Rotational Kinematics -06- Particle Rotating Inside a Conical Enclosure

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A particle is inside a smooth conical envelope rotating at a certain speed. If the particle is 45 centimeters from the axis of the cone, what is its velocity?  Watch the video about ROTATIONAL KINEMATICS in full: https://dailymotion.com/playlist/x8z9ke  #rotationalkinematics #rotationalmotion

Transcript

00:00Hi friends, a. Be a knowledgeable person who educates others.

00:05Be an intelligent person who makes others intelligent.

00:09A successful person who makes others successful.

00:13A conical shell rotates at a certain angular velocity.

00:18A particle is on the inner surface of the cone.

00:22We are asked to calculate the magnitude of the particle's velocity so that the particle does not move relative to its original position.

00:30Problems like this require imagination.

00:35How about we watch this animation for a moment?

00:40It is not easy to draw the inner surface of the conical shell.

00:45A particle is on the color boundary line.

00:50What is important to note is that the surface is smooth.

00:55Of course, the particle will move in a straight line toward the apex of the cone.

01:00In our case, the cone is rotating at a certain angular velocity.

01:08The particle on the surface also rotates with the cone.

01:14During this process, the particle remains on the boundary line.

01:17This means that the particle is not actually moving at all relative to its initial position.

01:26However, a stationary observer outside the cone sees the particle rotating about the cone's axis.

01:32This linear velocity is what we want to calculate.

01:35I think this animation is clear enough.

01:41Now we will identify the forces acting on the particle.

01:43Particles have mass and they are attracted toward the center of the earth by the force of gravity.

01:52The particle touches the surface of the cone and the normal force acts in the normal direction.

02:00It seems there are no other forces besides these two.

02:07The particle's trajectory is circular.

02:10We can say that the particle is in circular motion.

02:16The force that causes the particle to move around the axis is the centripetal force.

02:20However, unfortunately, there is no such force.

02:25We must decompose the normal force into its component vectors.

02:31It is clear that N sine theta is directed towards the center of the circle.

02:36This is the force that contributes to the centripetal force.

02:41Simply write N sine theta is equal to MV squared over R.

02:46R here is the distance from the particle to the axis of rotation.

02:50We don't yet know the values of N and V, so we need one more equation.

02:58Since the particle isn't moving in a vertical direction, the resultant force in that direction is 0.

03:05N cosine theta minus MG equals 0.

03:10We can write this equation as N cosine theta equals MG.

03:17To eliminate N, we can divide the two equations.

03:20Sine over cosine is tangent.

03:29From here, AV equals root RG tangent of theta.

03:35It looks like all the values on the right-hand side are already written on the worksheet.

03:39V is about 1.5 meters per second.

03:42Happy studying, everyone.

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