In Problem 6, what are the magnitudes of (a) the horizontal...

In Problem 6, what are the magnitudes of (a) the horizontal component and (b) the vertical component of the net force acting on the block at point $Q ?(\mathrm{c})$ At what height $h$ should the block be released from rest so that it is on the verge of losing contact with the track at the top of the loop? (On the verge of losing contact means that the normal force on the block from the track has just then become zero.) (d) Graph the magnitude of the normal force on the block at the top of the loop versus initial height $h$, for the range $h=0$ to $h=6 R$

Key Concepts

Force Decomposition

This concept involves breaking a force vector into its perpendicular components, typically horizontal and vertical. It is crucial in analyzing forces acting at an angle, allowing one to apply Newton’s laws separately in each direction to solve for net forces or accelerations.

Newton’s Second Law in Dynamics

Newton’s Second Law states that the net force on an object is equal to the mass multiplied by its acceleration. This principle is fundamental when determining the resultant force acting on an object by summing the contributions of various force components.

Circular Motion and Centripetal Force

In circular motion, an object moving in a curved path requires a centripetal force directed inward toward the center of the circle. The relationship between centripetal force, mass, velocity, and radius is essential when analyzing situations where a block traverses a loop, as it ensures the object follows the curved path.

Energy Conservation in Mechanical Systems

This principle is based on the idea that the sum of kinetic and potential energy remains constant in a closed system without non-conservative forces. It is used to relate the potential energy at a height to the kinetic energy at various points along a track, providing a way to determine speeds and forces at different positions.

Normal Force and the Condition for Loss of Contact

The normal force is the perpendicular force exerted by a surface on an object in contact with it. When an object is on the verge of losing contact with a circular track, the normal force becomes zero. This condition is pivotal for determining the minimum initial conditions, such as height, needed for the object to maintain continuous contact during its motion.

Transcript

00:01 All right, so the problem about energy conservation, but also about forces.

00:12 So in part a, first we do energy conservation.

00:17 So at the top, you have just potential energy, u -sof -p, no kinetic energy there, and that's equal to a q.

00:25 You have, and that is converted to potential energy at q and kinetic energy at q.

00:33 So mghh will be equal to mgr, because that's the height advises to plus.

00:44 Kinetic energy is one half mv squared.

00:50 So the m's cancel out here to give you a v, or a v squared of gh, minus gr times 2.

01:07 And then h is 5r, so you have 5gr minus gr, so that's 2 times 4gr, giving you 8 times gr for v squared.

01:22 Therefore, the force in the x direction, your horizontal force will be due to the centripetal force.

01:37 And the centripetal force will be directed horizontally, and its magnitude is given by mv squared over r.

01:45 And so you replace v squared here, 8gr over r, giving us 8mg for the magnitude.

02:02 So you replace values here, 0 .0 .2.

02:05 3 2 kilograms is mass 9 .8 meters per second squared is velocity, as gravitation acceleration, excuse me.

02:19 Therefore, f sub x is 2 .5 newtons, that's the magnitude, and it is directed to the left, because that's the direction of acceleration due to the centripetal force.

02:37 So ball will be feeling a force, net force, the left of 2 .5 newtons.

02:44 Part b is much simpler because the net vertical force is just gravity.

02:50 So that's mg .032 times 9 .8 meters per second squared, giving you 0 .31 newtons as the net vertical force.

03:04 And it will be directed, the net vertical force will be directed, directed a directed, directed a downwards, of course.

03:19 So in part c, we want to equate the centripetal force to the gravitational force.

03:27 This means that mv squared, v being the final velocity over r is equal to m g...

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