The pulley shown in figure ( 10-E 8 ) has a radius 10 cm and moment of inertia 0-5 kg-m^2 abou

The pulley shown in figure ( 10-E 8 ) has a radius 10 cm...

Key Concepts

Moment of Inertia

The moment of inertia measures an object's resistance to changes in its rotation and depends on both the mass of the object and its distribution relative to the axis. It plays a crucial role in determining how an applied torque will affect the angular acceleration of the system.

Newton's Second Law

Newton's second law is fundamental for analyzing both translational and rotational motion. It states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass, and in rotational systems, that the angular acceleration is proportional to the net torque and inversely proportional to the moment of inertia.

Rotational Dynamics

Rotational dynamics involves studying the motion of objects that rotate about an axis. It examines how torques result in angular accelerations, emphasizing the balance between applied torques and the rotational inertia of the object.

Relationship between Linear and Angular Motion

The relationship between linear and angular motion is encapsulated by the equation a = R?, which links the linear acceleration at the edge of a rotating object to its angular acceleration. This connection is essential for analyzing systems where rotational motion influences translational dynamics, such as in pulley systems.

Transcript

00:01 In this question we have given a pulley whose radius is given to us that is 10 semi and moment of inertia is given to us and it is told that the planes are frictionless and we have to calculate the acceleration of block this that is 4 kg block.

00:22 Now i assume that the mass of this block is m1 so m1 is given to us that is 2 kg and m2 is 4 kg we have given the moment of inertia of this pulley that is i is equal to 0 .5 kg meter square and radius of the pulley is 0 .1 meter so i'm going to show the forces acting on the system one is the component of this weight in this direction that is m1g sine theta and there also the component of weight in this direction is m2g sine theta which causes the acceleration of the block in this direction and tension in this string is let that is t1 and tension in this string is t2 now due to the motion of this block there is a rotation in this pulley that the angular acceleration is given by alpha and due to and now the force acting in this block is m2a and on this block this is m1a which causes this block to move in upward direction and this causes to move block in this direction now i'm going to solve this problem so from the fbd we can write that from free body diagram we can say that t1 minus m1g sine theta that is net force is equal to m1a this is our first this is the for block m1 similarly for pulley i can write t2 minus t1 is equal to i into a by r square this is our equation second and for block m2 i can write that m2g sine theta minus t 2 is equal to m2a this is our equation third now adding equation 1 and third so add equation 1 and 3rd we get m2g sine theta plus this is minus t 2 plus t 1 is a minus m 1g sine theta this equals to m1 plus m2 into a now simplify this equation so this is m2 minus m1 g sine theta is equal to sorry this is plus minus i upon r square into a is equals to m1 plus m2 into from here we get the value of acceleration a that is m2 minus m1 g sine theta upon this is m1 plus m2 plus i upon r square so this is the generalized equation now put the values in this equation we have all the known information so m2 is 4 kg m1 is 2 kg into this is 10 sine angle is 45 upon 4 plus sorry this is 2 plus 4 plus i is given to that is 0 .5 upon this is 0 .1 whole square after solving this we get acceleration that is 0 .248 or i can write it 0 .25 meter per second square...

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