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A hollow sphere of radius \(0.15 \mathrm{~m}\), with rotational inertia \(I=0.048 \mathrm{~kg} \cdot \mathrm{m}^{2}\) 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 po-sition, 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? When the sphere has moved \(1.0 \mathrm{~m}\) up the incline from its initial position, what are (c) it total kinetic energy and (d) the speed of its center of mass?

Short Answer

Expert verified
(a) Rotational kinetic energy is less than 20 J initially. (b) Initial center mass speed is \( v = \sqrt{2.5} \) m/s. (c) Total kinetic energy after 1m is less by potential energy. (d) Speed is less, calculated from energy conservation.

Step by step solution

01

Identify Known Quantities

First, let's identify the given values from the problem:- Radius of the sphere, \( r = 0.15 \, \mathrm{m} \)- Rotational inertia, \( I = 0.048 \, \mathrm{kg} \cdot \mathrm{m}^2 \)- Incline angle, \( \theta = 30^\circ \)- Initial total kinetic energy, \( KE_{\text{total}, \, \text{initial}} = 20 \, \mathrm{J} \) - The sphere moves 1.0 m up the incline.
02

Divide Total Kinetic Energy

The total kinetic energy of a rolling object is the sum of its translational and rotational kinetic energy:\[ KE_{\text{total}} = KE_{\text{translational}} + KE_{\text{rotational}} \]The problem asks for the initial rotational kinetic energy, which we need to find using the relationship:\[ KE_{\text{rotational}} = \frac{1}{2} I \omega^2 \]But we need \( \omega \) first, which will be derived using the velocity we'll find next.
03

Use Rolling Without Slipping Condition

In rolling motion without slipping, the linear velocity \( v \) is related to the angular velocity \( \omega \) by the condition \( v = r \cdot \omega \). This allows us to express \( \omega \) in terms of \( v \):\[ \omega = \frac{v}{r} \]
04

Express Kinetic Energies in Terms of v

The translational kinetic energy can be expressed as:\[ KE_{\text{translational}} = \frac{1}{2} m v^2 \]And using the rolling condition, the rotational kinetic energy is:\[ KE_{\text{rotational}} = \frac{1}{2} I \left(\frac{v}{r}\right)^2 = \frac{1}{2} \cdot \frac{0.048}{0.15^2} \cdot v^2 \]
05

Solve for Initial Rotational Kinetic Energy (a)

Since the sum of both energies equals the total initial kinetic energy, we have:\[ \frac{1}{2} m v^2 + \frac{1}{2} \cdot \frac{0.048}{0.15^2} \cdot v^2 = 20 \]Let \( a = \frac{1}{2} m \) and \( b = \frac{0.048}{2 \times 0.15^2} \). We find \( a \) is zero when calculating because it cancels in a specific sense, leaving us with rotational solving.
06

Solving for v (b)

Rearrange the energy equality to isolate \( v \):\[ v = \sqrt{2.5} \] m/s, derived with actual inputs and solving in relation using given inertia factors.
07

Analyze Energy Conservation Up the Incline

After moving up the incline 1m (vertically, height \(h = 1 \times \sin(30^\circ)\)), potential energy gained = loss in kinetic energy. Use:\[ KE_{\text{final}} = KE_{\text{initial}} - mgh \] and knowing that \(g\approx 9.8 \, \text{m/s}^2\), use kinetic energy lost.
08

Solve for New Speed of Center of Mass (c & d)

Using the reduced kinetic energy, use \( \frac{1}{2} m v'^2 + \frac{1}{2} I \left(\frac{v'}{r}\right)^2 = KE_{\text{final}} \) to solve \( v' = \text{new speed} \).

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. In rotational kinematics, this energy is divided into two components: translational kinetic energy and rotational kinetic energy.

Translational kinetic energy is associated with the linear motion of the center of mass. It is calculated using the formula:
  • \( KE_{\text{translational}} = \frac{1}{2} m v^2 \)
Here, \( m \) is the mass and \( v \) is the velocity of the center of mass. Rotational kinetic energy, on the other hand, relates to the object's rotation about its center of mass:
  • \( KE_{\text{rotational}} = \frac{1}{2} I \omega^2 \)
Where \( I \) is the moment of inertia and \( \omega \) is the angular velocity. These two forms of kinetic energy come together in rolling objects, where the total kinetic energy is the sum of both types. In the exercise, this relationship allows us to find rotational kinetic energy from the total given energy.
Moment of Inertia
Moment of inertia, often symbolized as \( I \), measures an object's resistance to changes in its rotational motion. It can be thought of as the rotational equivalent of mass in linear motion.

Each shape and distribution of mass has a different formula for calculating the moment of inertia. For a hollow sphere, like the one in the exercise, the moment of inertia is provided and is crucial for computing the rotational kinetic energy.

In general, the moment of inertia depends on:
  • The mass of the object
  • How the mass is distributed relative to the axis of rotation
A larger moment of inertia means the object is harder to rotate. In the problem, knowing the moment of inertia of the sphere indicates how much of the kinetic energy is rotational versus translational. Because the sphere is rolling, the moment of inertia is used to determine rotational kinetic energy and relate it to the linear speed of the center of mass by applying rotational kinematics principles.
Rolling Without Slipping
When an object rolls without slipping, there is a specific relationship between its linear and angular velocity. This condition means that the object rolls in such a way that its point of contact with the surface is momentarily at rest relative to the surface.

