A slender rod is 80.0 cm long and has mass 0.120 kg. A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.0500-kg sphere

Question

A slender rod is 80.0 cm long and has mass 0.120 kg. A small 0.0200-kg sphere is welded to one end of the rod, and a small 0.0500-kg sphere is welded to the other end. The rod, pivoting about a stationary, frictionless axis at its center, is held horizontal and released from rest. What is the linear speed of the 0.0500-kg sphere as it passes through its lowest point?

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Tryphena 5 months 2021-08-16T05:05:55+00:00 1 Answers 7 views 0

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    2021-08-16T05:07:53+00:00

    Answer:

    Explanation:

    Given that,

    Slender rod

    Length of rod=80cm=0.8m

    Mass of slender rod=0.12kg

    Sphere Bob at one end

    Mass M1=0.02kg

    Sphere Bod at the other end

    Mass M2 =0.05kg

    Linear speed of mass 2 at the lowest point

    We need to calculate the change in potential of the complete system. m2 and m3 are the masses at the rod ends. note the rod centre of mass neither gains nor loses potential.

    So, at the lowest point,

    ∆U = M2•g•y2 + M1•g•y1

    Note, at the lowest point, the mass 1 is 40cm (0.4m) form the midpoint, Also, the mass 2 is -40cm(-04m) from the midpoint

    ∆U = M2•g•y2 + M1•g•y1

    ∆U=0.05•9.81•(-0.4) + 0.02•9.82•0.4

    ∆U=-0.1962+0.07848

    ∆U=-0.11772 Nm

    Now, the moment of inertia of the rod is given as

    I=∫r²dm

    dm=2pdr

    I= 2p∫r²dr

    I= 2 × 0.12/0.8 ∫r²dr; from r=0 to 0.4

    I=0.3 [r³/3] from r=0 to 0.4

    I= 0.3 [ 0.4³/3 -0] ,from r=0 to 0.4

    I=0.3 × 0.02133

    I=0.0064kg/m².

    calculating of inertia of the end masses.

    I(1+2)=Σmr² = (m1+m2)r²

    I(1+2)=(0.02+0.05)0.4²

    I(1+2)=0.07×0.4²

    I(1+2)=0.0112 kg/m²

    Now, the Energy of the masses due to angular velocity is given as

    K.E=½ (I + I(1+2))w²

    K.E=½(0.0064+0.0112)w²

    K.E= 0.0088w²

    Using conservation of energy

    The potential energy is equal to the kinetic energy of the system

    K.E=P.E

    0.0088w²=0.11772

    Then, w²=0.11772/0.0088

    w²=13.377

    w=√13.377

    w=3.66rad/s

    Then, the relationship between linear velocity and angular velocity is given by

    v=wr

    v=3.66×0.4

    v=1.463m/s

    The required linear speed is 1.46m/s approximately

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