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## Rigid Body Dynamics I, Physics tutorial

:IntroductionThe rigid body is the idealization of the solid body in which deformation is neglected. In other words, distance between any two given points of the rigid body remains constant in time in spite of of external forces exerted on it. Although such an object can't physically exist because of relativity, objects can usually be assumed to be perfectly rigid if they are not moving near speed of light.

The rigid body is stated as the aggregate of point masses such that relative separation between any two of these always remains invariant, i.e., for any position of body, r

_{ik}= a constant.A rigid body is one which has the definite shape. It doesn't change even when deforming force is applied. In nature there is no perfectly rigid body as all real bodies experience some deformation when forces are exerted. So, the perfectly rigid body can only be idealized. For instance, deformation of the lawn tennis ball as it bounces off ground can be ignored.

Translational Motion of a Rigid Body:Assume you are traveling in the bus, then, during the certain interval of time, the displacement will be exactly equal to that of co-passenger given both of you don't move with respect to bus. This will also be true for any two objects joined to body of the bus, say a bulb and a switch. This is the feature of translational motion. The rigid body is said to perform pure translational motion if every particle in it goes through same displacement as every other particle in given interval of time.

Rotational Motion of Rigid Body:Let us assume the motion of earth. Every point on it moves in the circle (corresponding latitude), the centres of which lie on polar axis. Such motion is a case of the rotational motion. The rigid body is said to perform rotational motion if all particles in it move in circles, centres of which lie on the straight line known as the axis of rotation. When the rigid body rotates about the axis every particle in it remains at the fixed distance from axis.

General Motion of a Rigid Body:General motion of the rigid body is combination of translation and rotation. This can be understood by taking example of moving car. If you look at the tires of moving car you see that wheel is turning round and moving forward or backwards as case may be. So car changes position as wheels rotate.

:Moment of InertiaWe need now to measure the moment of inertia which takes into account the mass distribution of the body about the axis of rotation and which plays a role in rotational motion. This is analogous to that played by mass in linear motion.

Consider a rigid body rotating about a fixed axis through O with constant angular Velocity w. A particle A, of mass m

_{1}, at a distance r_{1}from O describes its own circular path and v_{1}is its linear velocity along the tangent of the path at the instant shown, thenv

_{1}= r_{1}wand the kinetic Energy of A = 1/2m

_{1}v_{1}^{2}= 1/2m

_{1}r^{2}ω^{2}The kinetic energy of the whole body is the sum of the kinetic energies of its component particles. Assuming these have masses m

_{1}, m_{2}, m_{3},..... m_{n}and are at distancesr

_{1}, r_{2}, r_{3}, ......r_{n}from O, then, as all particles have same angular velocity w, we haveTotal K.E for the whole body =

1/2m

_{1}r_{1}^{2}ω^{2}+ 1/2m_{2}r_{2}^{2}ω^{2}+......+∑

_{i=1}^{n}(1/2ω^{2}m_{i}r_{i}^{2}i.e. Total K.E =1/2ω

^{2}∑_{i=1}^{n}1/2m_{i}r_{i}^{2})∑

_{i=1}^{n}m_{i}r_{i}Where represents the sum of the m

_{i}r_{i}^{2}values for all the particles of the body. Note that the quantity Σm_{i}r_{i}^{2}depends on the mass and its distribution and it is a measure of the moment of inertia I of the body about the axis in question.So we define I as

I = ∑

_{i=1}^{n}m_{i}r_{i}We can then write the K.E. as

K.E of body =

1/2Iω

^{2}Comparing this with the kinetic energy for linear motion 1/2mv

^{2}we see that mass m is replaced by the moment of inertia I and the velocity v is replaced by the angular velocity w.The Unit of moment of inertia I is kg m

^{2}. Values of I for regular shapes of bodies can be determined using calculus.Radius of Gyration:No matter what shape of the body is, it is always possible to determine the radial distance from any given axis at which mass of body could be concentrated without changing moment of inertia of body about that axis. This distance is called as radius of gyration of body about given axis. It is signified by K. So if mass m of body really were concentrated at this distance, moment of inertia would be that of particle of mass m at a distance k from an axis, or mk

^{2}. But we see that this is equal to actual moment of inertia I, thusmk

^{2}= IDumbbell:Simplest rotating object which we can consider is a dumbbell. It comprises of two point masses m

_{1}and m_{2}connected by the massless rigid rod of length LConsider total mass M be m

_{1}+ m_{2}. Positioning mass m_{1}at the origin of x - axis and man m_{2 }at x = L. it could be shown that centre of mass is at x wherex = (m

_{1})(0) + (m_{2})(L)/(m_{1}+ m_{2})m

_{2}L/MIf we consider the case in which the axis of rotation goes through the centre of mass (C.M) (i.e through point x = m2 L/M, then the axis is taken perpendicular to the rod. So, measuring from the C.M, the coordinates of m1 and m2 will be -m2 L/M and L- (m2/m) = m1L/M respectively.

Now, the rotational inertia about an axis passing through the centre of mass and perpendicular to the axis of the dumbbell is given by I = m1(m2l/M)2 + m2(m1L/M)2

= L2(m

_{1}m_{2}^{2}+ m_{2}m_{1}^{2}/M^{2}) = m_{1}m_{2}L2(m_{2}+ m_{1}/(m_{1}+ m_{2})^{2})= (m

_{1}m_{2}/m_{1}+ m_{2})L^{2}:Moments and CouplesA force applied to the hinged or pivoted body changes rotation about hinge or pivot. Experience illustrates that turning effect or moment or torque of force is greater, the greater the magnitude of the force and the greater the distance of the point of application from pivot. Moment or torque of the force about the point is estimated by the product of force and perpendicular distance from line of action of force to point.

Thus in given Figure, if OAB is the trapdoor hinged at O and acted on by forces P and Q as given, then, Moment of P about O = P x OA and

Moment of Q about O = Q x OC

The particular distance should be taken. Alternatively we can resolve Q in components Q cosΘ perpendicular to OB and Q sin Θ along OB as shown in Figure.

The moment of latter about O is zero, its line of action passes through O. For former, we have Moment of Q cos Θ about O = Q cos Θ x OB 3.9a

= Q x OC

(as cosΘ = OC / OB),

Moments are estimated in Newton metres (Nm) and are given the positive sign if they are liable to generate clockwise rotation. The couple comprises of two equal and opposite parallel forces whose lines of action don't coincide. It always tends to change rotation. The couple is applied to the water tap to open it. We can say that moment or torque of couple P - P about O

= P x OA + P x OB (both are clockwise)

= P x AB

Therefore, moment of couple = one force x perpendicular distance between forces.

Equilibrium of Coplanar Forces

General conditions for equilibrium

If the body is acted on by the number of coplanar forces (i.e., forces in same plane) and is in equilibrium (that is there is rest or motion under constant speed) then

i) Components of forces in both of any two directions (generally taken at right angles) should balance.

(ii) Sum of the clockwise moments about the point equals sum of anticlockwise moments about same period.

First statement is a result of there being no translational motion in any direction and second follows as there is no rotation of body. In short, if the body is in equilibrium forces and moment should both balance.

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