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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 0.0800 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 0.0800 kg
sphere as its passes through its lowest point?

User Shoniisra
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1 Answer

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Answer:

The linear speed of the 0.0800 kg sphere as it passes through its lowest point is 0.989 m/s.

Step-by-step explanation:

To solve this problem, we can use the principle of conservation of energy, which states that the initial potential energy of the system is equal to the final kinetic energy of the system, assuming no energy is lost to friction or other non-conservative forces.

Initially, the system has only potential energy due to the height of the spheres above their lowest point. At the lowest point, all of this potential energy is converted to kinetic energy, which is shared between the two spheres. We can set the initial potential energy equal to the final kinetic energy and solve for the final speed of the 0.0800 kg sphere.

The initial potential energy of the system is given by:

U_i = mgh = (0.0200 kg)(9.81 m/s^2)(40.0 cm) + (0.0800 kg)(9.81 m/s^2)(40.0 cm) = 0.0782 J

where h is the height of the spheres above their lowest point, which is half the length of the rod.

The final kinetic energy of the system is given by:

K_f = (1/2)mv^2

where m is the mass of the 0.0800 kg sphere and v is its final speed.

Setting the initial potential energy equal to the final kinetic energy and solving for v, we get:

0.0782 J = (1/2)(0.0800 kg)v^2

v^2 = 0.97875 m^2/s^2

v = 0.989 m/s (to three significant figures)

User Fingerman
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