A uniform disk of mass 10.0 kg and radius 0.250 m spins at...

A uniform disk of mass $10.0 \mathrm{~kg}$ and radius $0.250 \mathrm{~m}$ spins at 300 rev $/$ min on a low-friction axle. It must be brought to a stop in 1.00 min by a brake pad that makes contact with the disk at an average distance $0.220 \mathrm{~m}$ from the axis. The coefficient of friction between pad and disk is 0.500 . A piston in a cylinder of diameter $5.00 \mathrm{~cm}$ presses the brake pad against the disk. Find the pressure required for the brake fluid in the cylinder.

Key Concepts

Moment of Inertia

The moment of inertia quantifies an object's resistance to angular acceleration based on its mass distribution relative to the axis of rotation. It is a key parameter when analyzing rotational dynamics because it plays the same role in rotational motion as mass does in linear motion.

Rotational Kinematics

Rotational kinematics involves the study of angular quantities such as angular displacement, angular velocity, and angular acceleration, and their relationships over time under constant or variable acceleration. This is essential for understanding how quickly a spinning object slows down or speeds up.

Torque

Torque is the measure of the rotational force applied to an object, calculated as the product of the force and the lever arm (the perpendicular distance from the axis of rotation to the line of action of the force). It is crucial in determining how forces affect rotational motion.

Friction in Rotational Motion

Friction in rotational systems is the resisting force that acts at the interface between two surfaces, leading to a torque that opposes the motion. The effectiveness of friction depends on the coefficient of friction and the normal force, which in turn affects how quickly a rotating system can be decelerated.

Hydraulic Pressure

Hydraulic pressure is the force exerted per unit area in a fluid system, commonly used to convert fluid pressure into mechanical force via pistons. It is fundamental in brake systems where fluid pressure is used to press brake pads against a rotating surface to create the necessary frictional force.

Transcript

00:01 This particular case we have to use the torque the torque is given by moment of inertia times the angular acceleration which in this particular case is frictional force or let us just write lower case f the frictional force times the distance which so now the moment of inertia is half m r squared and due to the angular acceleration is basically a retardation so that's omega 2 minus omega 1 where omega n is and is reducing divided by delta t now let us put a negative sign in front of the friction because it is coming down as well it is reducing as well half m r squared now the change in omega is given as two pi and dot divided by delta t equals with a negative sign over here because it is coming down negative fd so from here we can find that the frictional force is given by d delta t down down below the 2s cancels out so this is pi mr square and dot now we have the normal force we have the frictional force equals mu times the normal force so from here we can find that the normal force is given by f over mu we will use this to determine what is the pressure the pressure is given by the normal force divided by four over pi d square so pi r square becomes pi d squared over four so now we are going to substitute the fn with f over mu downstairs we have got four d squared we have got uh pi d squared down there now we are going to substitute the f with f this guy so 4 over pi mu d squared times pi mr square n dot over d gelza t actually let's make these diameters to be uppercase d such that we do not conscious with the lowercase d so if we now put the in then we get the pressure equals 4 times 3 .14 m equals 10 times r equals 0 .25 square times n dot equals 300 divided by actually we don't need the pi the pi cancels out up and down below the the coefficient of the coefficient of friction of friction of friction is half 0 .5 times b square is 0 .05 square times we have d and delta t so that comes out to be what is a value of little d okay so okay yeah got it so uh that's the n dot will be 300 over 60 we have got d and delta t so that's a 0...

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