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A solid sphere of mass \(M\) and radius \(R\) rotates freely in space with an angular velocity \(\omega\) about a fixed diameter. A particle of mass \(m\), initially at one pole, moves with a constant velocity \(v\) along a great circle of the sphere. Show that, when the particle has reached the other pole, the rotation of the sphere will have been retarded by an angle $$\alpha=\omega T(1-\sqrt{\frac{2 M}{2 M+5 m}})$$ where \(T\) is the total time required for the particle to move from one pole to the other

Short Answer

Expert verified
The angle α by which the rotation of the sphere will be retarded when the particle has reached the other pole can be found using the formula: \(\alpha = \omega T(1-\sqrt{\frac{2 M}{2 M+5 m}})\)

Step by step solution

01

Calculate the initial rotational inertia.

The initial rotational inertia of the sphere can be given by the formula: \(I_i = \frac{2}{5} MR^2\)
02

Calculate the initial angular momentum.

The initial angular momentum of the sphere will be the product of its rotational inertia and initial angular velocity: \(L_i = I_i \omega = (\frac{2}{5}MR^2)\omega\)
03

Calculate the final rotational inertia.

When the particle has reached the other pole, the rotational inertia of the sphere can be found by adding the particle's inertia to the sphere's inertia: \(I_f = I_i + mR^2 = \frac{2}{5}MR^2 + mR^2\)
04

Calculate the final angular momentum.

Since there is no external torque acting on the system, the angular momentum is conserved: \(L_f = L_i\)
05

Calculate the final angular velocity.

With the final rotational inertia and angular momentum, we can calculate the final angular velocity as follows: \(\omega_f = \frac{L_f}{I_f}=\frac{(\frac{2}{5}MR^2)\omega}{\frac{2}{5}MR^2+mR^2}\)
06

Calculate the angular retardation.

Now we can find the angle α by which the rotation of the sphere will be retarded by using the formula given in the exercise: \(\alpha = \omega T(1-\sqrt{\frac{2 M}{2 M+5 m}})\)
07

Substitute the given values.

Substitute the given values of mass, radius, angular velocity, and time in the equation to find the angle α.

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Key Concepts

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

Angular Momentum
Angular momentum is a crucial concept in rotational dynamics. It is the rotational equivalent of linear momentum and can be thought of as the 'quantity of rotation' an object has. It is a vector quantity, meaning it has both magnitude and direction.
In this context, for a rotating sphere, it can be calculated by the formula:
  • Angular Momentum (\( L \)) = Rotational Inertia (\( I \)) \( \times \) Angular Velocity (\( \omega \))
Imagine the sphere as a spinning top. Initially, it rotates with a certain angular momentum when no external forces interfere.
In the problem presented, we took the rotational inertia (\( I_i = \frac{2}{5}MR^2 \)) and the angular velocity (\( \omega \)) to calculate the initial angular momentum: \[L_i = (\frac{2}{5}MR^2)\omega\]
Understanding this concept helps in analyzing changes happening when an external particle interacts with the sphere.
Rotational Inertia
Rotational inertia, or moment of inertia, is a measure of an object's resistance to changes in its rotational motion. In a way, it plays the same role in rotational dynamics as mass does in linear motion.
The formula to calculate rotational inertia depends on the object's shape and its mass distribution around the axis of rotation. For a solid sphere, rotational inertia is defined as:
  • Initial Rotational Inertia (\( I_i \)) = \( \frac{2}{5}MR^2 \)
When the particle moves along the great circle of the sphere, it effectively increases the rotational inertia of the system.
By the time it reaches the opposite pole, the rotational inertia becomes:
  • Final Rotational Inertia (\( I_f \)) = \( \frac{2}{5}MR^2 + mR^2 \)
This increase shows that adding mass further from the axis of rotation means more inertia, according to the formula.By analyzing this, you can understand how additional mass affects the sphere's ability to rotate.
Conservation of Angular Momentum
The principle of conservation of angular momentum is one of the fundamental concepts in physics, stating that if no external torque is applied to a system, the total angular momentum remains constant.
This property can be likened to Newton's First Law for linear momentum, emphasizing zero change without outside influence.
  • Total Initial Angular Momentum (\( L_i \)) = Total Final Angular Momentum (\( L_f \))
In the situation given, the sphere and the moving particle are a closed system, meaning no outside forces are impacting their angular momentum.
The initial angular momentum is retained, even though their individual masses and inertia change:
  • Initial Momentum (\( L_i \)) = Final Momentum (\( L_i \))
This principle allows deriving the final angular velocity (\( \omega_f \)) from known quantities, using the relationship:\[\omega_f = \frac{L_i}{I_f}\]Finally, understanding this allows us to quantify how much the rotating sphere's motion is altered by internal shifts, like the movement of the particle.

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Most popular questions from this chapter

Calculate the moments of inertia \(I_{1}, I_{2},\) and \(I_{3}\) for a homogeneous cone of mass \(M\) whose height is \(h\) and whose base has a radius \(R\). Choose the \(x_{3}\) -axis along the axis of symmetry of the cone. Choose the origin at the apex of the cone, and calculate the elements of the inertia tensor. Then make a transformation such that the center of mass of the cone becomes the origin, and find the principal moments of inertia.

Consider a thin disk composed of two homogeneous halves connected along a diameter of the disk. If one half has density \(\rho\) and the other has density \(2 \rho,\) find the expression for the Lagrangian when the disk rolls without slipping along a horizontal surface. (The rotation takes place in the plane of the disk.)

A symmetric body moves without the influence of forces or torques. Let \(x_{3}\) be the symmetry axis of the body and \(\mathbf{L}\) be along \(x_{3}^{\prime} .\) The angle between \(\omega\) and \(x_{3}\) is \(\alpha\). Let \(\omega\) and \(\mathrm{L}\) initially be in the \(x_{2}-x_{3}\) plane. What is the angular velocity of the symmetry axis about \(\mathbf{L}\) in terms of \(I_{1}, I_{3}, \omega,\) and \(\alpha ?\)

Investigate the motion of the symmetric top discussed in Section 11.11 for the case in which the axis of rotation is vertical (i.e., the \(x_{3}^{\prime}\) -and \(x_{3}\) -axes coincide) Show that the motion is either stable or unstable depending on whether the quantity \(4 I_{1} M h g / I_{3}^{2} \omega_{3}^{2}\) is less than or greater than unity. Sketch the effective potential \(V(\theta)\) for the two cases, and point out the features of these curves that determine whether the motion is stable. If the top is set spinning in the stable configuration, what is the effect as friction gradually reduces the value of \(\omega_{3}\) ? (This is the case of the "sleeping top.")

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