A bar of mass m=5 kg is held as shown between four disks each of mass m^'=2 kg and radius r=75 m

step 1 Hello everyone as soon in the figure. There is a bar of mass 5 KG. It is held as soon between th

A bar of mass m=5 kg is held as shown between four disks...

A bar of mass $m=5 \mathrm{kg}$ is held as shown between four disks each of mass $m^{\prime}=2 \mathrm{kg}$ and radius $r=75 \mathrm{mm} .$ Knowing that the forces exerted on the disks are sufficient to prevent slipping and that the bar is released from rest, for each of the cases shown, determine the velocity of the bar after it has moved through the distance $h$

Key Concepts

Conservation of Mechanical Energy

This concept involves the principle that in the absence of non-conservative forces (or when they do not do work), the total mechanical energy (potential plus kinetic energy) of a system remains constant. In problems involving objects moving under gravity, the potential energy lost is converted into kinetic energy, which may include both translational and rotational energy components.

Translational Kinetic Energy

Translational kinetic energy is the energy associated with the linear motion of an object's center of mass. It is calculated using the object's mass and the square of its velocity. In problems where parts of a system move linearly, understanding how gravitational potential energy transforms into translational kinetic energy is key.

Rotational Kinetic Energy

Rotational kinetic energy refers to the energy possessed by a rotating object due to its angular velocity and moment of inertia. For objects that are rolling or rotating, a portion of the energy transfer involves converting potential energy into rotational kinetic energy, which needs to be accounted for when applying energy conservation.

Moment of Inertia

Moment of inertia quantifies an object's resistance to changes in its rotational motion. It depends on the mass distribution relative to the axis of rotation. In dynamics problems, knowing the moment of inertia is essential for calculating rotational kinetic energy and for understanding how energy is partitioned between rotational and translational motion.

Rolling Without Slipping

The rolling without slipping condition is a kinematic constraint that relates the linear speed of the center of mass of a rolling object to its angular speed. Specifically, it ensures that there is a fixed relationship between these two quantities (v = r?), meaning that the point of contact does not slide relative to the surface. This condition plays a critical role in coupling translational and rotational motions in energy balance calculations.

Transcript

00:01 Hello everyone as soon in the figure.

00:04 There is a bar of mass 5 kg.

00:12 It is held as soon between the four discs each having the mass 2 kg and radius 75mm that is 0 .075 m miter.

00:33 It is given that force accepted on the disk are sufficient to prevent sylphic.

00:40 So there would be pure rolling motion.

01:06 If the war is released from rest, that is omega 1 is 0, v1 is 0, we have to find the velocity of war when it falls as distance in each three cases as given in the diagram.

01:52 As you may, v is velocity of war in downward direction and v -des is velocity.

02:18 Of center of mass g of the upper left disk as shown in the diagram for all the three arrangements the magnitude of mass center velocity are same for all discs and likewise the angular speed are same for all moment of inertia of each disk will be m -des -dre square by 2 kinetic energy of the system t's skull to half m b square plus four times half m v square m dash square half i omega square substituting the value mass of the bar is 5 k g v square mass of each disk excuse me please into v -daz square half mr square mass is 2 excuse me please so t is culta you will get 2 .5b square for position 1 t 1 is 0 work done position 2 bar has moved down our distance edge all the distance all the distance moves up sorry move down by h desk work done from position 1 to 2 you may also write u1mg h plus 4 times m dash g h dash mass is 5 kg bar and that of each disk it is 2 so potential energy sorry, by gravity you will get 5g h plus 8 g h -des.

09:00 Kynematics and kinetic energy for case a here mass center of each disk not moving that is v -des is 0 and h -dess is 0 so we can write omega is equal to v upon r so total energy in second position for case 1, 2 .5b square plus 0 2b squared.

10:48 So it becomes 4 .5b square using equation 1, kinematics and kinetic energy for case b.

12:11 The spontaneous center c of a typical disk lies, the stintaneous center c, of a typical disc like at its point of contact with the fixed wall as so in the figure...

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