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A size-5 soccer ball of diameter \(22.6 \mathrm{~cm}\) and mass \(426 \mathrm{~g}\)rolls up a hill without slipping, reaching a maximum height of \(5.00 \mathrm{~m}\) above the base of the hill. We can model this ball as a thin-walled hollow sphere. (a) At what rate was it rotating at the base of the hill? (b) How much rotational kinetic energy did it have then? Neglect rolling friction and assume the system's total mechanical energy is conserved.

Short Answer

Expert verified
The rate at which the soccer ball is rotating at the base of the hill is around 14.5 radians per second. The rotational kinetic energy it had at that moment is approximately 4.89 Joules.

Step by step solution

01

Identify the necessary variables and constants

The mass \(m\) of the soccer ball is 0.426 kg, its radius \(r\) is 0.113 m, the height \(h\) it reaches is 5.00 m, and the acceleration due to gravity \(g\) is \(9.81 m/s^2\). The moment of inertia \(I\) for a thin-walled hollow sphere is given by \(\frac{2}{3} m r^2\). We know energy is conserved, so the initial kinetic energy plus initial potential energy equals final kinetic energy plus final potential energy.
02

Solve for the angular velocity

At the base of the hill, the final potential energy is zero. At the top of the hill, the final kinetic energy is zero. We know that at the base of the hill, the pre-roll state, there was kinetic energy both due to translational and rotational motion. For a rolling object, the total kinetic energy is the sum of translational and rotational kinetic energy, given by \(KE_{total} = KE_{translational} + KE_{rotational} = \frac{1}{2}mv^2 + \frac{1}{2}Iw^2\), where \(v\) is velocity and \(w\) is angular velocity. Using energy conservation, \(\frac{1}{2}mv^2 + \frac{1}{2}Iw^2 = mgh\). We also use the rolling without sliding condition \(v = wr\), and plug this into the formula for total kinetic energy.
03

Calculate the angular velocity at the base of the hill

After replacing \(v = wr\) in the energy conservation equation and simplifying, we get \(w = \sqrt{\frac{15gh}{7r}}\).
04

Calculate the rotational kinetic energy at the base of the hill

The rotational kinetic energy is given by \(\frac{1}{2}Iw^2\). Plug the values of \(I\) and \(w\) from previous steps and find the rotational kinetic energy.

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Key Concepts

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

Angular Velocity
Angular velocity, symbolized by \( \omega \), is a measure of how quickly an object rotates or revolves around a fixed axis. For objects rolling without slipping, such as a soccer ball moving up a hill, angular velocity can be related to the linear velocity \( v \) by the equation \( v = \omega r \), where \( r \) is the radius of the object. This relationship shows that the linear velocity at the outer edge of the rolling object is equal to the product of its angular velocity and radius. It is crucial in understanding how rotational motion translates into linear motion and vice versa.

In the scenario of a soccer ball rolling up a hill, we can determine the angular velocity at the base of the hill by considering the conservation of mechanical energy, which accounts for both translational and rotational motion. Angular velocity is vital for calculating the rotational aspect of the kinetic energy, giving us insight into the ball's speed and rotational dynamics as it moves upward.
Moment of Inertia
The moment of inertia, often represented by \( I \), is like mass for rotational motion. It determines how difficult it is to change an object's rotation rate around a specific axis. The moment of inertia depends both on the mass of the object and the distribution of that mass relative to the axis of rotation.

For example, a thin-walled hollow sphere, such as the soccer ball mentioned in our problem, has a moment of inertia calculated using the formula \( I = \frac{2}{3}mr^{2} \). This information is essential when determining the rotational kinetic energy, as it contributes to how much work is required to get the soccer ball spinning or to slow it down. Different shapes and mass distributions have different formulas for calculating the moment of inertia, which becomes pivotal when comparing the rotational behaviors of varied objects.
Conservation of Mechanical Energy
The conservation of mechanical energy principle states that if no external forces (like friction or air resistance) do work on a system, the total mechanical energy (the sum of kinetic and potential energy) remains constant throughout the motion. This is a fundamental principle in physics that allows us to relate the speed and height of an object at different points in time without having to know all the details of the motion in between.

