00 \mathrm{m}\) and weight \(80.0 \mathrm{N} .\) A cable, inclined at an angle of \(35^{\circ}\) with the boom, is attached at a distance of \(2.38 \mathrm{m}\) from the hinge at the wall. The weight of the sign is \(120.0 \mathrm{N} .\) What is the tension in the cable and what are the horizontal and vertical forces \(F_{x}\) and \(F_{y}\) exerted on the boom by the hinge? Comment on the magnitude of \(F_{y}\).
A crustacean (Hemisquilla ensigera) rotates its anterior limb to strike a mollusk, intending to break it open. The limb reaches an angular velocity of 175 rad/s in \(1.50 \mathrm{ms} .\) We can approximate the limb as a thin rod rotating about an axis perpendicular to one end (the joint where the limb attaches to the crustacean). (a) If the mass of the limb is \(28.0 \mathrm{g}\) and the length is \(3.80 \mathrm{cm}\) what is the rotational inertia of the limb about that axis? (b) If the ex-tensor muscle is 3.00 mm from the joint and acts perpendicular to the limb, what is the muscular force required to achieve the blow?
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