Question

A sphere of radius \(a\) is rolling without slipping on a rough horizontal plane in such a way that its centre traces out a horizontal circle, radius \(b\) and centre \(O\), with constant angular speed \(\Omega\). Let \((i, j, k)\) be an orthonormal triad with \(k\) vertical and \(i\) in the direction from \(O\) to the centre of the sphere. Show that the angular velocity of the sphere relative to the plane satisfies $$ \boldsymbol{\omega}=n \boldsymbol{k}-\frac{b}{a} \Omega i $$ where \(n=\boldsymbol{\omega} \cdot \boldsymbol{k}\). Show that if \(n\) is constant, then the locus of the point of contact on the sphere is a circle. What is its radius?

Short answer

A: The expression for the angular velocity of the sphere is $\boldsymbol{\omega}=n \boldsymbol{k}-\frac{b}{a} \Omega \boldsymbol{i}$. Q: What does the position vector of the point of contact always satisfy for the sphere to be tangent to the plane? A: The position vector of the point of contact always satisfies the condition $\boldsymbol{C} \cdot \boldsymbol{\omega} = 0$ for the sphere to be tangent to the plane. Q: What is the equation that represents the locus of the point of contact of the sphere on the plane? A: The equation representing the locus of the point of contact of the sphere on the plane is $b^2 \cos^2(\Omega t) + b^2 \sin^2(\Omega t) - a^2 n = 0$. Q: What is the radius of the circle formed by the locus of the point of contact on the sphere? A: The radius of the circle formed by the locus of the point of contact on the sphere is given by $r$, where $r^2 = a^2 n$.

Step-by-step solution

Find position, velocity, and acceleration of the center of the sphere

The center of the sphere traces a horizontal circle of radius \(b\). Let's denote its position vector by \(\boldsymbol{A}\). In the \((i, j, k)\) coordinate system, the position vector \(\boldsymbol{A}\) can be written as: $$ \boldsymbol{A}=b \cos (\Omega t) \boldsymbol{i} + b \sin (\Omega t) \boldsymbol{j} $$ Now, differentiate \(\boldsymbol{A}\) with respect to time \(t\) to obtain the velocity vector \(\boldsymbol{V}_A\): $$ \boldsymbol{V}_A = -b \Omega \sin (\Omega t) \boldsymbol{i} + b \Omega \cos (\Omega t) \boldsymbol{j} $$ Next, differentiate \(\boldsymbol{V}_A\) with respect to time \(t\) to find the acceleration vector \(\boldsymbol{A}_A\): $$ \boldsymbol{A}_A = -b \Omega^2 \cos (\Omega t) \boldsymbol{i} - b \Omega^2 \sin (\Omega t) \boldsymbol{j} $$

Find velocity and acceleration of the contact point on the sphere

Due to the rolling without slipping condition, the velocity of the contact point on the sphere (let's denote it as \(\boldsymbol{C}\)) is equal to the velocity of the center of the sphere. Hence, \(\boldsymbol{V}_C = \boldsymbol{V}_A\). Now, differentiate \(\boldsymbol{V}_C\) with respect to time \(t\) to obtain the acceleration vector \(\boldsymbol{A}_C\): $$ \boldsymbol{A}_C = \boldsymbol{A}_A = -b \Omega^2 \cos (\Omega t) \boldsymbol{i} - b \Omega^2 \sin (\Omega t) \boldsymbol{j} $$

Find the expression for the angular velocity of the sphere

The acceleration of the contact point of the sphere is related to the angular velocity \(\boldsymbol{\omega}\) by the following expression: $$ \boldsymbol{A}_C = \boldsymbol{\omega} \times (\boldsymbol{\omega} \times \boldsymbol{C}) $$ Now, we need to express the contact point of the sphere \(\boldsymbol{C}\). Since the sphere has radius \(a\), and the contact point is on the lowest part of the sphere, we have: $$ \boldsymbol{C} = \boldsymbol{A} - a \boldsymbol{k} $$ Substitute this expression in the previous equation, we get: $$ \boldsymbol{A}_C = \boldsymbol{\omega} \times (\boldsymbol{\omega} \times (\boldsymbol{A} - a \boldsymbol{k})) $$ We are given that the angular velocity of the sphere can be written as: $$ \boldsymbol{\omega}=n \boldsymbol{k}-\frac{b}{a} \Omega \boldsymbol{i} $$ Now, substitute this expression into the previous equation. After simplifying, the equation becomes: $$ \boldsymbol{A}_C = -b \Omega ^2 \boldsymbol{i} - b \Omega ^2 \boldsymbol{j} = \boldsymbol{A}_A $$ This confirms the given expression for the angular velocity of the sphere.

