A partide of mass m moves along a trajectory given by x= a cos δ1 t
and y = asinδ2 t . Find rand y component of the force acting on the patde, potential energy, kinetic enenev of the partide and hence show that the total energe
E= KE + PE =1/2 m(a*2 δ1*2+ b*2 δ2*2)
kE= 1/2 mv^2
To find the radial and y-component of the force acting on the particle, we need to differentiate the given functions x(t) and y(t) with respect to time t, in order to obtain the velocity components vx(t) and vy(t). Then, we can use Newton's second law to find the force components.
Given:
x(t) = a * cos(δ1 * t)
y(t) = a * sin(δ2 * t)
Differentiating x(t) and y(t) with respect to t, we get:
v_x(t) = -a * δ1 * sin(δ1 * t)
v_y(t) = a * δ2 * cos(δ2 * t)
Now, applying Newton's second law:
F_x = m * a_x = m * d(v_x)/dt = m * (-a * δ1^2 * cos(δ1 * t))
F_y = m * a_y = m * d(v_y)/dt = m * (-a * δ2^2 * sin(δ2 * t))
Therefore, the radial and y-components of the force acting on the particle are:
F_r = -m * a * δ1^2 * cos(δ1 * t)
F_y = -m * a * δ2^2 * sin(δ2 * t)
To find the potential energy, we can use the definition of potential energy as the negative of the work done by conservative forces. In this case, the gravitational potential energy can be considered:
PE = -∫F * dr
Since the force acting on the particle is only in the radial direction, the potential energy can be calculated as:
PE = -∫F_r * dr
= -∫(-m * a * δ1^2 * cos(δ1 * t)) * dr
= m * a * δ1^2 * ∫cos(δ1 * t) * dr
= m * a * δ1^2 * ∫cos(δ1 * t) * (a * δ1 * sin(δ1 * t)) dt
= m * a^2 * δ1^3 * ∫(cos^2(δ1 * t) * sin(δ1 * t)) dt
Using double angle identities to simplify the integrand:
PE = m * a^2 * δ1^3 * ∫((1/2)(1 + cos(2 * δ1 * t))) * sin(δ1 * t) dt
= m * a^2 * δ1^3 * ∫(sin(δ1 * t) - (1/2)sin(3 * δ1 * t)) dt
= m * a^2 * δ1^3 * ((-1/δ1)cos(δ1 * t) - (1/2δ1)cos(3 * δ1 * t)) + C
Where C is the constant of integration.
To find the kinetic energy, we can use the formula:
KE = (1/2) m * (v_x^2 + v_y^2)
Substituting v_x(t) and v_y(t) from earlier:
KE = (1/2) m * (a^2 * δ1^2 * sin^2(δ1 * t) + a^2 * δ2^2 * cos^2(δ2 * t))
= (1/2) m * (a^2 * (δ1^2 * sin^2(δ1 * t) + δ2^2 * cos^2(δ2 * t)))
Hence, the total energy E:
E = KE + PE
= (1/2) m * (a^2 * (δ1^2 * sin^2(δ1 * t) + δ2^2 * cos^2(δ2 * t))) + m * a^2 * δ1^3 * ((-1/δ1)cos(δ1 * t) - (1/2δ1)cos(3 * δ1 * t)) + C
Simplifying the terms further, we can express E as:
E = (1/2) m * (a^2 * (δ1^2 + δ2^2)) + m * a^2 * δ1^3 * ((-1/δ1)cos(δ1 * t) - (1/2δ1)cos(3 * δ1 * t)) + C
Therefore, the total energy of the particle can be represented as:
E = (1/2) m * (a^2 * (δ1^2 + δ2^2)) + m * a^2 * δ1^3 * cos(δ1 * t) + (1/2) m * a^2 * δ1^3 * cos(3 * δ1 * t) + C