Rotational Kinematics -08- String Tension in a Conical Swing

Rebiaz Studio

A 6 kg particle is attached to a vertical rod using two strings of equal length. When the system rotates about the rod's axis, the strings elongate as shown in the figure. What must be the tension in the lower string so that the tension in the upper string is 312 N?  Watch the video about ROTATIONAL KINEMATICS in full: https://dailymotion.com/playlist/x8z9ke  #rotationalkinematics #rotationalmotion

Transcript

00:00Hi friends, your future starts with you.

00:04If you want to be successful, you have to learn.

00:09We see a particle connected by two ropes.

00:14Now, we are asked to determine the tension in one rope when the tension in the other rope is known.

00:23This animation might help you understand the problem.

00:28A particle is in the XY plane.

00:33This particle is connected by a rope to a pivot on the Z axis.

00:38The upper rope and the lower rope.

00:43Shortly after, the particle rotates around the axis.

00:48Of course, both ropes are under tension.

00:52We will calculate the tension in the ropes.

00:56This system is easier to see in a two-dimensional view.

01:03As usual, we will identify what forces are acting on the particle.

01:09A particle with mass around the Earth will experience the force of Earth's gravity downward.

01:17The ropes are not broken or slack, meaning there is a tension force on the rope.

01:24The particle is rotating horizontally, but there is no force acting in that direction.

01:31We must decompose the tension force in the rope into its component vectors.

01:37Theta here is the angle formed by the rope and the horizontal line.

01:45From the problem sheet, the length of the string is 13 meters.

01:50The distance from the particle to the axis is 12 meters.

01:55This length is 10 meters.

01:59Since this system is symmetric, we can divide this length into two equal parts.

02:06We see that the particle does not move vertically at all.

02:11The resultant force in the vertical direction is zero.

02:15T1 sine theta minus T2 sine theta equals mg.

02:21We can see the mass and tension of the upper string from the problem sheet.

02:28The value of sine theta itself can be seen in the figure.

02:35This factor is 120.

02:40This is a simple calculation.

02:44T2 is approximately 156 newtons.

02:49It turns out that the tension in the upper string is greater than the tension in the lower string.

02:57Happy learning, everyone!

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