A uniform cylinder of mass m1 and radius r is pivoted on...

A uniform cylinder of mass m1 and radius R is pivoted on frictionless bearings. A massless string wrapped around the cylinder is connected to a block of mass m2 that is on a frictionless incline of angle θ as shown below. The system is released from rest when the block is at a vertical distance h above the bottom of the incline. (Use any variable or symbol stated above along with the following as necessary: g.) (a) What is the acceleration of the block? a = (b) What is the tension in the string? T = (c) What is the speed of the block as it reaches the bottom of the incline? v = (d) Evaluate your answers for the special case where θ = 90° and m1 = 0. Are your answers what you would expect for this special case? a = m/s^2 T = N v = m/s

Transcript

00:01 All right, hello, in this question we have this cylinder, which is fixed in place and is going to be caused to rotate by the string that connects it to this other block down here, which is on a frictionless incline at a distance h.

00:12 And we're asked to figure out what is the acceleration of the block for the first part.

00:16 So for this, we're going to need to go ahead and draw a free body diagram.

00:20 So we're gonna have the weight of the mass going down, we're gonna have some normal force going up, and then we're gonna have some tension in the string opposing it.

00:28 And the reason there's a tension is because we also need to accelerate this thing, it's gonna have some angular acceleration.

00:34 So let's go ahead and write some of the torques for that disc there.

00:38 We're going to have tension, and that's acting at a distance r from the pivot of radius r.

00:43 That's gonna equal the moment of inertia times the angular acceleration.

00:47 Well, what is this in terms of things we know? we know that our moment of inertia, because it's a uniform disc, is gonna be 1 1 2 m r squared.

00:56 And then we know that our tangential acceleration is equal to our angular acceleration times r.

01:00 So alpha is going to be our acceleration of this system divided by r.

01:05 Cancel those guys out, and we'll get 1 1 2 m 1 r a.

01:09 That's our sum of the torques.

01:10 Let's go ahead and look at a sum of the forces in the x direction for this here.

01:14 I'm gonna say that down the ramp is positive x and normal is positive y.

01:18 So we have two components here.

01:19 We have a component of our weight that accelerates it.

01:23 This will be our angle theta here.

01:24 And so we have our weight times just the sine of theta, just the x component that is being opposed by the tension, and that's gonna equal m times the acceleration.

01:34 So if i solve this equation up here for tension, i'm gonna get that the tension is equal to 1 1 2 m 1 times a.

01:42 And then i can plug that into here in order to figure out what the acceleration is gonna be.

01:50 So i have this equation here.

01:51 I wanna solve that for acceleration.

01:53 So i can go ahead and do that.

01:54 And i'll get that the acceleration is gonna be m g.

01:57 This is m 2 g, the mass of the block, times the sine of whatever that angle is divided by m 2 plus 1 1 2 m 1.

02:08 So that'll be our acceleration.

02:11 In order to figure out what our tension is, now that we know our acceleration, we can plug that back into here.

02:15 So that's part a, part b is what is the tension.

02:18 We'll just plug that back in.

02:19 And we'll just leave it like that because there's no really easy way to simplify that down...

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