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Ch 10: Dynamics of Rotational Motion
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
Chapter 10, Problem 19c

A 2.20-kg hoop 1.20 m in diameter is rolling to the right without slipping on a horizontal floor at a steady 2.60 rad/s. Find the velocity vector of each of the following points, as viewed by a person at rest on the ground: (i) the highest point on the hoop; (ii) the lowest point on the hoop; (iii) a point on the right side of the hoop, midway between the top and the bottom.

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First, understand that the hoop is rolling without slipping, which means the linear velocity of the center of mass is equal to the tangential velocity at the rim due to rotation. The linear velocity \( v \) of the center of mass can be calculated using \( v = r \cdot \omega \), where \( r \) is the radius of the hoop and \( \omega \) is the angular velocity.
Calculate the radius \( r \) of the hoop. Since the diameter is given as 1.20 m, the radius \( r \) is half of the diameter: \( r = \frac{1.20}{2} \) m.
Determine the linear velocity \( v \) of the center of mass using the formula \( v = r \cdot \omega \). Substitute the values: \( v = \frac{1.20}{2} \cdot 2.60 \) rad/s.
For the highest point on the hoop, the velocity vector is the sum of the linear velocity of the center of mass and the tangential velocity due to rotation. The tangential velocity at the highest point is in the same direction as the linear velocity, so the total velocity is \( v_{highest} = v + r \cdot \omega \).
For the lowest point on the hoop, the tangential velocity due to rotation is in the opposite direction to the linear velocity. Therefore, the velocity vector at the lowest point is \( v_{lowest} = v - r \cdot \omega \). For the point on the right side, midway between the top and bottom, the tangential velocity is perpendicular to the linear velocity, so use vector addition to find \( v_{midway} = \sqrt{v^2 + (r \cdot \omega)^2} \).

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Key Concepts

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

Rolling Without Slipping

Rolling without slipping is a condition where an object rolls on a surface without any relative motion between the point of contact and the surface. This means the linear velocity of the center of mass is equal to the angular velocity times the radius. For a hoop, this ensures that the translational and rotational motions are synchronized, crucial for analyzing the velocity of different points on the hoop.
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Velocity of Points on a Rotating Object

The velocity of a point on a rotating object is determined by both its translational motion and its rotational motion. For a hoop rolling without slipping, the velocity at any point is the vector sum of the translational velocity of the center of mass and the tangential velocity due to rotation. This concept helps in calculating the velocity vectors for specific points on the hoop.
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Reference Frames

A reference frame is a perspective from which motion is observed and measured. In this problem, the observer is at rest on the ground, which means the velocities of points on the hoop are measured relative to the ground. Understanding reference frames is essential to correctly interpret the velocities of the points on the hoop as they appear to a stationary observer.
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Related Practice
Textbook Question

A 2.00-kg textbook rests on a frictionless, horizontal surface. A cord attached to the book passes over a pulley whose diameter is 0.150 m, to a hanging book with mass 3.00 kg. The system is released from rest, and the books are observed to move 1.20 m in 0.800 s. What is the tension in each part of the cord?

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Textbook Question

A solid ball is released from rest and slides down a hillside that slopes downward at 65.0° from the horizontal. In part (a), why did we use the coefficient of static friction and not the coefficient of kinetic friction?

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Textbook Question

A 2.20-kg hoop 1.20 m in diameter is rolling to the right without slipping on a horizontal floor at a steady 2.60 rad/s. Find the velocity vector for each of the points in part (c), but this time as viewed by someone moving along with the same velocity as the hoop.

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Textbook Question

A 15.0-kg bucket of water is suspended by a very light rope wrapped around a solid uniform cylinder 0.300 m in diameter with mass 12.0 kg. The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls 10.0 m to the water. With what speed does the bucket strike the water?

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Textbook Question

A 12.0-kg box resting on a horizontal, frictionless surface is attached to a 5.00-kg weight by a thin, light wire that passes over a frictionless pulley (Fig. E10.16). The pulley has the shape of a uniform solid disk of mass 2.00 kg and diameter 0.500 m. After the system is released, find the acceleration of the box.

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Textbook Question

A playground merry-go-round has radius 2.40 m2.40\(\text{ m}\) and moment of inertia 2100 kg m22100\(\text{ kg m}\)^2 about a vertical axle through its center, and it turns with negligible friction. A child applies an 18.0 N18.0\(\text{ N}\) force tangentially to the edge of the merry-go-round for 15.0 s15.0\(\text{ s}\). If the merry-go-round is initially at rest, what is its angular speed after this 15.0 s15.0\(\text{ s}\) interval?

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