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ikadub [295]
4 years ago
15

A uniform rod of mass m has very tiny lead balls, each of mass m, welded to each end. If the rod hangs in empty space, and a for

ce F is applied to the rod, consider three cases: A) The force is applied to the upper lead ball; B) The force is applied halfway between the upper lead ball and the center of mass of the system; C) The force is applied at the center of mass of the system.
Physics
1 answer:
cluponka [151]4 years ago
7 0

Options:

1.Case

C2.It will be the same in all three cases.

3.Case A

4.Case B

Answer:C2. It will be the same in all three cases.

Explanation:In the above scenerio,

F, is the only net force that is acting on the System. Hence, the acceleration of the center of mass (a point in an object where a force can be applied to cause a linear acceleration without leading to an angular acceleration)is the same in each of the cases highlighted.The torque(the rotational equivalent of a linear force) depends on where the force acts.

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A thin rod of length 1.4 m and mass 140 g is suspended freely from one end. It is pulled to one side and then allowed to swing l
m_a_m_a [10]

Answer:

a The kinetic energy is  KE = 0.0543 J

b The height of the center of mass above that position is  h = 1.372 \ m    

Explanation:

From the question we are told that

  The length of the rod is  L = 1.4m

   The mass of the rod m = 140 = \frac{140}{1000} = 0.140 \ kg  

   The angular speed at the lowest point is w = 1.09 \ rad/s

Generally moment of inertia of the rod about an axis that passes through its one end is

                   I = \frac{mL^2}{3}  

Substituting values

               I = \frac{(0.140) (1.4)^2}{3}

               I = 0.0915 \ kg \cdot m^2

Generally the  kinetic energy rod is mathematically represented as

             KE = \frac{1}{2} Iw^2

                    KE = \frac{1}{2} (0.0915) (1.09)^2

                           KE = 0.0543 J

From the law of conservation of energy

The kinetic energy of the rod during motion =  The potential energy of the rod at the highest point

   Therefore

                   KE = PE = mgh

                        0.0543 = mgh

                             h = \frac{0.0543}{9.8 * 0.140}

                                h = 1.372 \ m    

                 

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