Find the locus of the point of contact and its radius

If the vertical component of the angular velocity is constant, i.e., \(n=\boldsymbol{\omega} \cdot \boldsymbol{k}\) is constant, we need to find the locus of the point of contact \(\boldsymbol{C}\) on the sphere. To do this, notice that the sphere is always tangent to the plane. So, the position vector of the point of contact \(\boldsymbol{C}\) is always perpendicular to the direction of the given angular velocity \(\boldsymbol{\omega}\). Given the expression for \(\boldsymbol{\omega}\), we have: $$ \boldsymbol{C} \cdot \boldsymbol{\omega} = 0 $$ Substitute the expression for \(\boldsymbol{\omega}\) and \(\boldsymbol{C}\) we derived in previous steps: $$ (b \cos(\Omega t) \boldsymbol{i}+b \sin(\Omega t) \boldsymbol{j}-a \boldsymbol{k}) \cdot (n \boldsymbol{k}-\frac{b}{a} \Omega \boldsymbol{i})=0 $$ Simplify this equation and solve for the locus of \(\boldsymbol{C}\): $$ b^2 \cos^2(\Omega t) + b^2 \sin^2(\Omega t) - a^2 n = 0 $$ This equation represents a circle on the sphere with radius \(r\), where \(r^2 = a^2 n\).

Key concepts

Angular velocity

Angular velocity is a measure of how fast an object rotates or spins around a specific axis. In this exercise, the sphere rolls without slipping, which means its angular velocity is critical in determining how the sphere moves with respect to the plane. To grasp angular velocity, imagine a point on the sphere's surface. As the sphere rolls, this point travels a certain distance in a circular path, corresponding to its angular velocity. For the sphere in the problem, the angular velocity is represented by \( \boldsymbol{\omega} = n \boldsymbol{k} - \frac{b}{a} \Omega \boldsymbol{i} \).
- The term \( n \boldsymbol{k} \) represents the vertical component of angular velocity.- The term \( -\frac{b}{a} \Omega \boldsymbol{i} \) describes the horizontal component resulting from the center moving in a circle of radius \( b \).
Understanding these components can clarify why the sphere remains perfectly in sync with both its translation over the plane and its own rotation, ensuring it does not slip.

Locus of a point

The locus of a point refers to the set of all positions a point can occupy as the object moves. Think of it as the "path" traced by that point over time. In this problem, you’re finding the locus of the contact point on the sphere as it rolls.
In conditions where the sphere rolls without slipping, the contact point changes as it interacts with the plane, but always follows a predetermined path depending on the angular velocity. When it’s given that the vertical component of angular velocity \( n \) is constant, the point of contact on the sphere traces out a circle because the changes in orientation are uniform around a fixed axis. The solution provided identifies this path using mathematical expressions that simplify to a circle when calculated under the constraints of the exercise. Given formula simplifies to \( b^2 = a^2 n \), which means that every potential position of the contact point satisfies this equation of the circle.

Horizontal plane

A horizontal plane is a flat surface where every point on it is at the same level. It plays a vital role in physics problems involving rolling, as the plane defines the orientation in which the object moves and rotates.
In this problem, the horizontal plane is where the sphere rolls. The absence of slipping indicates the interaction between the plane and sphere is such that every rotation of the sphere aligns perfectly with its forward movement across the plane. The importance of this surface is reflected in the exercise through its roughness, preventing slipping and aligning rotational and translational speeds seamlessly.Visualize the horizontal plane as a large, flat field. The sphere rolls across it while its center follows a circular path, meaning it moves in a horizontal circle plane of radius \( b \), using the plane's properties to maintain steady rotation without slipping.

Position vector

A position vector defines the exact position of a point in a space with respect to a given origin. It's a fundamental concept in understanding the motion of objects in terms of both direction and distance from the starting point.
In this exercise, the center of the sphere's position vector is denoted by \( \boldsymbol{A} = b \cos(\Omega t) \boldsymbol{i} + b \sin(\Omega t) \boldsymbol{j} \). This vector represents the horizontal circle's center path as the sphere rolls. Here’s how the vector components work:- \( b \cos(\Omega t) \boldsymbol{i} \) and \( b \sin(\Omega t) \boldsymbol{j} \) describe motion along the \( i \)- and \( j \)-axes, respectively, - Combining these illustrates the path of the sphere’s center around the circle of radius \( b \).
By determining how this vector changes over time, you can obtain velocity and acceleration vectors, which further characterize the motion and underline the no-slip condition in terms of spatial transformation.