## Weekly outline

### 30 October to 05 Nov

- Few Problems are provided to check your understanding, the topic is in progress and everyone needs few problems to keep moving and keep learning. The first part contains problems on Kinematics of Rigid Bodies. Solutions, if needed will be provided in class. However answers will be provided on weekend.
**1.**A plate moves along a horizontal surface. Components of the velocity for three corners are:x -component of \(v_{A}\) = 2m/sy- component of \(v_{B}\) = -3 m/sy- component of \(v_{C}\) = 5 m/sFind the angular speed of the plate and the velocity of corner D**2.**A bar moves in the plane of the page so that end A has a velocity of 7m/s and decelerates at a rate of 3.3. m/s^{2}. What are the velocity and acceleration of point C when BA is at 30^{0}to the horizontal?**3.**A rod is moving on a horizontal surface and is shown at time t. What is vy of end A and \(\omega\) of the rod at the instant below?**4.**A plate ABCD moves on a horizontal surface. At time t corners A and B have the following velocities:\(v_{A} = 3\vec{i} +2\vec{j}\) m/s\(v_{B} = (v_{B})_{x} \vec{i}+5\vec{j}\) m/sFind the location of the instantaneous axis of rotation.**5.**A cone is rolling without slipping such that its centerline rotates at the rate \(\omega_{1}\) of 5 revolutions per second about the z axis. What is the angular velocity \(\omega\) of the body relative to the ground? What is the angular acceleration for the body? An inextensible, flexible, heavy rope is hung at two pegs of the same height as shown in the figure. The rope remains is in equilibrium. The length of the rope hanging between the pegs is ℓ, and the vertical distance of the bottom of the hung part of the rope and the peg level is h as shown in figure. Find the total length of the rope? (Friction is negligible.)

**Solution provided in the class.**The distributive property of vector products can be used to determine the moment of the resultant of several concurrent forces. If several forces \(\vec{F_{1}},\vec{F_{1}}\) . . . are applied at the same point A (Fig.), and if we denote by \(\vec{r}\) the position vector of A, it follows immediately from that

In words, the moment about a given point O of the resultant of several concurrent forces is equal to the sum of the moments of the various forces about the same point O. This property, which was originally established by the French mathematician Varignon (1654–1722) long before the introduction of vector algebra, is known as Varignon’s theorem.

Chasles’ theorem asserts that it is always possible to represent an arbitrary displacement of a rigid body by a translation of its center of mass plus a rotation around its center of mass. The proof in this note is rather detailed and an understanding of it is not necessary for following the development of the lecture. However, the result is interesting and its proof provides a nice exercise in vector methods for those interested.

To avoid algebraic complexities, we consider here a simple rigid body consisting of two masses \(m_{1}\) and \(m_{2}\) joined by a massless rigid rod of length l. The position vectors of \(m_{1}\) and \(m_{2}\) are \(\vec{r_{1}}\) and \(\vec{r_{2}}\), respectively, as shown in the sketch. The position vector of the center of mass of the body is \(\vec{R}\), and \(\vec{r'_{1}}\) and \(\vec{r'_{2}}\) are the position vectors of \(m_{1}\) and \(m_{2}\) with respect to the center of mass. The vectors \(\vec{r'_{1}}\) and \(\vec{r'_{2}}\) are back to back along the line joining the masses.

In an arbitrary displacement of the body, \(m_{1}\) is displaced by \(d\vec{r_{1}}\) and \(m_{1}\) is displaced by \(d\vec{r_{2}}\). Because the body is rigid, \(d\vec{r_{1}}\) and \(d\vec{r_{2}}\) are not independent, and we begin our analysis by finding their relation. The distance between \(m_{1}\) and \(m_{2}\) is fixed and of length l. Therefore

or

Taking differentials of Eq. (1), and recalling thatEquation (2) is the “rigid body condition” we seek. There are evidently two ways of satisfying Eq. (2): either \(d\vec{r_{1}}\) = \(d\vec{r_{2}}\), or \(d\vec{r_{1}} - \(d\vec{r_{2}}\)\) is perpendicular to\(\vec{r_{1}}- \vec{r_{2}}\).We now turn to the translational motion of the center of mass. By definition,Therefore, the displacement d\(\vec{R}\) of the center of mass isIf we subtract this translational displacement from \(d\vec{r_{1}}\) and \(d\vec{r_{2}}\), the residual displacements \(d\vec{r_{1}}\) - \(d\vec{R}\) and \(d\vec{r_{2}}\) - \(d\vec{R}\) should give a pure rotation around the center of mass. Before investigating this point, we notice that sincethe residual displacements are................(4)Using Eq. (3) in Eq. (4) we haveandNote that if \(d\vec{r_{1}}\) = \(d\vec{r_{1}}\), the residual displacements \(d\vec{r'_{1}}\) and \(d\vec{r'_{2}}\) are zero and the rigid body translates without rotating.We must now show that the residual displacements represent a pure rotation around the center of mass to complete the proof. The sketch shows what a pure rotation would look like.First we show that

where we have used Eq. (5) and the rigid body condition Eq. (2).

