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A solid uniform sphere and a thin-walled, hollow sphere have the same mass \(M\) and radius \(R .\) If they roll without slipping up a ramp that is inclined at an angle \(\beta\) above the horizontal and if both have the same \(v_{\mathrm{cm}}\) before they start up the incline, calculate the maximum height above their starting point reached by each object. Which object reaches the greater height, or do both of them reach the same height?

Short Answer

Expert verified
Both solid and hollow spheres reach the same height. This is because the kinetic energy is the same for both before they start up the incline (given by the problem), even though their masses and moments of inertia are different. Since all kinetic energy is converted to potential energy, both will reach the same maximum height.

Step by step solution

01

Identify the variables and knowns

In this problem, the knowns are the mass of the spheres \(M\), the radius \(R\), and the speed of the centre of the sphere \(v_{\text{cm}}\) before starting up the incline.
02

Determine the moment of inertia and initial kinetic energy for both type of spheres

The moment of inertia \(I\) for a hollow sphere is given by \(I_{\text{hollow}} = \frac{2}{3}MR^2\) and for solid sphere it's \(I_{\text{solid}} = \frac{2}{5}MR^2\). The kinetic energy before starting up the incline is shared between translation and rotation and is given by \(K.E_{\text{initial}} = \frac{1}{2}Mv_{\text{cm}}^2 + \frac{1}{2} I \omega^2\), where \(\omega = v_{\text{cm}}/R\) is the angular velocity.
03

Determine the potential energy at the maximum height for both type of spheres

As the spheres are rolling without slipping and assuming there's no friction, all of the initial kinetic energy is converted into potential energy at the maximum height. So we can equate the initial kinetic energy to \(MgH\), where \(H\) is the maximum height.
04

Write the equation in terms of \(H\) and solve

We equate initial kinetic energy to \(MgH\): \(MgH = K.E_{\text{initial}}\) to find \(H_{\text{hollow}}\) and \(H_{\text{solid}}\). Comparing the two heights will answer which sphere reaches higher.

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Key Concepts

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

Moment of Inertia
Moment of inertia, often symbolized as "I", represents the resistance of a body to angular acceleration when a torque is applied. It depends on the mass distribution of the object around its axis of rotation. For objects with the same mass and shape, like in our exercise, the distribution of that mass can alter their moment of inertia.
In our problem, we have two spheres: a hollow one and a solid one. For the solid sphere, the moment of inertia is calculated as \( I_{\text{solid}} = \frac{2}{5}MR^2 \), and for the hollow sphere, it is \( I_{\text{hollow}} = \frac{2}{3}MR^2 \).
Both these formulas indicate that the hollow sphere has a higher moment of inertia compared to the solid one. This means the hollow sphere has more mass distributed far from its axis of rotation, making it less willing to change its rotational motion.
Understanding moment of inertia is crucial because it directly affects how objects behave when they roll on inclined planes, influencing their kinetic energy dynamics.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. It has two components for rolling motion: translational kinetic energy and rotational kinetic energy.
Translational kinetic energy is given by \( \frac{1}{2} M v_{\text{cm}}^2 \) where \( M \) is the mass and \( v_{\text{cm}} \) is the linear speed of the center of mass. Rotational kinetic energy is \( \frac{1}{2} I \omega^2 \), with \( I \) being the moment of inertia and \( \omega \) the angular velocity \( \left( \omega = \frac{v_{\text{cm}}}{R} \right) \).
In our exercise, both spheres convert the sum of these kinetic energies into potential energy as they roll up the inclined plane. The distinct moments of inertia of the hollow and solid spheres mean their initial distribution of kinetic energy differs, influencing how much they can convert into potential energy later.
Potential Energy
Potential energy in this context refers to the gravitational potential energy gained as an object climbs an incline. It's calculated using the formula \( PE = M g H \), where \( H \) is the height above the starting point, \( g \) is the acceleration due to gravity, and \( M \) is the mass of the object.
As the spheres roll up the inclined plane, the initial kinetic energy is converted to potential energy. This energy conservation principle allows us to set \( MgH = \text{Initial Kinetic Energy} \) to determine how high each sphere can ascend before their motion stops.
The maximum height achieved by each sphere is deeply influenced by their initial kinetic energy configuration, impacted by their respective moments of inertia.
Inclined Plane
An inclined plane is a flat surface tilted at an angle to the horizontal. It's a fundamental concept in physics, helping to study forces and motion easily.
In our exercise, both spheres roll up an inclined plane at an angle of \( \beta \). As they move, the gravitational force acts on them, trying to pull them back down the incline. The incline affects how kinetic energy is converted into potential energy.
Without slipping, a frictional force that allows the spheres to roll without sliding plays a crucial role. This involves complex dynamics beyond simple sliding objects, as rolling incorporates rotational aspects tied to each sphere's moment of inertia.
  • For spheres with the same mass and initial speed, how high they go depends on how their kinetic energy transitions to potential energy.
  • An inclined plane acts as an essential tool in understanding how different shapes and mass distributions behave under gravity.
Altogether, these principles showcase the fascinating ways physics governs objects in motion.

