A 9-in.-radius cylinder of weight 18 lb rests on a 6-lb...

Name: Problem 73, Chapter 17: Vector Mechanics for Engineers: Statics and Dynamics
Uploaded: 2020-08-14T02:21:29-08:00
Duration: 6 min 45 s
Channel: Aatish Gupta
Description: A 9-in.-radius cylinder of weight 18 lb rests on a 6-lb carriage. The system is at rest when a force P of magnitude 2.5 lb is applied as shown for 1.2 s. Knowing that the cylinder rolls without sliding on the carriage and neglecting the mass of the wheels of the carriage, determine the resulting velocity of (a) the carriage, (b) the center of the cylinder.

A 9-in.-radius cylinder of weight 18 lb rests on a 6-lb carriage. The system is at rest when a force P of magnitude 2.5 lb is applied as shown for 1.2 s. Knowing that the cylinder rolls without sliding on the carriage and neglecting the mass of the wheels of the carriage, determine the resulting velocity of (a) the carriage, (b) the center of the cylinder.

Key Concepts

Rolling Without Slipping

This concept refers to the condition where an object, such as a cylinder, rolls on a surface without sliding. It imposes the constraint that the point of contact between the rolling object and the surface has zero relative velocity, thereby linking the object's translational velocity to its angular velocity through the relationship v = ?r. This idea is essential when analyzing the kinematics and forces in systems where rolling motion is involved.

Impulse-Momentum Theorem

This theorem states that the impulse applied to an object or system (the product of force and the time interval over which it acts) is equal to the change in its momentum. It is a key concept in dynamics for determining the resulting velocities of bodies when forces act over short time intervals, making it particularly useful for problems where an impulse is applied to a mechanical system.

System Analysis in Dynamics

In many mechanics problems involving interacting bodies, analyzing the system as a whole can simplify the problem. By considering all internal forces as canceling out, one can focus on external forces and impulses acting on the overall system. This approach is crucial for understanding how the motion of one component affects another in coupled systems, such as a rolling cylinder on a movable carriage.

Constraints in Coupled Motion

When dealing with systems where components are interconnected through conditions like rolling without slipping, additional constraints are imposed that relate the motions of different parts. These constraints ensure that the translational and rotational motions remain consistent with each other, allowing for the derivation of relationships that are essential to solve for the individual motions and velocities in the system.

Transcript

00:01 So we have a cylinder being pulled on top of a cart by a force p.

00:06 And we have all these givens here, which i've arranged.

00:08 So we have an r of nine inches, which becomes three quarters of a foot.

00:13 We have 18 pounds, which is the mass of the cylinder.

00:17 That becomes 0 .559 pounds second squared per foot.

00:22 And we did the same thing for weight b.

00:24 And the pulling force is 2 .5 pounds.

00:28 So the first thing we're going to look at is the row.

00:31 Rolling force, the rolling force along, we're going to look at the cylinder itself.

00:45 So on this cylinder we have a force or an impulse rather that's being created around this point c.

00:57 So we're going to look at our moments about c.

01:03 So our initial, our initial angular momentum is zero since it's starting from rest.

01:09 We have a this we have this impulse being created pt times r and we have the resulting translational rotational rotational momentum and we have our rotational angular angular momentum as well so we have ptr so that's 2 .5 times 1 .2 times the radius which is 0 .75 is equal to 0 .55 and this is the m of a by the way 559 this is equal to 1 half m r squared so we have 0 .559 times v times r which is 0 .75 times v a is equal to this is plus 1 .75 a is equal to this is plus 1 1 1 half f .m .a so that's 0 .559 times 0 .75 squared times omega.

02:40 So now we have 2 .25 is equal to 0 .559 times 0 .75 which is 0 .419 va plus 0 .5 times 0 .49 v .a plus 0 .5 times 0 .5 times.

03:11 0 .559 times .75 squared is 0 .157 omega.

03:23 So now we have one of our simultaneous equations that we will use to solve this problem.

03:34 So we're going to highlight this guy.

03:37 We're going to move them up here...

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