Ch. 11 - Angular Momentum; General Rotation

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. 11 - Angular Momentum; General Rotation

Problem 79a

Chapter 11, Problem 79a

A particle of mass m uniformly accelerates as it moves counterclockwise along the circumference of a circle of radius R: $\vec{r}$ = î R cos θ + ĵ R sin θ with θ = ω₀t + (1/2)αt² , where the constants ω₀ and α are the initial angular velocity and angular acceleration, respectively. Determine the object’s tangential acceleration $\vec{a}$ _tan and determine the torque acting on the object using $\vec{τ} = \vec{r} \times F$ .

Verified step by step guidance

Step 1: Understand the problem. The particle is moving along a circular path with a given position vector $ \mathbf{r} = \hat{i} R \cos \theta + \hat{j} R \sin \theta $, where $ \theta = \omega_0 t + \frac{1}{2} \alpha t^2 $. The goal is to find the tangential acceleration $ \mathbf{a}_{\text{tan}} $ and the torque $ \mathbf{\tau} $ acting on the particle using $ \mathbf{\tau} = \mathbf{r} \times \mathbf{F} $.

Step 2: Calculate the tangential acceleration $ \mathbf{a}_{\text{tan}} $. The tangential acceleration is related to the angular acceleration $ \alpha $ by the formula $ a_{\text{tan}} = R \alpha $. Since $ \alpha $ is given as a constant, substitute it into the formula to express $ \mathbf{a}_{\text{tan}} $ as a vector in the tangential direction.

Step 3: Express the force $ \mathbf{F} $ acting on the particle. The force is related to the tangential acceleration by Newton's second law: $ \mathbf{F} = m \mathbf{a}_{\text{tan}} $. Substitute the expression for $ \mathbf{a}_{\text{tan}} $ into this equation to find $ \mathbf{F} $.

Step 4: Compute the torque $ \mathbf{\tau} $ using $ \mathbf{\tau} = \mathbf{r} \times \mathbf{F} $. Use the given position vector $ \mathbf{r} $ and the force vector $ \mathbf{F} $ to calculate the cross product. Recall that the cross product in two dimensions can be simplified using the determinant of a matrix.

Step 5: Simplify the expression for $ \mathbf{\tau} $. After performing the cross product, simplify the resulting expression to find the torque in terms of the given quantities $ m, R, \omega_0, \alpha, $ and $ t $. Ensure the units are consistent and the result is expressed in vector form.

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Key Concepts

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

Tangential Acceleration

Tangential acceleration refers to the rate of change of the tangential velocity of an object moving along a circular path. It is directly related to the angular acceleration (α) of the object and can be calculated using the formula a_tan = R * α, where R is the radius of the circle. This acceleration acts along the direction of the motion and is responsible for changing the speed of the particle as it moves along the circular path.

Torque

Torque is a measure of the rotational force applied to an object, which causes it to rotate about an axis. It is calculated using the cross product of the position vector (r) and the force vector (F), expressed as τ = r × F. The magnitude of torque depends on the angle between the position vector and the force, as well as the distance from the axis of rotation, making it a crucial concept in understanding rotational dynamics.

Angular Kinematics

Angular kinematics describes the motion of objects in rotational motion, analogous to linear kinematics for straight-line motion. It involves parameters such as angular displacement (θ), angular velocity (ω), and angular acceleration (α). The equations of motion for angular kinematics allow us to relate these quantities over time, such as θ = ω₀t + (1/2)αt², which is essential for analyzing the motion of the particle in the given problem.

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Related Practice

Textbook Question

The position of a particle with mass m traveling on a helical path (see Fig. 11–48) is given by $\vec{r}$ = R cos (2πz/d) î + R sin (2πz/d) ĵ + zk̂ where R and d are the radius and pitch of the helix, respectively, and z has time dependence z = v_𝓏t where v_𝓏 is the (constant) component of velocity in the z direction. Determine the time-dependent angular momentum $\vec{L}$ of the particle about the origin.

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Textbook Question

A boy rolls a tire along a straight level street. The tire has mass 8.0 kg, radius 0.32 m and moment of inertia about its central axis of symmetry of 0.83 kg·m². The boy pushes the tire forward away from him at a speed of 2.1 m/s and sees that the tire leans 12° to the right (Fig. 11–49). How will the resultant torque due to gravity and the normal force $\vec{F_{N}}$ affect the subsequent motion of the tire?

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Textbook Question

A radio transmission tower has a mass of 76 kg and is 12 m high. The tower is anchored to the ground by a flexible joint at its base, but it is secured by three cables 120° apart (Fig. 11–52). In an analysis of a potential failure, a mechanical engineer needs to determine the behavior of the tower if one of the cables breaks. The tower would fall away from the broken cable, rotating about its base. Determine the speed of the top of the tower as a function of the rotation angle θ. Start your analysis with the rotational dynamics equation of motion d $\vec{L}$ /dt = ${\vec{τ_{e x t}}}_{}$ . Approximate the tower as a tall thin rod.

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Textbook Question

Water drives a waterwheel (or turbine) of radius R = 3.0 m as shown in Fig. 11–50. The water enters at a speed v₁ = 7.0m/s and exits from the waterwheel at a speed v₂= 3.8 m/s. If the water causes the waterwheel to make one revolution every 6.0 s, how much power is delivered to the wheel?

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Textbook Question

A baseball bat has a sweet spot where a ball can be hit with almost effortless transmission of energy. A careful analysis of baseball dynamics shows that this special spot is located at the point where an applied force would result in pure rotation of the bat about the handle grip. Determine the location of the sweet spot, xₛ, of the bat shown in Fig. 11–53. The linear mass density of the bat is given roughly by (0.61 + 3.3x²) kg/m, where x is in meters measured from the end of the handle. The entire bat is 0.84 m long. The desired rotation point should be 5.0 cm from the thin end where the bat is held. [Hint: Where is the cm of the bat?]

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