Review problem. A uniform disk of mass 10.0 kg and ra dius 0.250 m spins at 300 rev/min on a low

Review problem. A uniform disk of mass 10.0 kg and ra dius...

Review problem. A uniform disk of mass $10.0 \mathrm{kg}$ and $\mathrm{ra}$ dius $0.250 \mathrm{m}$ 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 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

Rotational Kinematics

Rotational kinematics deals with the motion of objects rotating about an axis, including quantities such as angular velocity, angular acceleration, and time. In problems involving deceleration of rotating bodies, it is important to convert rotations per minute to radians per second and to determine the angular acceleration necessary to bring the motion to a stop within a specified time.

Rotational Dynamics

Rotational dynamics focuses on the relationship between torque, moment of inertia, and angular acceleration. The net torque acting on a rotating object is equal to the product of its moment of inertia and its angular acceleration. This concept is crucial when determining the braking torque required to decelerate a spinning disk.

Moment of Inertia

The moment of inertia quantifies how an object's mass is distributed relative to an axis of rotation and determines the resistance of the object to angular acceleration. For geometries like a uniform disk, the moment of inertia has a standard formula that is used to calculate the torque needed to alter its state of motion.

Friction Force and Friction Torque

Friction is the force that opposes relative motion between surfaces in contact. The friction force is given by the product of the normal force and the coefficient of friction. When this force is applied at some distance from the axis of rotation, it creates a frictional torque that works to decelerate the rotating object.

Hydraulic Pressure and Force

Pressure in a hydraulic system is defined as force per unit area. Using the relationship between pressure, force, and the area of a piston, one can determine the necessary fluid pressure to produce a given normal force. This normal force, in turn, is responsible for the frictional force that provides the braking torque.

Transcript

00:01 Hello, so in this problem, we have a disk that's spinning, and we're applying a brake pad to the disc to stop it from spinning.

00:14 We're trying to figure out the pressure of this disc pad or this brake cylinder we're applying in order to stop this disc from spinning.

00:27 Here are some of the numbers.

00:28 So our disk has a mass of 10 kilograms.

00:34 It has a total radius.

00:42 The disc has a radius of 0 .25 meters.

00:49 The brake pad is on average a distance 0 .22 meters away from the center.

00:59 The diameter of the cylinder we're working with is the brake cylinder working with is 5 centimeters.

01:11 And the coefficient of friction, the kinetic friction, is equal to 0 .5.

01:18 And finally, the disc is rotating at an angular velocity of omega of 300 revolutions per minute.

01:34 Okay, so how do we tackle this problem? well, there are a few steps.

01:41 We can start out by figuring out the torque applied onto the disk using two methods.

01:52 So we have two equations that's available for us for torque.

01:55 The first is r cross f, as we all know.

02:00 So let's write that down.

02:01 We have r cross f.

02:03 So what's r cross f? in this case, the radius vector, as you can see right here, is always perpendicular to the force applied to this disk, which is going to be friction force, so it's just going to be rf since sine of 90 degrees is just one.

02:26 And i said two methods of figuring out our torque.

02:31 What was the other method? the other method was analogous to f equals ma.

02:38 We have torque equals i alpha.

02:42 So we have the moment of inertia multiplied by the angular exploration.

02:49 How do we, let me rewrite this capital f as lowercase f to show that this is the force due to friction.

02:57 Okay, now we can figure out what i is.

03:01 Well, the moment of inertia of a disk is one half, one half mr squared.

03:09 So we got that figured out, that's pretty easy.

03:13 And for figuring out alpha, we can use the analogous form of the of the equations of motions for anglers.

03:27 For anglers.

03:30 So we have the final angular velocity equals the initial angular velocity plus our angular acceleration multiply by time.

03:40 We're trying to find when this stops.

03:42 Oh, also another thing we need to add is that t is one minute.

03:48 So we're trying to stop this disk in one minute.

03:54 Okay, and we're trying to stop it in one minute.

03:57 So our final angle of velocity should equal 0.

04:00 And for initial angular velocity, we have 300 revolutions per minute.

04:07 But since i said that rf is positive, we are directing the vector f in the positive direction, meaning that our w here is going to be negative...

Please give Ace some feedback