Beginning from rest when θ=20^∘, a 35 -kg child slides with negligible friction down the sliding

Beginning from rest when θ=20^∘, a 35 -kg child slides with...

Key Concepts

Centripetal Acceleration in Circular Motion

For any object moving along a curved path, especially on a circular arc, there is a centripetal (radial) acceleration directed towards the center of the curvature. This acceleration, given by the formula v²/r, requires a net inward force that is typically provided by the normal force in addition to any components of other forces acting towards the center.

Newton's Second Law in Curvilinear Coordinates

Applying Newton's second law in the context of curvilinear (circular) motion involves analyzing the forces in tangential and radial directions separately. In the tangential direction, the net force (such as the component of gravity) produces the tangential acceleration. In the radial direction, the net inward force (comprising components of the normal and gravitational forces) produces the centripetal acceleration, ensuring the object stays on the circular path.

Force Decomposition

This concept involves breaking down the gravitational force into components that are tangential and normal (or radial) to the path of motion. In the context of an object moving along a curved (circular arc) incline, the tangential component (mg sin?) drives the motion along the path, while the normal component (mg cos?) influences the centripetal or normal acceleration required to keep the object moving along the curved track.

Conservation of Energy

When dealing with frictionless motion, the principle of energy conservation is key. It states that the loss in gravitational potential energy as the object descends is converted into kinetic energy. This principle allows one to compute the speed of the object at a given position along the path by equating the change in potential energy to the gain in kinetic energy.

Transcript

00:01 We have mass of the child m equals to 35 kg and initial speed v0 equals to 0 meter per second and keta not equals to 20 degree and radius of the track is equal to 2 .5 meter.

00:18 Okay.

00:18 So now first of all we can draw the diagram for the condition.

00:24 So this is our tangential line.

00:26 Okay.

00:27 And this is our normal line.

00:30 And this is the vertical line okay these are three different different lines okay this is our nor vertical line and this is our tangential line and this is our normal line so now there is a gravitational force along this direction which is m g vector and there is a normal force along this direction which is n vector okay and this angle is theta and this angle is also equals to theta okay so now we can write in the tangential direction that m g cost theta will be equals to m a t okay mass multiplied by tangential acceleration so from here we get a t equals to g cost theta okay this is say our equation number one now we know that dv by d t is equal to acceleration and it can be rewritten as dv by d t d b d s multiplied by d s by d t so it can be written as v dv by r d theta.

01:42 And it is also equals to g cos theta which is a tangential acceleration and tangential acceleration is equal to g cost theta.

01:51 So we get this equation.

01:53 So from here we get vdv integration from v0 to v which is equals to rg cata t theta integration from theta knot to theta okay so from here after solving we get v square by 2 minus v0 square by 2 that is equals to r g sine theta minus sine theta not okay and as we have v not equals to 0 so we will get that v square or v equals to under root of 2 g sine theta minus sine theta not is equal to 20 degrees so 20 degree okay say this is our equation number two now from the equation of motion in the normal direction we can write that m minus m g sine theta which will give us a centipital force so m v square by r okay so from here normal force n comes out to be m g sine theta plus v square by r say this is our equation number 3.

03:07 So now moving to the part a of the problem in which we have theta equals to 30 degrees.

03:13 So now we will calculate all the values of equation 1, 2 and 3.

03:18 So now substituting theta equals to 30 degree in equation 1, we get 80 equals to 9 .81 cost 30 degree.

03:32 So this will be equals to 8 .5 g.

03:36 Meter per second square okay and after substituting value in equation two we will get v equals to under root of two multiplied by r is 2 .5 and g is 9 .81 bracket sine 30 degree minus sine 20 degree okay so from here after solving we will get velocity v equals to or speed v equals to 2 .78 meter per second okay and similarly uh in the equation third, after substituting, we will get n equals to m is 35 kg, g is 9 .81, sine 30 degree plus v is equal to 2 .78...

Please give Ace some feedback