Question

A \(24-\mathrm{cm}\) -long pen is tossed up in the air, reaching a maximum height of \(1.2 \mathrm{~m}\) above its release point. On the way up, the pen makes 1.8 revolutions. Treating the pen as a thin uniform rod, calculate the ratio between the rotational kinetic energy and the translational kinetic energy at the instant the pen is released. Assume that the rotational speed does not change during the toss.

Short answer

Answer: The ratio between rotational kinetic energy and translational kinetic energy at the instant the pen is released is approximately 0.383.

Step-by-step solution

Find the velocity of the pen at the release point

To find the velocity of the pen when it is released, we can use the following kinematic equation: \(v^2 = u^2 + 2as\) Here, \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration due to gravity, and \(s\) is the maximum height reached. The final velocity at the maximum height will be 0, and the acceleration due to gravity is -9.81 \(m/s^2\). So, the equation can be written as: \(0 = u^2 - 2(9.81)(1.2)\) Now, we can solve for the initial velocity \(u\): \(u = \sqrt{2(9.81)(1.2)}\approx 4.84 \;\mathrm{m/s}\)

Calculate angular velocity

We know that the pen makes 1.8 revolutions on its way up. To find the angular velocity of the pen, we will first convert it to radians. \(1.8\;\mathrm{revolutions} = 1.8\times 2\pi\;\mathrm{radians} = 3.6\pi\;\mathrm{radians}\) Now, we will use the equation for angular velocity: \(\omega = \dfrac{\theta}{t}\) To find the time \(t\), we can use the equation of motion for vertical displacement: \(h = ut - \frac{1}{2}gt^2\) Here, \(h\) is the maximum height reached, \(u\) is the initial-vertical velocity and \(g\) is the acceleration due to gravity. We've already calculated the initial-vertical velocity, \(u\approx 4.84 \;\mathrm{m/s}\)and \(h = 1.2\;\mathrm{m}\), and \(g= 9.81\;\mathrm{m/s^2}\): \(1.2 = 4.84 t - \frac{1}{2}(9.81)t^2\) Solving the quadratic equation for \(t\), we get two possible values, one of which is the time taken to reach the maximum height \(t\approx 0.494\;\mathrm{s}\). Now, we can calculate the angular velocity: \(\omega = \dfrac{3.6\pi}{0.494}\approx 22.9\;\mathrm{rad/s}\)

Calculate the rotational and translational kinetic energies

Now that we have the initial velocity \(u\) and angular velocity \(\omega\), we can calculate the rotational and translational kinetic energies. The translational kinetic energy is given by: \(K_t = \frac{1}{2}mv^2 = \frac{1}{2}m(4.84)^2\) The rotational kinetic energy of a rotating rod is given by: \(K_r = \frac{1}{2}I\omega^2\) Here, \(I\) denotes the moment of inertia of the rod about an axis perpendicular to the rod and passing through its center of mass. For a uniform rod of length \(L\) and mass \(m\), the moment of inertia is given by: \(I = \frac{1}{12}mL^2 = \frac{1}{12}m(0.24)^2\) Now, substituting the value of \(I\) in the equation for rotational kinetic energy: \(K_r = \frac{1}{2}\left(\frac{1}{12}m(0.24)^2\right)(22.9)^2\)

Calculate the ratio between rotational and translational kinetic energies

We are now ready to calculate the ratio between the rotational and translational kinetic energies. Notice that the mass \(m\) will cancel out in the ratio: \(\dfrac{K_r}{K_t} = \dfrac{\frac{1}{2}\left(\frac{1}{12}(0.24)^2\right)(22.9)^2}{\frac{1}{2}(4.84)^2} \approx 0.383\) Hence, the ratio between the rotational kinetic energy and the translational kinetic energy at the instant the pen is released is approximately 0.383.

Key concepts

Moment of Inertia

The moment of inertia is a crucial concept when dealing with rotational motion. It is essentially the rotational equivalent of mass for linear motion. The moment of inertia quantifies how much torque is needed for a desired angular acceleration around an axis. This value depends not only on the mass but also on the distribution of that mass with respect to the axis of rotation. For a thin, uniform rod like the pen in the exercise, the moment of inertia about an axis through its center is given by the formula \( I = \frac{1}{12}mL^2 \).

This formula arises because each infinitesimal mass element along the rod is multiplied by the square of its distance from the axis before summing these contributions.

• "\( I \)" represents the moment of inertia.
• "\( m \)" is the mass of the rod.
• "\( L \)" is the length of the rod.

Understanding moment of inertia helps explain why it is easier to spin an object with mass concentrated closer to the axis of rotation, compared to one with the same mass spread further away.

Angular Velocity

Angular velocity describes how quickly an object rotates around an axis. Similar to linear velocity, which deals with the rate of change of an object’s position, angular velocity focuses on rotation. It is measured in radians per second.

In the exercise, the pen makes 1.8 revolutions. To convert this into angular displacement in radians, we use the relationship that one complete revolution equals \( 2\pi \) radians, giving us an angular displacement of \( 3.6\pi \) radians.

• Angular velocity \( \omega \) is calculated using the formula \( \omega = \frac{\theta}{t} \). Here, "\( \theta \)" is the angular displacement and "\( t \)" is the time taken for the displacement.

Calculating angular velocity allows us to understand the rotational speed of the object. In the pen’s case, the calculated angular velocity was approximately \( 22.9 \) rad/s, indicating how quickly it spins.

Kinematic Equations

Kinematic equations are used to describe the motion of objects. They relate various motion parameters like displacement, time, initial velocity, final velocity, and acceleration. These are instrumental in solving problems in both linear and rotational motion.

In linear motion, the equation \( v^2 = u^2 + 2as \) was used to find the pen's initial velocity as it was released. Here:

• "\( v \)" is the final velocity, which is zero at the maximum height.
• "\( u \)" is the initial velocity.
• "\( a \)" is the acceleration due to gravity, \( -9.81 \; m/s^2 \).
• "\( s \)" is the displacement, the maximum height reached.

The solution showed that the initial velocity was calculated to be approximately \( 4.84 \; m/s \). Understanding these relationships helps dissect and predict the motion of the pen as it ascends and rotates.