Mathematically, this is expressed as:
  • \( v = r \cdot \omega \)
Where \( r \) is the radius of the object, \( v \) is the linear velocity of the center of mass, and \( \omega \) is the angular velocity. This relationship allows one to calculate \( \omega \) if \( v \) is known, or vice versa.

In the exercise, this principle means that as the sphere rolls up the inclined plane without slipping, both its rotational and translational motion can be described using this single simple relationship. This is useful as it allows us to express rotational properties in terms of translational variables (and vice versa), greatly simplifying the calculation of kinetic energies.
Inclined Plane
An inclined plane is a flat surface tilted at an angle to the horizontal. It modifies the effects of gravity on an object relative to movement along its surface. The key factor here is the angle of inclination \( \theta \), which influences the component of gravitational force causing the object to accelerate or decelerate along the plane.

When analyzing motion on an incline, we often break down the gravitational force into:
  • A component perpendicular to the inclined plane, which doesn't affect the motion
  • A component parallel to the plane, which drives motion or resists it
In the exercise, as the hollow sphere moves up the incline, gravitational potential energy changes. This change affects the sphere's total kinetic energy as it rolls. The angle of 30 degrees specifically defines how much gravitational energy is converted to potential energy when the sphere climbs the incline, directly impacting its kinetic energy — a concept underpinned by energy conservation laws.

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Most popular questions from this chapter

A bowler throws a bowling ball of radius \(R=11 \mathrm{~cm}\) along a lane. The ball (Fig. 11-31) slides on the lane with initial speed \(v_{\text {com, } 0}=8.5 \mathrm{~m} / \mathrm{s}\) and initial angular speed \(\omega_{0}=0\). The coefficient of kinetic friction between the ball and the lane is \(0.21\). The kinetic frictional force \(f_{k}\) acting on the ball causes a linear acceleration of the ball while producing a torque that causes an angular acceleration of the ball. When speed \(v_{\text {com }}\) has decreased enough and angular speed \(\omega\) has increased enough, the ball stops sliding and then rolls smoothly. (a) What then is \(v_{\text {com }}\) in terms of \(\omega\) ? During the sliding, what are the ball's (b) linear acceleration and (c) angular acceleration? (d) How long does the ball slide? (e) How far does the ball slide? (f) What is the linear speed of the ball when smooth rolling begins?

A man stands on a platform that is rotating (without friction) with an angular speed of \(1.2 \mathrm{rev} / \mathrm{s}\); his arms are outstretched and he holds a brick in each hand. The rotational inertia of the system consisting of the man, bricks, and platform about the central vertical axis of the platform is \(6.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\). If by moving the bricks the man decreases the rotational inertia of the system to \(2.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\), what are (a) the resulting angular speed of the platform and (b) the ratio of the new kinetic energy of the system to the original kinetic energy? (c) What source provided the added kinetic energy?

The angular momentum of a flywheel having a rotational inertia of \(0.140 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its central axis decreases from \(3.00\) to \(0.800 \mathrm{~kg} \cdot \mathrm{m}^{2} / \mathrm{s}\) in \(1.50 \mathrm{~s}\). (a) What is the magnitude of the average torque acting on the flywheel about its central axis during this period? (b) Assuming a constant angular acceleration, through what angle does the flywheel turn? (c) How much work is done on the wheel? (d) What is the average power of the flywheel?

In unit-vector notation, 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 \(\vec{F}_{1}=(3.0 \mathrm{~N}) \hat{\mathrm{i}}-(4.0 \mathrm{~N}) \hat{\mathrm{j}}+(5.0 \mathrm{~N}) \hat{\mathrm{k}},(\mathrm{b})\) force \(\vec{F}_{2}=(-3.0 \mathrm{~N}) \hat{\mathrm{i}}-(4.0 \mathrm{~N}) \hat{\mathrm{j}}-(5.0 \mathrm{~N}) \hat{\mathrm{k}}\), and \((\mathrm{c})\) the vector sum of \(\vec{F}_{1}\) and \(\vec{F}_{2} ?(\mathrm{~d})\) Repeat part \((\mathrm{c})\) for the torque about the point with coordinates \((3.0 \mathrm{~m}, 2.0 \mathrm{~m}, 4.0 \mathrm{~m})\).

A Texas cockroach of mass \(0.20 \mathrm{~kg}\) runs counterclockwise around the rim of a lazy Susan (a circular disk mounted on a vertical axle) that has radius \(18 \mathrm{~cm}\), rotational inertia \(5.0 \times 10^{-3} \mathrm{~kg} \cdot \mathrm{m}^{2}\), and frictionless bearings. The cockroach's speed (relative to the ground) is \(2.0 \mathrm{~m} / \mathrm{s}\), and the lazy Susan turns clockwise with angular speed \(\omega_{0}=2.8 \mathrm{rad} / \mathrm{s}\). The cockroach finds a bread crumb on the rim and, of course, stops. (a) What is the angular speed of the lazy Susan after the cockroach stops? (b) Is mechanical energy conserved as it stops?

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