Two uniform cylinders, each of weight W=14 lb and radius r=5 in. are connected by a belt as show

Two uniform cylinders, each of weight W=14 lb and radius r=5...

Two uniform cylinders, each of weight $W=14$ lb and radius $r=5 \mathrm{in.}$ are connected by a belt as shown. If the system is released from rest, determine (a) the angular acceleration of each cylinder, $(b)$ the tension in the portion of belt connecting the two cylinders, (c) the velocity of the center of the cylinder $A$ after it has moved through $3 \mathrm{ft.}$

Two uniform cylinders, each of weight $W=14$ lb and radius $r=5 \mathrm{in.}$ are connected by a belt as shown. If the system is released from rest, determine (a) the angular acceleration of each cylinder, $(b)$ the
tension in the portion of belt connecting the two cylinders, (c) the velocity of the center of the cylinder $A$ after it has moved through $3 \mathrm{ft.}$

Key Concepts

Moment of Inertia

The moment of inertia quantifies an object's resistance to change in its rotational motion. For a uniform cylinder, it is determined by its mass distribution relative to its axis. This concept is essential in predicting how a cylinder will accelerate under a given torque, and it plays a crucial role in solving rotational dynamics problems.

Torque and Angular Acceleration

Torque is the rotational equivalent of force and produces an angular acceleration according to Newton's second law for rotation (torque equals moment of inertia times angular acceleration). This relationship allows the calculation of angular acceleration when a net torque is applied, forming the foundation of analyzing the dynamics of rotating systems.

Belt and Pulley Dynamics

In systems where cylinders are connected by a belt, the belt transmits forces between the rotating bodies. Understanding belt dynamics involves analyzing the tension in the belt, which can differ in various segments due to the applied forces and rotational motions. Properly accounting for the belt tension is critical when connecting the rotational behavior of individual cylinders in a composite system.

Linear and Angular Motion Relationship

The relationship between linear and angular quantities is key in systems where rotation results in translational movement. This relation, given by v = ?r and a = ?r, links the angular acceleration or velocity of a rotating body to the linear acceleration or speed of points on its circumference. It is especially important when determining how far a cylinder moves linearly based on its rotational behavior.

Transcript

00:01 Hello everyone in this problem it is given the two uniform cylinders each weighing 14 pound and radius 5 inch are connected by a belt as shown in the figure.

00:22 If the system is released from rest that is omega not is zero u is zero we have to calculate angular acceleration of each cylinder tension in the portion of vent connected to the two cylinder.

00:47 So tension a b and c the velocity of center of the cylinder a after it is moved three fit so tension velocity of center of a sorry for moving down by three fit that acceleration of a is downward drawing its geometrical figure only like this this is the center this is the tension in the belt between the two cylinder, a and v.

02:08 This is the tension in the belt at p point.

02:19 Weight will act vertically downward.

02:24 Acceleration is downward.

02:28 This is g.

02:38 Let acceleration of a downward.

02:52 Acceleration of v is aav downward.

03:22 Angular acceleration of a in cropped by direction and acceleration of p acceleration of a b liquid written as a a plus r into alpha a so it becomes twice of r into alpha a here you may write acceleration of a is equal to alpha into r a and that will be into alpha v at the point of contact say this is equation 1 this is 2 moment of inertia both having the same mass mass of a is equal to mass of v already written moment of inertia of both cylinder will be the same mr square by 2 for disk a moment of force about b point counter clock by taken positive so r into omega a sorry r into weight of a minus two r tension a v is equal to r m a acceleration of a plus moment of inertia of a into angular acceleration of a so this is equation five now for disk v first we will draw the geometrical diagram of it weight vertically downward, tension in the belt at the point of contact, reaction, by and bx moving clockwise.

07:23 So for disk v, you can find moment of force about b point.

07:40 This is the center, about b point, if we find it would be r into tension av, it's called to iv, v, alpha v...

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