Two particles A and B of respective masses M(B) and M(A) are connected together by means of an inextensible string of negligible mass. The string passes over the groove of a pulley of negligible mass.
At instant to=0 the system S(pulley,mass,string) is left without initial velocity with particle B being at distance h from the axis of the pulley .The reference level of gravitational potential energy is that of a horizontal plane containing the axis of the pulley and particle A. (neglect friction)
a) calculate at to=0 the mechanical energy of the system in terms of M(B), g, and h.
b) calculate the mechanical energy of the system after covering a distance x .
c) calculate by applying the principle of the conservation of mechanical energy ,the velocity v of A or B in terms of x, M(A), M(B), and g. Deduce the acceleration of the motion.
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To solve this problem, let's break it down into parts:
a) To calculate the mechanical energy of the system at t=0, we need to consider the potential energy and the kinetic energy. At t=0, both particles A and B are at rest, so there is no kinetic energy. The only form of energy present is potential energy due to the gravitational field.
Particle A is at a reference level, so its potential energy is zero. The potential energy of particle B can be calculated as the product of its mass, M(B), the acceleration due to gravity, g, and the height from the reference level, h.
So, the mechanical energy at t=0 is given by the potential energy of particle B: E = M(B)gh.
b) To calculate the mechanical energy of the system after covering a distance x, we need to consider the change in potential energy and any change in kinetic energy. The potential energy of particle B decreases as it moves down by a distance x, given by M(B)gx. There is no change in the potential energy of particle A because it remains at the reference level.
Since the string is inextensible, particle A will move up by the same distance x. As a result, particle A acquires potential energy of M(A)gx. However, the string is inextensible, so the decrease in potential energy of particle B is compensated by the increase in potential energy of particle A. Therefore, the total change in potential energy is zero.
Since the potential energy remains constant, the change in mechanical energy is also zero. So, the mechanical energy of the system after covering a distance x is the same as at t=0, i.e., E = M(B)gh.
c) According to the principle of conservation of mechanical energy, the total mechanical energy of the system remains constant. So, the mechanical energy of the system after covering a distance x is equal to the mechanical energy at t=0.
Using this information, we can write the equation:
M(A)gx + (1/2)M(A)v^2 = M(B)gh , where v is the velocity of particle A or B.
From this equation, we can solve for v in terms of x, M(A), M(B), and g. Rearranging the equation, we get:
(1/2)M(A)v^2 = M(B)gh - M(A)gx
v^2 = 2(M(B)gh - M(A)gx) / M(A)
v = sqrt(2(M(B)gh - M(A)gx) / M(A))
The acceleration can be obtained by taking the derivative of velocity with respect to time:
a = dv/dt = d/dt(sqrt(2(M(B)gh - M(A)gx) / M(A)))
= -sqrt(2(M(B)g) / M(A)) * dx/dt
= -sqrt(2(M(B)g) / M(A)) * dx/dt
So, the acceleration of the motion is given by -sqrt(2(M(B)g) / M(A)) times the rate of change of distance x with respect to time.