Ch. 10 - Rotational Motion

Giancoli Douglas - Physics for Scientists and Engineers 5th edition

Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Not the one you use?Change textbook

Giancoli Douglas 5th edition

Ch. 10 - Rotational Motion

Problem 84

Chapter 10, Problem 84

A cyclist accelerates from rest at a rate of 1.00 m/s². How fast will a point at the top of the rim of the tire (diameter = 68.0 cm) be moving after 2.75 s? [Hint: At any moment, the lowest point on the tire is in contact with the ground and is at rest—see Fig. 10–69.]

Verified step by step guidance

Step 1: Identify the given values and the quantities to find. The cyclist accelerates from rest, so the initial velocity (v₀) is 0 m/s. The acceleration (a) is 1.00 m/s², and the time (t) is 2.75 s. The diameter of the tire is 68.0 cm, which can be converted to meters (0.68 m). The goal is to find the speed of a point at the top of the rim of the tire after 2.75 s.

Step 2: Calculate the linear velocity of the center of the wheel after 2.75 s using the kinematic equation for velocity: v = v₀ + at. Since v₀ = 0, the equation simplifies to v = at. Substitute the given values for acceleration and time to find the linear velocity of the center of the wheel.

Step 3: Understand the motion of the point at the top of the rim. The top of the rim has a velocity relative to the ground that is the sum of the linear velocity of the center of the wheel and the rotational velocity of the rim. This is because the top of the rim moves forward faster than the center due to the wheel's rotation.

Step 4: Relate the rotational velocity of the rim to the linear velocity of the center. The rotational velocity at the rim is equal to the linear velocity of the center because the wheel rolls without slipping. Therefore, the velocity of the top of the rim relative to the ground is 2v, where v is the linear velocity of the center.

Step 5: Combine the results. Multiply the linear velocity of the center (v) by 2 to find the velocity of the top of the rim relative to the ground. This gives the final answer for how fast the point at the top of the rim is moving after 2.75 s.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Linear Acceleration

Linear acceleration refers to the rate of change of velocity of an object over time. In this scenario, the cyclist accelerates from rest at a constant rate of 1.00 m/s². This means that for every second, the cyclist's speed increases by 1.00 m/s, which is crucial for calculating the final speed after a given time.

Rotational Motion

Rotational motion describes the movement of an object around a central axis. In this case, the tire of the bicycle rotates as the cyclist moves forward. The relationship between linear speed and rotational speed is important, as the point on the rim of the tire will have a linear speed that depends on both the angular velocity and the radius of the tire.

Kinematics Equations

Kinematics equations are mathematical formulas that describe the motion of objects. They relate displacement, initial velocity, final velocity, acceleration, and time. For this problem, the equation v = u + at (where v is final velocity, u is initial velocity, a is acceleration, and t is time) will be used to determine the speed of the cyclist after 2.75 seconds, which can then be translated to the speed of the point on the tire.

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Related Practice

Textbook Question

The 1100-kg mass of a car includes four tires, each of mass 35 kg (including wheels) and diameter 0.80 m. Assume each tire and wheel combination acts as a solid cylinder. Determine the total kinetic energy of the car when traveling 95 km/h.

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Textbook Question

Bicycle gears:How is the angular velocity ωᵣ of the rear wheel of a bicycle related to the angular velocity ωբ of the front sprocket and pedals? Let Nբ and Nᵣ be the number of teeth on the front and rear sprockets, respectively, Fig. 10–71, and Rբ and Rᵣ their respective radii. The teeth are spaced the same on both sprockets and the rear sprocket is firmly attached to the rear wheel. Evaluate the ratio ωᵣ / ωբ when the front and rear sprocketshave 42 and 28 teeth.

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Textbook Question

How is the angular velocity ωᵣ of the rear wheel of a bicycle related to the angular velocity ωբ of the front sprocket and pedals? Let Nբ and Nᵣ be the number of teeth on the front and rear sprockets, respectively, Fig. 10–71, and Rբ and Rᵣ their respective radii. The teeth are spaced the same on both sprockets and the rear sprocket is firmly attached to the rear wheel.

On a 12.0-cm-diameter audio compact disc (CD), digital bits of information are encoded sequentially along an outward spiraling path. The spiral starts at radius R₁ = 2.5 cm and winds its way out to radius R₂ = 5.8 cm. To read the digital information, a CD player rotates the CD so that the player’s readout laser scans along the spiral’s sequence of bits at a constant linear speed of 1.25 m/s. Thus the player must accurately adjust the rotational frequency ƒ of the CD as the laser moves outward. Determine the values for ƒ (in units of rpm) when the laser is located at R₁ and when it is at R₂.

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Textbook Question

The 1100-kg mass of a car includes four tires, each of mass 35 kg (including wheels) and diameter 0.80 m. Assume each tire and wheel combination acts as a solid cylinder. Determine the fraction of the kinetic energy in the tires and wheels.

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