5. CAPÍTULO CONCLUSIONES Y LÍNEAS FUTURAS
5.1 CONCLUSIONES Y LÍNEAS FUTURAS
θ
F l
m
Q. 1 An object of mass m is tied to a string of length l and a variable force F is applied on it which brings the string gradually at angle θ with the vertical. Find the work done by the force F. [Solving using potential energy method]
Q. 2 A body is dropped from a certain height. When it lost an amount of P.E. ‘U’, it acquires a velocity ‘v’. The mass of the body is :
(a) 2U v/ ² (b) 2 /v U2 (c) 2v/U (d) U2/2 .v
Q. 3 A particle of mass m is attached to a light string of length l, the other end is fixed. Initially the string is kept horizontal and the particle is given an upward velocity v. The particle is just able to complete a circle.
(a) The string becomes slack when the particle reaches its highest point (b) The velocity of the particle becomes zero at the highest point (c) The kinetic energy of the ball in initial position was 1 2 .
2mv =mgl (d) The particle again passes through the initial position.
Q. 4 A small block of mass m slides along the frictionless loop-the-loop track shown in fig. (a) If it starts from rest at P, what is the resultant force acting on it at Q? (b) At what height above the bottom of the loop should the block be released so that the force it exerts against the track at the top of the loop is equal to its weight?
R Q
5R P
Q. 5 A simple pendulum of length l, the mass of whose bob is m, is observed to have a speed v0 when the cord makes the angle θ0 with the vertical (0< <θ0 π/2), as in fig. In terms of g and the foregoing given quantities, determine (a) the speed v1 of the bob when it is at its lowest position; (b) the least value v2 that v0 could have if the cord is to achieve a horizontal position during the motion.
θ l
m
0
v0
Q. 6 A chain of length l and mass m lies on the surface of a smooth hemisphere of radius R > l with one end tied to the top of the hemisphere. Find the gravitational potential energy of the chain with respect to the given P.E. level.
R
ZERO P. E. LEVEL
Q. 7 A smooth sphere of radius R is made to translate in a straight line with a constant acceleration a. A particle kept on the top of the sphere is released from there at zero speed with respect to the sphere. Find the speed of the particle with respect to the sphere as a function of the angle θ it slides. [Solve using potential energy method]
Q. 8 A block rests on an inclined plane as shown in figure. A spring to which it is attached via a pulley is being pulled downward with gradually increasing force. The value of µs is known. Find the potential energy U of the spring at the moment when the block begins to move.
m
k
θ F µs
Q. 9 The particle m in figure is moving in a vertical circle of radius R inside a track. There is no friction. When m is at its lowest position, its speed is v0. (a) What is the minimum value vm to v0 for which m will go completely around the circle without losing contact with the track? (b) Suppose v0 is 0.775vm. The particle will move up the track to some point at P at which it will lose contact with the track and travel along a path shown roughly by the dashed line. Find the angular position θ of point P.
R P
v0
m θ
Q. 10 A chain of length l and mass m lies on the surface of a smooth sphere of radius R > l with one end tied to the top of the sphere.
(a) Find the gravitational potential energy of the chain with reference level at the centre of the sphere.
(b) Suppose the chain is released and slides down the sphere. Find the kinetic energy of the chain, when it has slide through an angle θ.
(c) Find the tangential acceleration dv
dt of the chain when the chain starts sliding down.
Q. 11 A spherical ball of mass m is kept at the highest point in the space between two fixed, concentric spheres A and B (see figure).
The smaller sphere A has a radius R and the space between the two spheres has a width d. The ball has a diameter very slightly less than d. All surfaces are frictionless. The ball is given a gentle push (towards the right in the figure). The angle made by the radius vector of the ball with the upward vertical is denoted by θ (shown in the figure)
θ
O
R
Sphere B
Sphere A d
(a) Express the total normal reaction forces exerted by the spheres on the ball as a function of angle θ. (b) Let NA and NB denote the magnitudes of the normal reaction forces on the ball exerted by the spheres A and B, respectively. Sketch the variations of NA and NB as functions of cosθ in the range 0≤ ≤θ π by drawing two separate graphs, taking cosθ on the horizontal axes. Also sketch the variations of NA and NB as functions of θ.
Section - 6 MORE ON CONSERVATIVE FORCES Conservative forces and potential energy again
The interaction of a particle with surrounding bodies can be described in two ways : by means of forces or through the use of the notion of potential energy. In classical mechanics both ways are extensively used. The first approach, however, is more general because of its applicability to forces in the case of which the potential energy is impossible to introduce (i.e., non conservative forces). As to the second method, it can be utilized only in the case of conservative forces.
Our objective is to establish the relationship between potential energy and the force of the conservative field, or putting it more precisely, to define the conservative field of forces F r! !( ) from a given potential energy U r( )!
as a function of a position of a particle in the field.
