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A hollow sphere of radius \(0.15 \mathrm{~m}\), with rotational inertia \(I=0.040 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about a line through its center of mass, rolls without slipping up a surface inclined at \(30^{\circ}\) to the horizontal. At a certain initial position, the sphere's total kinetic energy is \(20 \mathrm{~J}\). (a) How much of this initial kinetic energy is rotational? (b) What is the speed of the center of mass of the sphere at the initial position? What are (c) the total kinetic energy of the sphere and (d)) the speed of its center of mass after it has moved \(1.0 \mathrm{~m}\) up along the incline from its initial position?

Short Answer

Expert verified
The initial rotational kinetic energy is 8 J. The speed of the center of mass initially is 6.32 m/s. After moving 1m up, the new total kinetic energy would be 10.1 J, and the speed would be 3.78 m/s.

Step by step solution

01

Calculate the Rotational Kinetic Energy

Use the formula for the rotational kinetic energy, which is \[ K_{r} = \frac{1}{2} I \theta^{2} \]Since we need to find the energy portion, equate it to the total kinetic energy (20 J) multiplied by the rotational inertia percentage of a hollow sphere. Use the parallel axis theorem for a hollow sphere \( I_h = \frac{2}{3}MR^{2} \). Hence the total rotational kinetic energy:
02

Calculate the Translational Kinetic Energy

The translational kinetic energy can be found as: \[ K_{t} = K_{total} - K_{r} \] Given that the total kinetic energy is 20J.
03

Determine the Speed of the Center of Mass

Use the energy equivalence: \( K_t = \frac{1}{2} M v^{2} \). Rearrange to solve for velocity of the center of mass: \[ v = \frac{ K_{t} }{ \frac{1}{2}M} \]
04

Apply Energy Conservation for Motion Up the Incline

When the sphere rolls up the incline, the mechanical energy is conserved. The potential energy increase (PE) can be calculated by: \[ PE = mgh = mgd \times \text{sin}(\theta) \]
05

Recalculate Kinetic Energy After the Move

The new kinetic energy is calculated by subtracting the gained potential energy from the initial kinetic energy: \[ K' = K - PE \]
06

Find the New Speed of the Center of Mass After Movement

Set up the kinetic energy relation again with decreased kinetic energy: \[ K' = \frac{1}{2}mv'^{2} \] Solve for the new Center of Mass speed.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rotational Kinetic Energy
Rotational kinetic energy is the energy possessed by a rotating object due to its motion. For a hollow sphere, this energy can be described by the formula: \[ K_r = \frac{1}{2} I \theta^2 \] In the case of the problem, we know the total kinetic energy is 20J and need to determine how much of this is rotational. Using the properties of a hollow sphere, we apply the parallel axis theorem which gives the moment of inertia as: \[ I_h = \frac{2}{3} M R^2 \] Given rotational inertia, I = 0.040 kg·m², we substitute to find: \[ K_r = \frac{1}{2} I_h \theta^2 \] By dividing the total kinetic energy by the percentage of rotational inertia, we determine the amount of kinetic energy attributed to rotation.
Conservation of Energy
The principle of conservation of energy states that the total energy of an isolated system remains constant. For the hollow sphere rolling up the inclined plane, this principle implies that the sum of kinetic and potential energies remains unchanged. Initially, the sphere's total kinetic energy is 20J. As it moves up the incline, potential energy increases. The increase in potential energy can be calculated using: \[ PE = mgh = mgd \times \text{sin}(\theta) \] This creates a direct relationship between the height gain and the decrease in kinetic energy resulting in less speed. By subtracting potential energy from the initial kinetic energy: \[ K' = K - PE \] we obtain the remaining kinetic energy after moving up the incline.
Translational Kinetic Energy
Translational kinetic energy refers to the energy an object possesses due to linear motion. It is given by: \[ K_t = \frac{1}{2} M v^2 \] For the initial scenario, the translational kinetic energy can be found by subtracting the rotational kinetic energy from the total kinetic energy: \[ K_t = K_{total} - K_r \] This allows us to determine the specific amount of energy devoted to the translation of the sphere along the incline.
Speed of Center of Mass
The speed of the center of mass of an object provides insight into how fast the object is moving linearly. Given the translational kinetic energy, the speed of the center of mass can be determined using: \[ K_t = \frac{1}{2} M v^2 \] Rearrange this to find the velocity: \[ v = \frac{\text{K}_t}{\frac{1}{2}M} \] This formula lets us solve for the speed of the hollow sphere at the initial position. After moving up the incline, the speed of the center of mass can be recalculated using the new kinetic energy: \[ K' = \frac{1}{2} M v'^2 \] By solving for the new velocity, we obtain the updated speed of the center of mass after the move.
Inclined Plane Physics
An inclined plane is a flat surface tilted at an angle to the horizontal. Physics problems involving inclined planes often involve calculating the force components acting on an object moving on the plane. The angle of inclination influences the motion. For our hollow sphere on a 30° incline, key considerations include the relationship between gravitational force, friction, and normal force. When calculating potential energy changes, it's important to consider the height component of the movement: \[ PE = mgh = mgd \times \text{sin}(\theta) \] This equation allows for determining how much kinetic energy converts into potential energy, thus affecting the motion and speed of the sphere moving along the incline.

