In Fig. 9.38 , the cylinder and pulley turn without friction...

In Fig. 9.38 , the cylinder and pulley turn without friction about stationary horizontal axles that pass through their centers. A light rope is wrapped around the cylinder, passes over the pulley, and has a $3.00-\mathrm{kg}$ box suspended from its free end. There is no slipping between the rope and the pulley surface. The uniform cylinder has mass 5.00 $\mathrm{kg}$ and radius 40.0 $\mathrm{cm} .$ The pulley is a uniform disk with mass 2.00 $\mathrm{kg}$ and radius 20.0 $\mathrm{cm} .$ The box is released from rest and descends as the rope unwraps from the cylinder. Find the speed of the box when it has fallen 1.50 $\mathrm{m} .$

In Fig. 9.38 , the cylinder and pulley turn without friction about stationary horizontal axles that pass through their centers. A light rope is wrapped around the cylinder, passes over the pulley,
and has a $3.00-\mathrm{kg}$ box suspended from its free end. There is no slipping between the rope and the pulley surface. The uniform cylinder has mass 5.00 $\mathrm{kg}$ and radius 40.0 $\mathrm{cm} .$ The pulley is a uniform disk with mass 2.00 $\mathrm{kg}$ and radius 20.0 $\mathrm{cm} .$ The box is released from rest and descends as the rope unwraps from the cylinder. Find the speed of the box when it has fallen 1.50 $\mathrm{m} .$

Key Concepts

Conservation of Energy

This principle involves equating the loss in gravitational potential energy to the gain in kinetic energy of the system, which is essential when analyzing systems where potential energy is converted into kinetic energy. In such problems, the total mechanical energy (considering both translational and rotational kinetic energies) is conserved when only conservative forces are doing work.

Rotational Dynamics

Rotational dynamics deals with the motion of bodies that can rotate about an axis, considering both angular acceleration and applied torques. It is crucial in problems involving pulleys or cylinders where parts of the system rotate, and one must consider the rotational kinetic energy along with the forces causing the rotation.

Moment of Inertia

The moment of inertia is a property of a body that quantifies its resistance to angular acceleration about an axis. Understanding and calculating the moment of inertia for different shapes (like disks and cylinders) is key in analyzing the rotational motion in the system and determining how the distribution of mass affects the dynamics.

Kinematic Constraints in Non-slipping Systems

In systems with no slipping between contacts, such as between a rope and a pulley, there is a fixed relationship between the linear and angular motions. These constraints ensure that the linear speed of the edge of the rotating object is equal to the speed of the connected mass, linking rotational motion to translational motion.

Transcript

00:01 So here in this problem we are given that the cylinder and the pulley it turns without friction about a horizontal axis that passes through the center and we are also given that a light rope is wrapped around the cylinder and it also passes over the pulley that has a mass of 3kg which is suspended at the free end and we are given that the uniform mass of the cylinder have 5kg so we can write that the mass of the box let it be m or we can note with mb is given as 3kg and the mass of uniform cylinder let it be m1 or we can write mc is equals to 5kg and the mass of the disk we can write m2 or we can write md it is given as 2kg and we are given that the radius of cylinder is 20 cm or we can take 0.

01:01 So this is 40 centimeters or 20 so this is 40 centimeter or we can write 0 .4 meter.

01:12 Now we have to calculate the speed of the box when it falls from a height of 1 .50 meter.

01:18 That means h we can take it as 1 .50 meter.

01:23 This is 1 .5 meter.

01:26 Now we're going to apply the law of conservation here to calculate the speed of the box when it falls.

01:31 From this particular height so according to the conservation of energy we can write that the change in gravitational potential energy would be equals to the change in kinetic energy of the cylinder plus the change in kinetic energy of the disk plus the kinetic energy of the release box so we can write this expression m g h is equals to half i1 omega 1 square plus i 2 1 by 2 i 2 i 2 2 square plus half mv square where v is the velocity of the speed or velocity of the box or the speed of the box now i1 is the moment of inertia of the cylinder i 2 is the moment of energy of the disk so we can write this expression as m g h is equals to 1 by 2 i 1 will be equals to m 1 r1 square and omega 1 square we can write it as v over r1 square so it's become whole square plus 1 by 2 i 2 will be at the moment of inertia of disk it will be equals to 1 by 2 m2 r2 square so 1 by 2 m 2 r2 square and omega 2 we can write as v over r2 squared plus 1 by 2 mv square so from here we're going to simplify this and we'll get the expression of mgh will be equals to m1 plus m2 plus 2m divide by 4 v square...

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