SimilarlyFinally, we require that the residual displacements correspond to rotation through the same angle Δθ. With reference to the sketch, this condition in vector form isNote thatby definition of the center of mass. Using Eqs. (5) and (6), we havecompleting the proof.

### 2 October - 8 October

The message of the principle of relativity, whether it is Newtonian or Einsteinian relativity, is that all inertial frames are equivalent as far as the laws of mechanics are concerned. However, in practice, many problems can be solved more readily by a judicious choice of frame. In collision problems the best choice is usually the centre of momentum frame. This is the frame in which the total momentum of the system is zero.

Consider the following two collisions between a pair of cars.

Collision (i): A car travelling at 60 mph (96 km h

^{- 1}) collides with a stationary car. Collision(ii): Two cars each travelling at 30 mph (48 km h

^{- 1}) collide head on. It is clear that the single moving car in (i) has more kinetic energy than the two moving cars in (ii). But the need to conserve momentum in collision (i) means that not all the kinetic energy is available to cause damage, whereas in collision (ii) all the kinetic energy can go into causing damage and momentum will still be conserved. So the message is clear: to determine the energy that can be made available we have to transfer to the centre of momentum frame. Bearing this in mind which of the above collisions, assuming they are both inelastic, causes the most damage?Solution : In collision (i) not all the \(\frac{1}{2}mv^{2}\) of kinetic energy is available to cause damage since conservation of momentum implies both cars must be moving after the collision. In collision (ii) the cars come to rest after the collision so the kinetic energy available is \(2 \frac{1}{2}m (v /2)^{2} = \frac{1}{4}mv^{2}\). The amount of damage cannot depend on the frame from which we view the event. So in (i) look at the collision in the centre of mass frame: it is exactly the same as collision (ii)! Thus the damage must be equal.

- (a) The photoelectric effect is the absorption of a photon by an electron: It IS responsible for ejecting electrons from an illuminated metal surface and for the ionisation of atoms. However photoelectric absorption by isolated electrons is never observed. Why not?(b) How are energy and momentum conserved in the ionisation of an atom, which can take place in isolation?(c) The production of an electron-positron pair from a photon can occur when the photon energy exceeds the combined rest energy of the pair. The process is observed when gamma rays interact with matter. Pair production by isolated gamma rays is never observed. Explain why this is.Solution : (a) Consider the putative absorption of the photon by the lone electron. The process is shown in the laboratory frame in figure

(a) Consider the putative absorption of the photon by the lone electron. The process is shown in the laboratory frame in figureand transformed to the CM frame in figureIn the CM frame it is clear that momentum is conserved but relativistic energy is not since, comparing the energy before and after, hv'+ \( \gamma \)(mc^{2}> mc^{2}as \( \gamma \)> 1.(b) When an isolated atom absorbs a photon its rest mass changes. This is not possible for the electron since, being elementary, it has no excited states.(c) In the laboratory frame the putative process is shown in the figure. In the final state the total momentum is zero in the centre of mass frame. However, for the initial state, there is no frame in which the momentum is zero because a single photon cannot have a rest frame. It is therefore obvious that momentum cannot be conserved so the process cannot occur.

### 9 October - 15 October

Galileo is credited with establishing that all bodies fall with the same acceleration under gravity. He did this not by dropping bodies from the leaning tower of Pisa, as legend has it, but by rolling balls down an inclined plane. This has the advantage of diluting gravity which makes it easier to measure the time of fall. However, Galileo was fortunate in the shapes of bodies he chose to compare.

(a) Show that a sphere of lead and a sphere of wood rolling down an inclined plane cover the same distance in the same time. [Hint: Write down the conservation of energy for a rolling ball and differentiate it with respect to time to find the acceleration.]

(b) Does the size of the balls make any difference?

(c) Does the shape of the rolling object make any difference?

(d) Would Galileo have made his discovery if he had compared the rolling motion of spheres and cylinders?