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Most popular questions from this chapter

Two spheres are rolling without slipping on a horizontal floor. They are made of different materials, but each has mass \(5.00 \mathrm{~kg}\) and radius \(0.120 \mathrm{~m} .\) For each the translational speed of the center of mass is \(4.00 \mathrm{~m} / \mathrm{s}\). Sphere \(A\) is a uniform solid sphere and sphere \(B\) is a thin-walled, hollow sphere. How much work, in joules, must be done on each sphere to bring it to rest? For which sphere is a greater magnitude of work required? Explain. (The spheres continue to roll without slipping as they slow down.

A block with mass \(m\) is revolving with linear speed \(v_{1}\) in a circle of radius \(r_{1}\) on a frictionless horizontal surface (see Fig. E10.42). The string is slowly pulled from below until the radius of the circle in which the block is revolving is reduced to \(r_{2}\). (a) Calculate the tension \(T\) in the string as a function of \(r\), the distance of the block from the hole. Your answer will be in terms of the initial velocity \(v_{1}\) and the radius \(r_{1} .\) (b) Use \(W=\int_{r_{1}}^{r_{2}} \overrightarrow{\boldsymbol{T}}(r) \cdot d \overrightarrow{\boldsymbol{r}}\) to calculate the work done by \(\vec{T}\) when \(r\) changes from \(r_{1}\) to \(r_{2}\). (c) Compare the results of part (b) to the change in the kinetic energy of the block.

A \(12.0 \mathrm{~kg}\) box resting on a horizontal, frictionless surface is attached to a \(5.00 \mathrm{~kg}\) 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 \(2.00 \mathrm{~kg}\) and diameter \(0.500 \mathrm{~m} .\) After the system is released, find (a) the tension in the wire on both sides of the pulley, (b) the acceleration of the box, and (c) the horizontal and vertical components of the force that the axle exerts on the pulley.

When an object is rolling without slipping, the rolling friction force is much less than the friction force when the object is sliding; a silver dollar will roll on its edge much farther than it will slide on its flat side (see Section 5.3 ). When an object is rolling without slipping on a horizontal surface, we can approximate the friction force to be zero, so that \(a_{x}\) and \(\alpha_{z}\) are approximately zero and \(v_{x}\) and \(\omega_{z}\) are approximately constant. Rolling without slipping means \(v_{x}=r \omega_{z}\) and \(a_{x}=r \alpha_{z}\). If an object is set in motion on a surface without these equalities, sliding (kinetic) friction will act on the object as it slips until rolling without slipping is established. A solid cylinder with mass \(M\) and radius \(R\), rotating with angular speed \(\omega_{0}\) about an axis through its center, is set on a horizontal surface for which the kinetic friction coefficient is \(\mu_{\mathrm{k}}\). (a) Draw a free-body diagram for the cylinder on the surface. Think carefully about the direction of the kinetic friction force on the cylinder. Calculate the accelerations \(a_{x}\) of the center of mass and \(\alpha_{z}\) of rotation about the center of mass. (b) The cylinder is initially slipping completely, so initially \(\omega_{z}=\omega_{0}\) but \(v_{x}=0\) Rolling without slipping sets in when \(v_{x}=r \omega_{z} .\) Calculate the distance the cylinder rolls before slipping stops. (c) Calculate the work done by the friction force on the cylinder as it moves from where it was set down to where it begins to roll without slipping.

A solid cylinder with radius \(0.140 \mathrm{~m}\) is mounted on a frictionless, stationary axle that lies along the cylinder axis. The cylinder is initially at rest. Then starting at \(t=0\) a constant horizontal force of \(3.00 \mathrm{~N}\) is applied tangential to the surface of the cylinder. You measure the angular displacement \(\theta-\theta_{0}\) of the cylinder as a function of the time \(t\) since the force was first applied. When you plot \(\theta-\theta_{0}\) (in radians) as a function of \(t^{2}\left(\right.\) in \(\left.\mathrm{s}^{2}\right),\) your data lie close to a straight line. If the slope of this line is \(16.0 \mathrm{rad} / \mathrm{s}^{2},\) what is the moment of inertia of the cylinder for rotation about the axle?

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