In the context of the soccer ball rolling up the hill, we assume that there are no external forces doing work, such as rolling friction. This assumption means that the mechanical energy at the base of the hill, when the ball is rotating quickly, will be the same as the mechanical energy at the top of the hill, despite the soccer ball coming to a stop. This principle simplifies the problem significantly and allows us to use the ball's height and mass to find its initial rotational kinetic energy and angular velocity.
Rolling Without Slipping
Rolling without slipping is a term that describes a particular type of motion where an object, like a wheel or a ball, rolls on a surface in such a way that the velocity of the point of contact with the ground is zero relative to the ground. This means there's no sliding or skidding as it rolls; every point on the rim contacts the surface momentarily without moving across it.

When an object rolls without slipping, its angular velocity \( \omega \) and its linear velocity \( v \) are related by the previously mentioned formula \( v = \omega r \). This formula is a result of the no-slip condition and allows us to relate the translational motion of the ball to its rotation. In the problem at hand, this condition helps us determine the angular velocity based on the ball's known linear displacement up the hill. It's a crucial concept for solving many real-world problems involving rolling motion, as it connects the rotation of the object with its path or trajectory.

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Most popular questions from this chapter

Example 10.7 discusses a uniform solid sphere rolling with- out slipping down a ramp that is at an angle \(\beta\) above the horizontal. Now consider the same sphere rolling without slipping up the ramp. (a) In terms of \(g\) and \(\beta\), calculate the acceleration of the center of mass of the sphere. Is your result larger or smaller than the acceleration when the sphere rolls down the ramp, or is it the same? (b) Calculate the friction force (in terms of \(M, g,\) and \(\beta\) ) for the sphere to roll without slipping as it moves up the incline. Is the result larger, smaller, or the same as the friction force required to prevent slipping as the sphere rolls down the incline?

The flywheel of an engine has moment of inertia \(1.60 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about its rotation axis. What constant torque is required to bring it up to an angular speed of 400 rev \(/ \min\) in 8.00 s, starting from rest?

A hollow, thin-walled sphere of mass \(12.0 \mathrm{~kg}\) and diameter \(48.0 \mathrm{~cm}\) is rotating about an axle through its center. The angle (in radians) through which it turns as a function of time (in seconds) is given by \(\theta(t)=A t^{2}+B t^{4},\) where \(A\) has numerical value 1.50 and \(B\) has numerical value \(1.10 .\) (a) What are the units of the constants \(A\) and \(B ?\) (b) At the time \(3.00 \mathrm{~s}\), find (i) the angular momentum of the sphere and (ii) the net torque on the sphere.

Block \(A\) rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block \(B\) is suspended from the free end of the rope. The light rope that connects the two blocks does not slip over the surface of the pulley (radius \(0.080 \mathrm{~m}\) ) because the pulley rotates on a frictionless axle. The horizontal surface on which block \(A\) (mass \(2.50 \mathrm{~kg}\) ) moves is frictionless. The system is released from rest, and block \(B\) (mass \(6.00 \mathrm{~kg}\) ) moves downward \(1.80 \mathrm{~m}\) in \(2.00 \mathrm{~s}\). (a) What is the tension force that the rope exerts on block \(B ?\) (b) What is the tension force on block \(A ?\) (c) What is the moment of inertia of the pulley for rotation about the axle on which it is mounted?

The mechanism shown in Fig. \(\mathbf{P} \mathbf{1 0 . 6 4}\) is used to raise a crate of supplies from a ship's hold. The crate has total mass \(50 \mathrm{~kg} .\) A rope is wrapped around a wooden cylinder that turns on a metal axle. The cylinder has radius \(0.25 \mathrm{~m}\) and moment of inertia \(I=2.9 \mathrm{~kg} \cdot \mathrm{m}^{2}\) about the axle. The crate is suspended from the free end of the rope. One end of the axle pivots on frictionless bearings; a crank handle is attached to the other end. When the crank is turned, the end of the handle rotates about the axle in a vertical circle of radius \(0.12 \mathrm{~m},\) the cylinder turns, and the crate is raised. What magnitude of the force \(\vec{F}\) applied tangentially to the rotating crank is required to raise the crate with an acceleration of \(1.40 \mathrm{~m} / \mathrm{s}^{2} ?\) (You can ignore the mass of the rope as well as the moments of inertia of the axle and the crank.)

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