An object of mass m1 = 0.005 kg moving with a velocity v1 = 20 m/s strikes and b

Question

An object of mass m1 = 0.005 kg moving with a velocity v1 = 20 m/s strikes and

becomes embedded in the pendulum as shown in the figure, where L =1.2 m. Assume that the

pendulum is a slender, uniform rod whose mass is 4.5 kg . Find the resulting motion ?(t), where ?

is assumed to be a small angle from vertical.

An object of mass m1 = 0.005 kg moving with a velocity v1 = 20 m/s strikes and becomes embedded in the pendulum as shown in the figure, where L = 1.2 m. Assume that the pendulum is a slender, uniform rod whose mass is 4.5 kg . Find the resulting motion ?(t), where? is assumed to be a small angle from vertical.

Explanation / Answer

Moment of Inertia of rod about O = M*(2L)^2 / 3

= 4.5*(2.4^2) / 3

= 8.64 Kgm^2

Total moment of inertia when moving mass gets embedde

= 8.64 + [m1 * (2L)^2]

= 8.64 + [0.005 * (2.4)^2]

I = 8.668 Kgm^2

By applying the conservation of angular momentum we can find the angular velocity just after impact

m1*v1*(2L) = I * omega

omega = (m1*v1*2L) / I

= (0.005*20*2.4) / 8.668

= 0.02768 rad/sec

Total enrgy of system after impact = 0.5*I*omega^2

= 0.5*8.668*0.02768^2

= 3.3219*10^-3 Joules

We can now find the amplitude of the system by applying conservation of energy,

under which the above calculated energy will be completely converted to potential energy

at its amplitude

If amplitude is a degrees, then,

h1 = L*[1 - cos(a)] = 1.2*[1 - cos(a)] ( height gained by cetre of mass of rod)

h2 = 2L*[1 - cos(a)] = 2.4*[1 - cos(a)] (height gained by mass m1)

So, potential energy = M*g*h1 + m1*g*h2

= 4.5*9.81*1.2*[1 - cos(a)] + 0.005*9.81*2.4* [1 - cos(a)]

= 53.091*[1 - cos(a)]

Equating to initial calculated energy,

53.091*[1 - cos(a)] = 3.3219*10^-3

a = 0.64 degrees

Now,

restoring torque T = m1*g*sin(theta)*2L + M*g*sin(theta)*L

T = 53.091*sin(theta)

As, theta is ver small so, sin(theta) = theta

T = I*alpha (alpha is angular acceleration)

53.091 * (theta) = 8.668 * alpha

theta = 0.1632 *(alpha)

If now omega is the angular frequency of oscillations, then

omega = sqrt(0.1632) = 0.404

So, equation of motion can be written as

theta(t) = 0.64 *sin(0.404*t) [

  

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