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Chapter 5: Problem 68

Two gears \(A\) and \(B\) are in contact. The radius of gear \(A\) is twice that of gear \(B\). (a) When \(A\) 's angular velocity is \(6.00 \mathrm{Hz}\) counterclockwise, what is \(B^{\prime}\) s angular velocity? (b) If \(A\) 's radius to the tip of the teeth is \(10.0 \mathrm{cm},\) what is the linear speed of a point on the tip of a gear tooth? What is the linear speed of a point on the tip of \(B\) 's gear tooth?

Short Answer

Expert verified
Answer: The linear speed of a point on the tip of Gear B's gear tooth is \(1.2\pi \mathrm{m/s}\).

Step by step solution

01

Understand the relationship between gears A and B and find a ratio

Since the radius of gear A is twice that of gear B, their circumferences have the same ratio because the circumference is directly proportional to the radius. The ratio is 2:1 and since the gears are in contact, they must have the same linear velocity.
02

Calculate angular velocity of gear B

Angular velocity of gear A (\(\omega_A\)) is given as \(6.00 \mathrm{Hz}\). To find the angular velocity of gear B (\(\omega_B\)), we divide gear A's angular velocity by the ratio (2:1) found in the previous step. $$ \omega_B = \frac{6.00 \mathrm{Hz}}{\frac{2}{1}} = 3.00 \mathrm{Hz} $$
03

Calculate the linear speed of the tip of Gear A

Now that we found the angular velocity of Gear A, we can calculate the linear speed (\(v_A\)) of a point on the tip of the gear tooth. We use the following equation: $$ v_A = r_A \cdot \omega_A $$ where, \(r_A\) is the radius of Gear A, and \(\omega_A = 6.00 \mathrm{Hz}\) Given Gear A's radius is \(10.0 \mathrm{cm}\) or \(0.1 \mathrm{m}\), we can calculate the linear speed: $$ v_A = 0.1 \mathrm{m} \cdot 6.00 \text{ 2}\pi \mathrm{rad/s} = 1.2\pi \mathrm{m/s} $$
04

Calculate the linear speed of the tip of Gear B

Since gears A and B are in contact and have the same linear velocity (as found in step 1), the linear speed of a point on the tip of Gear B's gear tooth is equal to the linear speed of Gear A which is \(1.2\pi \mathrm{m/s}\).

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Most popular questions from this chapter

What is the average linear speed of the Earth about the Sun?
A pendulum is \(0.80 \mathrm{m}\) long and the bob has a mass of 1.0 kg. At the bottom of its swing, the bob's speed is \(1.6 \mathrm{m} / \mathrm{s} .\) (a) What is the tension in the string at the bottom of the swing? (b) Explain why the tension is greater than the weight of the bob.
Objects that are at rest relative to the Earth's surface are in circular motion due to Earth's rotation. (a) What is the radial acceleration of an object at the equator? (b) Is the object's apparent weight greater or less than its weight? Explain. (c) By what percentage does the apparent weight differ from the weight at the equator? (d) Is there any place on Earth where a bathroom scale reading is equal to your true weight? Explain.
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