Explain why the bending of cyclist while taking a turn Bending of cyclist in curves derivation derive an expression for class 11 ?
Solving Problems in Mechanics by Constraint Relation Method
(1) Constraint Relation for a Moveable Pulley
Consider two blocks 1 and 2 attached at the ends of a string going over a moveable pulley. Let xp be the displacement of the pulley and x1 and x2 be the displacements of the blocks.
The constraint relation is as follows. The displacement of the pulley = average displacement of the blocks, i.e.
(a) If the blocks are displaced in the direction of the displacement of the pulley, the constraint relation is
(b) If block 1 moves up and block 2 moves down as shown in Fig. 3.41(b), the constraint relation is
(c) If one ends of the string is connected to a fixed end, as shown in Fig. (c), then x2 = 0 and the constraint relation is
(2) Constraint Relation for a Moveable Wedge (or Inclined Plane)
Consider a block of mass m moving with a certain velocity on a wedge (inclined plane) of inclination θ as shown in Fig. . The wedge in moved with a velocity u as shown.
To find the velocity v of the block perpendicular to the contact plane AB, we use the following constraint relation: The relative velocity of the block (with respect to the wedge) in a direction perpendicular to the contact plane is always zero. i.e. velocity v of the block perpendicular to AB = component of velocity v of the block perpendicular to AB. From Fig. (b) it follows that
v = u sin θ
Note : If the wedge is immoveable or is at rest, u = 0 then v = 0, i.e. the block cannot move perpendicular to AB. This happens because the component mg cos θ of the weight of the block balances with the normal reaction.
(3) Constraint Relation when the Distance Between Two Points Always Remains Constant
In cases when the distance between two points always remains constant, we use the following constraint relation. The relative velocity of one point of an object with respect to any other point on the same object in the direction of the line joining them always remains equal to zero, i.e. the velocity of one point on the object = the components of velocity of any other point along the line joining them.
If a body moves in a circle at a constant speed, it is said to be in uniform circular motion. In such a motion, the magnitude of the velocity (i.e. speed) is constant but the direction of the velocity vector is continually changing. Thus the velocity is changing with time. Hence the motion of the body is accelerated (see Fig. ). The acceleration is directed towards the centre of the circle and is called centripetal acceleration. The magnitude of the centripetal acceleration is given by
where w is the angular velocity (or angular frequency) and v is the speed along the circle. Since v = rw, we have
where r is the radius of the circular path. The angular frequency is related to time period T and frequency n as
Therefore, centripetal acceleration is also given by (since v = 2πr/T)
Banking of Round Tracks
When a car (or some other vehicle) negotiates a curved level road, the centripetal force required to keep the car in motion around the curve is provided by the friction between the road and the tyres. The weight of the car is supported by the normal reaction due to the earth. If the surface of the road is very rough, it provides a large amount of friction and hence the car can successfully negotiate the bend with a fairly high speed. If F is the total frictional force between the tyres and the road, then
when m is the mass of the car, v its speed around the curve and R is the radius of the curved track. The higher the value of F, the faster is the speed at which the bend can be negotiated. If µ is the coefficient of friction between the tyres and the road, then F ≤ µN; N = normal reaction = mg. The maximum speed which friction can sustain is
The large amount of friction between the tyres and the road would damage the tyres. To minimize the wearing out of tyres the road bed is banked, i.e. the outer part of the road is raised a little so that the road slopes towards the centre of the curved track. Suppose a car of mass m is moving around a banked track in a circular path of radius R as shown in Fig. . Let N1 and N2 be the reaction at each tyre due to the road. Then the total reaction is N = N1 + N2 acting in the middle of the car. If θ is the angle of the banking, the vertical component N cos θ supports the weight mg of the car while the horizontal component N sin θ provides the necessary centripetal force
where F is the force of friction acting radially inwards on the car. These equations give
The first equation determines the proper banking angle for given v, R and µ, and the second equation the maximum speed at which the car can successfully negotiate the curve for given R, µ and θ. For given θ and R, there is an optimum (best) speed for negotiating a banked curve at which there will be the least wear and tear, i.e. when friction is not needed at all (µ = 0). If µ = 0, this speed is
The car will not skid if the angle of banking of the track satisfies the relation
A Cyclist Negotiating a Curved Level Road
While negotiating a curved level (unbanked) road, a cyclist has to lean inwards which provides the necessary centripetal force which prevents him from falling down. Figure 3.51 shows a cyclist leaning at an angle θ with the vertical. N is the normal reaction which is given by
N = mg
where m is the mass of the cyclist plus the bicycle. The force of friction between the road and the tyres is
F = µN = µmg
The cyclist will skid if the centripetal force mv2 /R exceeds the frictional force F, i.e. if
where R is the radius of the curved road. Thus skidding occurs if
Motion in a Vertical Circle
Figure shows an object of mass m whirled with a constant speed v in a vertical circle of centre O with a string of length R. When the object is at top A of the circle, let us say that the tension (force) in the string is T1. Since the weight mg acts vertically downwards towards the centre O, we have,
At the point B, where OB is horizontal, the weight mg has no component along OB. Thus, if the tension in the string is T2 at B, we have
Force towards centre
At C, the lowest point of motion, the weight mg acts in the opposite direction to the tension T3 in the string. Thus at C we have,
Force towards centre,
From (i), (ii) and (iii), we see that the maximum tension occurs at lowest point C of the motion. Here the tension T3 must be greater than mg by mv2 /R to keep the object in a circular path. The minimum tension is given by (i) when the object is at the highest point A of the motion. Here part of the required centripetal force is provided by the weight and the rest by T1. In order to keep a body of mass m in a circular path, the centripetal force, at the highest point A, must at least be equal to the weight of the body. Thus
gives the minimum speed the body must have at the highest point so that it can complete the circle. Then the minimum speed vC the body must have at the lowest point C is given by
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