A rests on a horizontal tabletop. A light horizontal rope is...

A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block $B$ is suspended from the free end of the rope. The light rope that connects the two blocks does not slip over the surface of the pulley (radius $0.080 \mathrm{~m}$ ) because the pulley rotates on a frictionless axle. The horizontal surface on which block $A$ (mass $2.50 \mathrm{~kg}$ ) moves is frictionless. The system is released from rest, and block $B$ (mass $6.00 \mathrm{~kg}$ ) moves downward $1.80 \mathrm{~m}$ in $2.00 \mathrm{~s}$. (a) What is the tension force that the rope exerts on block $B$ ? (b) What is the tension force on block $A ?$ (c) What is the moment of inertia of the pulley for rotation about the axle on which it is mounted?

A rests on a horizontal tabletop. A light horizontal rope is attached to it and passes over a pulley, and block $B$ is suspended from the free end of the rope. The light rope that connects the two blocks does not slip over the surface of the pulley (radius $0.080 \mathrm{~m}$ ) because the pulley rotates on a frictionless axle. The horizontal surface on which block $A$ (mass $2.50 \mathrm{~kg}$ ) moves is frictionless. The system is released from rest, and block $B$ (mass $6.00 \mathrm{~kg}$ ) moves downward $1.80 \mathrm{~m}$ in $2.00 \mathrm{~s}$. (a) What is the tension force that the rope exerts on block $B$ ?
(b) What is the tension force on block $A ?$ (c) What is the moment of inertia of the pulley for rotation about the axle on which it is mounted?

Key Concepts

Kinematics

Kinematics involves using relationships between displacement, time, velocity, and acceleration. In problems with constant acceleration, kinematic equations allow us to determine unknown quantities such as acceleration from given displacements and time intervals, independent of the forces causing the motion.

Newton's Second Law

Newton's Second Law states that the net force acting on an object is equal to the product of its mass and acceleration (F = ma). This law is fundamental for analyzing the translational motion of objects and helps in setting up equations of motion for each mass in the system.

Rotational Dynamics

Rotational dynamics extends the principles of Newton's laws to rotating systems. It involves analyzing torques, which are the rotational equivalent of forces, and angular accelerations, relating them through the moment of inertia of the rotating body, analogous to mass in linear motion.

Inextensible Rope Constraint

The inextensible rope constraint means that the rope does not stretch, ensuring that the linear accelerations of the masses connected by the rope are directly related to the angular acceleration of the pulley. This relationship is crucial for coupling the translational and rotational motions in such systems.

Moment of Inertia

The moment of inertia is a measure of an object's resistance to changes in its rotational motion, much like mass in linear dynamics. It depends on the mass distribution relative to the axis of rotation and plays a key role in determining how a rotating object responds to an applied torque.

Transcript

00:01 In this problem, we're using our knowledge of torques to find the moment of inertia of a pulley.

00:06 In this system, we have two blocks in a frictionless world, where block b is suspended by a rope over the pulley, and then the rope attaches to a block on a frictionless surface behind it.

00:22 The mass of block b is big m, 6 kilograms.

00:27 The mass of block a is small m, 2 .5 kilograms.

00:30 The radius of the pulley is 0 .08 meters, and after is released from rest, the block m falls 1 .8 meters in two seconds.

00:45 So first, what we need to find is the tension one in this rope that is pulling up, that is holding up b.

00:56 So we know, of course, that the total force acting on b is going to be equal to the two forces that are pulling on it.

01:05 So fg and the tension pulling the other way.

01:12 So this full force on b is going to be equal to fg minus t1.

01:23 So already we can isolate t just by adding it to both sides and subtract.

01:29 Fb, t is equal to fg minus fb.

01:37 Now we know what both of these are.

01:39 Fg is of course equal to mg, and fb is going to be equal to ma.

01:44 This is the total force acting on it.

01:48 So i can pull out the m, g minus a.

01:53 So now we just need to figure out what a is, and we'll be able to find out the tension, which we can do easily, using r.

02:02 H equals v0 t plus one half, a, t squared, starting from rest, so that's nothing.

02:12 And then we can divide by t, multiply by two.

02:18 And we'll get a is equal to 2h over t squared.

02:24 We plug in our numbers, 2 times 1 .8 over 2 squared.

02:30 We will get 0 .9 as our acceleration.

02:37 So i just need to plug that into here.

02:39 Our mass, this should be big m, i'm sorry, 6 times our g 9 .81 minus 0 .9 .9 .9.

02:57 Now we'll get us our first tension force 53 .46 or 53 .5 newtons.

03:11 Okay...

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