Question

A slender rod is 90.0 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.

Answer 1

The student's question deals with determining the moment of inertia and analyzing the rotational motion of a rod with weights attached to its ends, a typical problem in high school physics involving concepts of rotational dynamics.

Step-by-step explanation:

The question provided is a physics problem that involves calculating the moment of inertia of a system and understanding rotational dynamics.

In such a setup with a rod and masses attached to each end, we often calculate the moment of inertia about a pivot point, which in this case is the center of the rod.

The problem may also involve using the parallel axis theorem if the pivot is not at the center of mass, and applying principles from rotational dynamics, such as conservation of angular momentum, to describe the motion of the system after it is released from rest.

Answer 2

Speed of 0.08 kg mass when it will reach to the bottom position is 1.94 m/s

Step-by-step explanation:

When rod is released from rest then due to unbalanced torque about the hinge the system will rotate

Now moment of inertia of the system is given as

`I = (ML^2)/(12) + (m_1L^2)/(4) + (m_2L^2)/(4)`

now we have

`M = 0.120 kg`

`m_1 = 0.02 kg`

`m_3 = 0.08 kg`

now we have

`I = (0.120(0.90)^2)/(12) + (0.02(0.90)^2)/(4) + (0.08(0.90)^2)/(4)`

so we have

`I = 8.1 * 10^[-3} + 4.05 * 10^[-3} + 0.0162`

`I = 0.02835`

now by energy conservation we can say work done by gravity must be equal to change in kinetic energy

so we have

`(1)/(2)I\omega^2 = m_1g (L)/(2) - m_2 g(L)/(2)`

`(1)/(2)(0.02835)\omega^2 = (0.08 - 0.02)(9.81)(0.45)`

`\omega = 4.32 rad/s`

Now speed of 0.08 kg mass when it reaches to bottom point is given as

`v = \omega (L)/(2)`

`v = 4.32 (0.45)`

`v = 1.94 m/s`