We have learnt by now that the work performed by conservative forces on a particle during the displacement of the particle from one point in the field to another may be described as the decrease of the potential energy of the particle, that is,
con .
w = −∆U
The same can be said about the elementary displacement dr!
as well:
is the elementary length covered along the path and Ft is the tangential component of F!, we shall rewrite equation (37) as
t .
i.e., the projection of the conservative force at a given point in the direction of the displacement dr!
equals the derivative of the potential energy U with respect to a given direction, taken with the opposite sign. The designation of a partial derivative ∂ ∂/ s emphasizes the fact of differentiating with respect to a finite direction.
The displacement dr!
can be resolved along any direction and, specifically, along the x, y, z coordinate axes. For example, if displacement dr!
is parallel to the x axis, it may be described as dr!=dxiˆ.
The work performed by the conservative force F!
over the displacement dr!
parallel to the x axis is ( ˆ) x ,
F dr!⋅ ! = ⋅F dxi! = ⋅F dx where Fx is the x-component of the force F!.
Substituting the last expression into equation (38), we get
x = F U
x
−∂
∂ ...(39)
where the partial derivative symbol implies that in the process of differentiating U x y z( , , ) should be considered as a function of only one variable, x, while all other variables are assumed constant. It is obvious that the equations for Fy and Fz are similar to that for Fx. So, having reversed the sign of the partial derivatives of the function U
The quantity in parentheses is referred to as the scalar gradient of the function U and is denoted by grad U or
∇
∇∇
∇∇U. Generally the second, more convenient, designation where ∇(“nabla”) signifies the operator
iˆ ˆj kˆ
x y z
∂ ∂ ∂
∇ = + +
∂ ∂ ∂
is used. Consequently, we can write,
! =
F −∇U ..(41)
i.e., the conservative force F! is equal to the potential energy gradient, taken with the minus sign. Put simply, the conservative force F! is equal to the antigradient of potential energy.
Example – 19
The potential energy of a particle in certain conservative field has the following form:
(a) U(x, y) = – αxy, where α is a constant;
(b) U r( )!
= a r! !⋅ ,
where a!
is a constant vector and r!
is the position vector of the particle in the field.
Find the conservative field force corresponding to each of these cases.
Solution: (a) We have,
Therefore,
U = ⋅a r! !
; then
x y z
a x a y a z
= ⋅ + ⋅ + ⋅ ˆ ˆ ˆ
U U U
F i j k
x y y
∂ ∂ ∂
= − ∂ + ∂ + ∂
!
ˆ ˆ ˆ (a ix a jy a kz )
= − + +
= −a!
Example – 20
A conservative force F(x) acts on a 1.0 kg particle that moves along the x-axis. The potential energy U(x) is given as U x( )= + −a (x b)2 where a = 20 J, b = 2 m. and x is in meters. At x = 5.0 m the particle has a kinetic energy of 20 J. It is known that there is no other force acting on the system. Based upon this information, answer the following questions:
(a) What is the mechanical energy of the system?
(b) What is the range of x in which the particle can move?
(c) What is the maximum kinetic energy of the particle and the position where it occurs?
(d) What is the equilibrium position of the particle?
Solution: We have,
( ) ( )2
U x = + −a x b 20 (x 2)2
= + −
We know that at x = 5.0, K.E. = 20 J, therefore, mechanical energy of the particle, E= Potential energy, U + kinetic energy, k
=U(atx= +5) k(atx=5) =20 (5 2)+ − 2+20
=49 .J [Ans.(a)]
We know that, E= U + k
⇒ k = −E U
49 20 (x 2)2
= − + − 29 (x 2)2
= − −
As the particle moves on the x-axis, its kinetic energy can never be negative, therefore, we have, 0
k≥
⇒ 29 (− −x 2)2≥0
⇒ (x−2)2≤29
⇒ x2− + −4x 4 29≤0
⇒ x2−4x−25≤0
⇒ x∈ −
[
3.38, 7.38]
[Ans (b)]When the particle has maximum kinetic energy, we have, dk 0
dx =
⇒
29 ( 2)2
0
d x
dx
− −
=
⇒ 0− ⋅ − ⋅2 (x 2) (1)=0
⇒ x=2
i.e., at x=2, the particle has maximum kinetic energy which is equal to 29 J. [Ans. (c)]
As the conservative force is antigradient of potential energy, we have, ( ) dU
F x = − dx
20 ( 2)2
d x
dx
+ −
= −
[
0 2 (x 2) (1)]
= − + ⋅ − ⋅ 4 2x
= − .
When the particle is in equilibrium, net force on it must be zero. As the only force acting on the particle is F(x), in equilibrium position
( ) 0 F x =
⇒ 4 2− x=0
⇒ x=2 .m [Ans.(d)]