Question Video: Finding the Magnitude of Force Acting on a Uniform Rod about a Fixed Horizontal Axis | Nagwa Question Video: Finding the Magnitude of Force Acting on a Uniform Rod about a Fixed Horizontal Axis | Nagwa

Question Video: Finding the Magnitude of Force Acting on a Uniform Rod about a Fixed Horizontal Axis

A uniform rod of length 2𝑙 and mass π‘š is free to rotate in a horizontal plane about a vertical axis passing through its center. A horizontal force of constant magnitude is applied to one of the ends of the rod in a direction perpendicular to the rod causing the rod to rotate with an angular acceleration of 24𝑔/𝑙, where 𝑔 is the acceleration due to gravity. Find the magnitude of the force applied.

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Video Transcript

A uniform rod of length two 𝑙 and mass π‘š is free to rotate in a horizontal plane about a vertical axis passing through its center. A horizontal force of constant magnitude is applied to one of the ends of the rod in a direction perpendicular to the rod, causing the rod to rotate with an angular acceleration of 24𝑔 over 𝑙, where 𝑔 is the acceleration due to gravity. Find the magnitude of the force applied.

Let’s start with a diagram of the situation. We have a uniform rod of length two 𝑙 and mass π‘š. A uniform rod refers to one-dimensional object with length but no width. But let’s draw it with some width just to make things a little clearer. So this rod has length two 𝑙 and mass π‘š. And it’s rotating about a vertical axis through its center with an angular acceleration of 24𝑔 over 𝑙. And we have a force 𝐅 applies to the end of the rod, which causes this angular acceleration. This is the quantity we wish to find.

We begin with Newton’s second law of motion in its rotational form. The sum of all the torques Ξ€ on a body is equal to the moment of inertia of that body 𝐼 times the angular acceleration πœ” prime. We need to find an expression for 𝐅 from this equation. A torque Ξ€ is given by the product of the perpendicular component of the force 𝐅 times its distance from the axis of rotation 𝑑.

In our case, the force applied 𝐅 is already perpendicular to the axis of rotation. Since the axis of rotation is about the center of the rod and the rod is of length two 𝑙, the distance 𝑑 is just equal to 𝑙. So the force 𝐅 is given by Ξ€ over 𝑙.

Substituting in the expression for Ξ€ from Newton’s second law gives us πΌπœ” prime over 𝑙. We already know that πœ” prime, the angular acceleration, is 24𝑔 over 𝑙. So our expression for 𝐅 becomes 24𝑔𝐼 over 𝑙 squared. We now have only one unknown in our expression for 𝐅, the moment of inertia of the rod 𝐼.

For a one-dimensional object such as this rod, the moment of inertia 𝐼 is given by the integral between the two endpoints π‘₯ one and π‘₯ two of the square of the distance from the axis of rotation π‘₯ with respect to the mass element dπ‘š.

Going back to our diagram of the rod, consider a small piece of the rod, shown here in orange. It has mass dπ‘š and length dπ‘₯. We want to integrate with respect to the length dπ‘₯. So we need to find an expression for dπ‘š in terms of dπ‘₯.

Since density is mass divided by volume, or in this case linear density is mass divided by length, then the linear density 𝜌 is equal to dπ‘š over dπ‘₯. Therefore, dπ‘š equals 𝜌 times dπ‘₯. But this rod is uniform, so its density is the same everywhere. Therefore, the density at any given point is the same as the overall density of the rod. Therefore, the density 𝜌 is equal to π‘š over two 𝑙. Therefore, dπ‘š is equal to π‘šdπ‘₯ over two 𝑙.

Therefore, the moment of inertia 𝐼 is given by the integral between π‘₯ one and π‘₯ two of π‘₯ squared times π‘š over two 𝑙 dπ‘₯. π‘š over two 𝑙 is just a constant, so we can move it outside the integration. The limits π‘₯ one and π‘₯ two are just the start and endpoints of the rod with respect to the axis of rotation. The axis of rotation has an π‘₯-value of zero. So the start point has an π‘₯-value of negative 𝑙 and the endpoint has a value of 𝑙. So we have π‘š over two 𝑙 times the integral between negative 𝑙 and 𝑙 of π‘₯ squared dπ‘₯.

Integrating with respect to π‘₯ gives us one-third π‘₯ cubed. Evaluating and simplifying gives us π‘šπ‘™ squared over three. Now, substituting this back into our equation for 𝐅, this gives us 24𝑔 over 𝑙 squared times π‘šπ‘™ squared over three. And simplifying gives us our answer: 𝐅 equals eight π‘šπ‘”.

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