Physics in Depth

A particle is in uniform circular motion if it travels around a circle on a circular arc at uniform (constant) speed. Although the speed does not vary , the particle is accelerating because the velocity changes in direction. fig. below shows the relationship between the velocity and acceleration vectors at various stages during uniform circular motion. Both vectors have constant magnitude, but their directions change continuously. The velocity is always directed tangent to the circle in the direction of motion. The acceleration is always directed radially inward. Because of this, the acceleration associated with uni form circular motion is called a centripetal (meaning "center seeking") acceleration. Centripetal acceleration Consider a particle 'p' moves at constant speed 'v' around a circle of radius r. At the instant shown in figure above, p has a coordinate x and y. \(\theta\) be the angular displacement of p.

Work, Energy and Power Work Work is said to be done by a force when the force produces a displacement on a body on which it acts. \[W=\vec{F}.\vec{S}\] Here, W is the work done by the force \(\vec{F}\) while displacing a body through \(\vec{s}\). Work is a scalar quantity; it has no property of direction but only magnitude. If the angle between the displacement vector and force vector is \(\theta\) as in fig., then the work done is, \[W=Fscos\theta \hspace{0.1cm}(\because \vec{a}.\vec{b}=abcos\theta)\] Equivalently, W= component of force along the displacement \(\times\) the displacement Special cases : When \(\theta=0^\circ\), W=Fs ( maximum work done ). When \(\theta=90^\circ\), W=0 ( no work done ) When \(\theta < 90^\circ\) ( positive work done ) When \(90^\circ < \theta \leq 180^\circ\) ( negative work done ) \(\implies\) When a coolie travels on a platform with a load on his head, he exerts a vertical force on the load as in fig

Mass and Pulley system Equation of motion for mass m is, \[\begin{align} F&=T-mg \nonumber \\ ma&=T-mg ... (i) \end{align}\] Equation of motion for mass M is, \[\begin{align} F&=T-Mg \nonumber \\ Ma&=T-Mg ... (ii) \end{align}\] Adding eqns (i) and (ii), \[\begin{align*} ma+Ma&=Mg-mg\\ a&=\frac{(M-m)g}{(M+m)} ... (iii) \end{align*}\] Substituting the vaue of a in eqn (i), \[\begin{align*} m\left(\frac{(M-m)g}{M+m}\right)&=T-mg\\ \frac{m(M-m)g}{M+m}+mg&=T\\ T&=\frac{m(M-m)g+mg(M+m)}{M-m}\\ \therefore T&=\frac{2Mm}{M+m}g ... (iv) \end{align*}\] Two masses 7 kg and 12 kg are connected at the two ends of a light inextensible string that passes over a frictionless pulley. Using free body diagram method, find the acceleration of masses and the tension in the string, when the masses are released. Here, m = 7 kg and M = 12 kg. Prior to using the formula directly

Newton's Laws of Motion Image source : Britannica In 1687, Newton formulated his laws of motion and published it in his book Philosophiae Naturalis Principia Mathematica . Isaac Newton first understood the relation between a force and the acceleration and study of this relation is given a name , Newtonian Mechanics . Though, there are some restrictions to Newtonian Mechanics such as: If the speeds of the interacting bodies are very large - an appreciable fraction of the speed of light- we must replace Newtonian mechanics with Einstein's special theory of relativity. If the interacting bodies are on the scale of atomic structure (for example, electrons) we must replace Newtonian mechanics with quantum mechanics. , it still applies to the motion of objects ranging in size from the very small (almost on the scale of atomic structure) to astronomical (galaxies and clusters of galaxies). Since, theory of relativity and quantum mechanics are beyond the