## UY1: More About Rolling Motion

Sphere On An Incline From sphere on an incline, we know that the frictional force and the acceleration of the centre of mass is given by: \begin{aligned} f &= \frac{2}{5} Ma_{\text{CM}} \\ a_{\text{CM}} &= \frac{5}{7} g \sin{\theta} \end{aligned} Combining the two, we have: $$f = \frac{2}{7} Mg \sin{\theta}$$ If we increase $\theta$ until the frictional force is larger than the …

## UY1: Sphere On An Incline

The figure above shows a sphere rolling down an incline. We will analyze this rolling motion. Important facts about accelerated rolling motion: Accelerated rolling motion is possible only if a frictional force is present. There is no loss of mechanical energy because the contact point (of the sphere with the ramp) is at rest at any instance. From the energy …

## UY1: Rolling Motion

For a rolling object, the axis of rotation is not fixed in space. The axis of rotation travels with the rolling object. There are several assumptions to simplify the analysis of rolling objects: Homogeneous rigid body High degree of symmetry (E.g. Cylinder, sphere, hoop) Pure rolling motion (rolling without slipping) Rolling motion along a flat surface (inclined, horizontal or even …

## UY1: Work, Energy & Power Of Rotating Object

Consider a point rotating through a distance $ds$. Let’s calculate the work done by the force $\vec{F}$. \begin{aligned} dW &= \vec{F}.d\vec{s} \\ &= \left( F \sin{\phi} \right) r \, d \theta \end{aligned} From the definition for torque, we know that $rF \sin{\theta} = \tau$. Hence, $$dW = \tau \, d \theta$$ From the equation above, we can see that only …

## UY1: Torque & Angular Acceleration

Torque Consider a force $\vec{F}$ acting on a rigid body at point $P$ in the direction as shown. The rod is pivoted so that it can only rotate about point $O$. In this case, the torque $\tau$ of force $\vec{F}$ about $O$ is defined as: \begin{aligned} \tau &= Fl \\ &= rF \sin{\theta} \\ &= F_{\text{tan}}r \end{aligned} ,where $l$ is …

## UY1: Rocket Propulsion

The operation of a rocket depends upon the law of conservation of linear momentum as applied to a system of particles, where the system is the rocket plus the ejected fuel. Please do not believe the urban folklore that the lift of a rocket is generated by the exhaust gases “pushing” against the ground. Because the gases are given momentum …

## UY1: Motion Of A System Of Particles

We have calculated the centre of mass of a system of particles. How does it related to the movement of a system of particles? The centre of mass of a system of particles: $$\vec{r}_{CM} = \frac{\sum\limits_{i} m_{i}\vec{r}_{i}}{M}$$ The velocity of the centre of mass will be: \begin{aligned} \vec{v}_{CM} &= \frac{d \vec{r}_{CM}}{dt} \\ &= \frac{1}{M} \sum\limits_{i} m_{i} \frac{d\vec{r}_{i}}{dt} \\ &= \frac{\sum\limits_{i} …