A small particle of mass m is pulled to the top of a...

A small particle of mass $m$ is pulled to the top of a frictionless half-cylinder (of radius $R )$ by a light cord that passes over the top of the cylinder as illustrated in Figure P7.29. (a) Assuming the particle moves at a constant speed, show that $F=m g \cos \theta .$ Note: If the particle moves at constant speed, the component of its acceleration tangent to the cylinder must be zero at all times. (b) By directly integrating $W=\int \overrightarrow{\mathbf{F}} \cdot d \overrightarrow{\mathbf{r}},$ find the work done in moving the particle at constant speed from the bottom to the top of the half-cylinder.

Key Concepts

Constant Speed Motion

Constant speed motion implies that there is no net acceleration in the direction of the path, which means that even though forces may act on the object, their components along the direction of motion cancel each other out. This condition simplifies the analysis of forces because it necessitates that any applied force's component along the path exactly counteracts the opposing forces, such as the gravitational components along the same direction.

Force Decomposition in Curved Motion

When dealing with motion along a curved path, forces must be decomposed into components that act tangentially and normally to the path. This decomposition is crucial for understanding the effect of each force component on the motion of the object, especially under constant speed conditions where the tangential acceleration is zero, allowing for the derivation of relationships between forces and motion parameters.

Work as the Integral of Force Dot Displacement

Work is defined as the integral of the dot product between the force vector and the displacement vector along the path of motion. This formulation is fundamental in mechanics because it accounts for both the magnitude and direction of the force relative to the displacement, enabling the calculation of energy transfer or work done in moving an object along a path.

Integration along a Curved Path

In scenarios where the path of motion is curved, such as in circular or arc-like trajectories, it is necessary to parameterize the path (often using an angular or arclength parameter) and then integrate over this parameter to evaluate the total work done by a force. This approach ensures that contributions from variations in the force and the geometry of the path are properly taken into account.

Transcript

00:01 In this numerical two tasks are given.

00:02 In the first, we have to prove f is equal to mg cost data.

00:07 So let's move on that.

00:08 First of all, we will calculate over here the total force acting on a particle of mass m over here.

00:18 The total force are the first one is gravitational force which is acting in the downward direction.

00:24 The other force is normal force which is acting perpendicular to the surface.

00:30 The third force is called applied force acting in the direction of the particle.

00:36 Now, we will calculate the total force with the help of newton's second law, which is f is equal to m .a.

00:47 That means force is equal to mass into acceleration.

00:52 Over here, force is a vector quantity as well as acceleration is also vector quantity.

00:58 Now we will take the summation of a force which is acting in a horizontal direction.

01:07 So the total force sigma f x will be is equal to f minus mg cos theta.

01:24 Here the total force we have calculated only in the x direction.

01:31 The reason is particle is moving only in an x direction it is not moving in a y or z direction that's why we are not taking the y and z component into the consideration moreover the force f which is a positive quantity because it is acting in the direction of the particle while the force m g because theta is a negative quantity because it is in the reverse direction of the particle movement.

02:02 Now moving further here the condition is also given that this particle is moving with the constant speed.

02:12 So its acceleration will be zero.

02:17 So i must say f is equal to m a is equal to zero and therefore sigma f of x will be also zero.

02:26 So so what i can write that is f minus m g cost theta which is is equal to zero.

02:39 And so our force will be f is equal to m g cos theta and the unit of force is newton.

02:52 So here our first part is proven.

03:01 Now we will go to the second part...

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