Video: Determining the Angular Speed of a Physical Pendulum Released from above the Horizontal

A uniform rod of mass 1.0 kg and length 2.0 m is free to rotate about one of its ends, as shown in the accompanying diagram. If the rod is released from rest at an angle of 60Β° above the horizontal, what is the speed of the tip of the rod as it passes the horizontal position?

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Video Transcript

A uniform rod of mass 1.0 kg and length 2.0 meters is free to rotate about one of its ends, as shown in the accompanying diagram. If the rod is released from rest at an angle of 60 degrees above the horizontal, what is the speed of the tip of the rod as it passes the horizontal position?

As we solve this problem, we’ll assume that the rod can rotate about its axis without any energy loss due to friction. We’ll also assume that 𝑔, the acceleration due to gravity, is exactly 9.8 meters per second squared. Let’s highlight some of the vital information given in the statement.

We’re told the rod has a mass of 1.0 kilograms and a length of 2.0 meters. The initial angle above the horizontal, that which the rod is released, is 60 degrees. We want to solve for the speed of the tip of the rod as it passes the horizontal position. We’ll call that speed 𝑣 sub 𝑑.

We can approach this problem under the assumption that the law of conservation of energy holds. Let’s see how that approach works itself out in this exercise. We can state the law of conservation of energy in equation form. Given a system of objects, the initial energy of the objects in that system, 𝐸 sub 𝑖, is equal to the final energy of the objects in that same system, 𝐸 sub 𝑓.

We can apply this general principle to our particular scenario. Let’s consider what is the initial energy 𝐸 sub 𝑖 of the rod. At the start, we’re told that the rod is released from rest, meaning its initial speed is zero. The rod has no kinetic energy initially, but it does have potential energy due to gravity.

If we establish the zero value of our height as the dotted horizontal axis, since the center of mass of a rod is above β„Ž equals zero, that means our rod does have potential energy thanks to gravity. We’re told that the overall length of a rod, which we’ll call capital 𝐿, is 2.0 meters. And that the mass of a rod, which we’ll call π‘š, is 1.0 kilograms.

Because this is a uniform rod, the center of mass of the rod is that its exact center: one meter from either end. To find the initial energy of the rod, we want to find the height of that center of mass point just as the rod is being released. By drawing a dotted vertical line from that point down to the horizontal axis, we create a right triangle where the length of the hypotenuse of that triangle is capital 𝐿 over two.

As we look at the corners of this triangle, we recognize that for a right triangle like this one the sine of the angle, in this case 60 degrees, is equal to the opposite side, β„Ž, divided by the hypotenuse 𝐿 over two.

Since 𝐿 equals 2.0 meters, and we’re dividing that value by two, we end up with the value on the right-hand side of our equation of β„Ž divided by 1.0 meters. Since any number divided by one is equal to that original number, we find that the height, β„Ž, simply equals the sine of 60 degrees meters.

Now that we know the initial height of the center of mass of the rod, we can write out the initial energy of our system, 𝐸 sub 𝑖, in terms of gravitational potential energy. That term will be π‘š, the mass of the rod, times 𝑔, the acceleration due to gravity, times β„Ž, which is the sine of 60 degrees.

Now that we’ve covered the initial energy of our system, we want to solve for the final energy of the system. That occurs at the moment when the rod is in a horizontal position having fallen through an angular extent of 60 degrees. At that instant in time, the rod will have no gravitational potential energy, because its center of mass will have a height of zero. However, the rod will have rotational kinetic energy.

Let’s recall the equation for rotational kinetic energy. Rotational kinetic energy, which we abbreviate KE sub π‘Ÿ, is equal to one half the moment of inertia of our object multiplied by its angular speed squared. Let’s apply this equation to our particular scenario. As we look at this equation, 𝐼, the moment of inertia, applies to a rod rotating about an axis at the end of the rod.

For an object like that rotating about an axis like that, the moment of inertia is equal to one third the mass of the rod times its length squared. When we plug this equation in for 𝐼 in our equation for rotational kinetic energy, we find an expression of one half times one third π‘šπΏ squared multiplied by πœ” squared.

The πœ” or rotational speed noted in this equation only applies at one instant in time, at the instant when the rod is in the horizontal position parallel to the horizontal axis. That’s why it’s true that we can take nearly this rotational kinetic energy term and insert that for the final energy of our system. Because at that moment when the road is in that position, potential energy due to gravity of the rod is zero.

We can now replace 𝐸 sub 𝑓, the final energy of our system, with our expression for rotational kinetic energy of the rod. We end up with this energy balance equation for our system. The initial gravitational potential energy of the rod when it’s released is equal to the rotational kinetic energy of the rod at the moment that the rod is horizontal.

We want to solve this equation for πœ”, the angular speed of the rod at that moment. If we can solve for πœ”, then we can use the relationship between 𝑣, the linear speed, and πœ”, the angular speed to solve for 𝑣 sub 𝑑.

As we move towards isolating πœ”, let’s first notice that this problem is mass independent. The mass of the rod appears on both sides and, therefore, can be canceled out. If we now multiplied both sides of the equation by six divided by 𝐿 squared, then the six and 𝐿 squared terms cancel out on the right-hand side of our equation.

We can then take the square root of both sides of our equation and see that the squared term and square root term on the right side cancel out. We’re left with an expression for πœ”; πœ” equals the square root of six divided by 𝐿 squared times 𝑔 times the sine of 60 degrees. 𝐿, we know, is 2.0 meters and 𝑔 is 9.8 meters per second squared. When we plug those values in to this equation, then we are now ready to enter these numbers into our calculator to solve for πœ”. And when we do that, we find that πœ” is equal to 3.568 radians per second.

Now we’re ready to take advantage of a relationship between linear velocity and rotational velocity. That relationship is that linear velocity 𝑣 is defined as π‘Ÿ times πœ”, where π‘Ÿ is the distance from the point being considered to the axis of rotation and πœ” is the angular speed. This equation applies to our situation, because we want to solve for 𝑣 sub 𝑑, the speed of the tip of the rod as it passes the horizontal line.

We know πœ” and π‘Ÿ is the length of the rod 𝐿. So let’s write this linear to rotational conversion equation relative to our scenario. 𝑣 sub 𝑑, the speed of the tip of the bar, is equal to the length of the bar multiplied by the angular speed with which the bar rotates.

When we plug in the values we have for 𝐿 and πœ” and then multiply these two numbers together, we find that the tip of this rod as it passes the horizontal line has a linear speed of 7.1 meters per second. This is the speed at which the end of the bar is moving at that moment in time.

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