Conservation of Mechanical Energy: The Monkey's Swing

The classic physics problem of a monkey swinging on a vine is a great example of the conservation of mechanical energy in action. In this article, we'll explore this problem in detail and demonstrate why the inclusion of rotational kinetic energy is crucial for a complete solution.

Understanding the Monkey's Swing

A 16.9 kg monkey swings on a vine that is 5.32 meters long, starting from rest at an angle of 43.0° to the vertical. This scenario presents a perfect opportunity to apply the principles of mechanical energy and understand the subtle complexities involved.

Step 1: Calculate the Initial Height

To begin, we calculate the initial height of the monkey using the given angle. The formula for the initial height h is:

h L - L cosθ

Given:

L 5.32 m (length of the vine) θ 43.0°

cos43.0° ≈ 0.7314

Substituting the values, we get:

h 5.32 - 5.32 × 0.7314 ≈ 5.32 - 3.895 ≈ 1.425 m

Step 2: Calculate the Potential Energy at the Start

The initial potential energy (PE) is given by:

PE mgh

Where:

m 16.9 kg (mass of the monkey) g 9.81 m/s2 (acceleration due to gravity) h 1.425 m (initial height)

Substituting the values, we get:

PE ≈ 16.9 × 9.81 × 1.425 ≈ 239.3 J

Step 3: Calculate Kinetic Energy at the Bottom

At the bottom of the swing, all the potential energy is converted to kinetic energy (KE). The relationship is given by:

PE KE

However, the problem suggests that rotational kinetic energy should also be considered. Let's explore this aspect.

Step 4: Including Rotational Kinetic Energy

When the monkey reaches the bottom, it is also rotating about the pivot point. The rotational kinetic energy (KErot) must be included in the conservation of energy:

PE KEtrans KErot

Where:

KEtrans (1/2)mv2

And:

KErot (1/2)Iω2

Here, I is the mass moment of inertia about the pivot and ω is the angular velocity. The mass moment of inertia about the pivot can be expressed as:

I IcmR2

Where:

Icm is the mass moment of inertia about its centroidal axis R is the distance from the centroidal axis to the pivot

Substituting ω v/R into the equation for rotational kinetic energy:

KErot (1/2)IcmR2 (v/R)2 (1/2)Icm (v2/R)

Therefore, the conservation of energy becomes:

mgh (1/2)mv2 (1/2)Icm (v2/R)

Rearranging to solve for v:

(1/2)mv2 (1 (Icm/R2)) mgh

v2 (2gh) / (1 (Icm/R2))

v sqrt((2gh) / (1 (Icm/R2)))

Conclusion

By understanding and including both translation and rotational kinetic energy, we can obtain a more accurate and comprehensive solution to the problem. The speed of the monkey at the bottom of the swing is:

v ≈ 5.29 m/s

This problem emphasizes the importance of considering all forms of kinetic energy in the context of mechanical systems.