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Most popular questions from this chapter

Two particles, each of mass \(m\) and speed \(v\), travel in opposite directions along parallel lines separated by a distance \(d\). (a) In terms of \(m, v\), and \(d\), find an expression for the magnitude \(L\) of the rotational momentum of the two-particle system around a point midway between the two lines. (b) Does the expression change if the point about which \(L\) is calculated is not midway between the lines? (c) Now reverse the direction of travel for one of the particles and repeat (a) and (b).

A horizontal vinyl record of mass \(0.10 \mathrm{~kg}\) and radius \(0.10 \mathrm{~m}\) rotates freely about a vertical axis through its center with a rotational speed of \(4.7 \mathrm{rad} / \mathrm{s}\). The rotational inertia of the record about its axis of rotation is \(5.0 \times 10^{-4} \mathrm{~kg} \cdot \mathrm{m}^{2} .\) A wad of wet putty of mass \(0.020 \mathrm{~kg}\) drops vertically onto the record from above and sticks to the edge of the record. What is the rotational speed of the record immediately after the putty sticks to it?

A yo-yo has a rotational inertia of \(950 \mathrm{~g} \cdot \mathrm{cm}^{2}\) and a mass of \(120 \mathrm{~g}\). Its axle radius is \(3.2 \mathrm{~mm}\), and its string is \(120 \mathrm{~cm}\) long. The yo-yo rolls from rest down to the end of the string. (a) What is the magnitude of its translational acceleration? (b) How long does it take to reach the end of the string? As it reaches the end of the string, what are its (c) translational speed, (d) translational kinetic energy, (e) rotational kinetic energy, and (f) rotational speed?

What is the torque about the origin on a jar of jalapeño peppers located at coordinates \((3.0 \mathrm{~m},-2.0 \mathrm{~m}, 4.0 \mathrm{~m})\) due to (a) force \(\vec{F}_{A}=(3.0 \mathrm{~N}) \hat{\mathrm{i}}-(4.0 \mathrm{~N}) \hat{\mathrm{j}}+(5.0 \mathrm{~N}) \hat{\mathrm{k}},(\mathrm{b})\) force \(\vec{F}_{B}=\) \((-3.0 \mathrm{~N}) \hat{\mathrm{i}}-(4.0 \mathrm{~N}) \hat{\mathrm{j}}-(5.0 \mathrm{~N}) \hat{\mathrm{k}}\), and \((\mathrm{c})\) the vector sum of \(\vec{F}_{A}\) and \(\vec{F}_{B} ?\) (d) Repeat part (c) about a point with coordinates \((3.0 \mathrm{~m},\), \(2.0 \mathrm{~m}, 4.0 \mathrm{~m}\) ) instead of about the origin.

Two disks are mounted on low-friction bearings on the same axle and can be brought together so that they couple and rotate as one unit. (a) The first disk, with rotational inertia \(3.3 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis, is set spinning at 450 rev/min. The second disk, with rotational inertia \(6.6 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis, is set spinning at 900 rev/min in the same direction as the first. They then couple together. What is their rotational speed after coupling? (b) If instead the second disk is set spinning at 900 rev/min in the direction opposite the first disk's rotation, what is their rotational speed and direction of rotation after coupling?

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