Problem 54 - College Physics

Question

You wish to accelerate a small merry-go-round from rest to a rotational speed of one-third of a revolution per second by pushing tangentially on it. Assume the merrygo-round is a disk with a mass of \(250 \mathrm{~kg}\) and a radius of \(1.50 \mathrm{~m} .\) Ignoring friction, how hard do you have to push tangentially to accomplish this in \(5.00 \mathrm{~s}\) ? (Use energy methods and assume a constant push on your part.)

Step-by-Step Solution

Answer

The required force is approximately 65.64 N.

Step 1: Determine final rotational speed

Convert the given rotational speed from revolutions per second to radians per second. We know that one revolution is equal to \(2\pi\) radians. So, a speed of \(\frac{1}{3}\) revolution per second can be converted as follows:\[\omega_f = \frac{1}{3} \times 2\pi \approx \frac{2\pi}{3} \, \text{rad/s}\]

Step 2: Calculate the moment of inertia for the disk

The moment of inertia \(I\) for a solid disk rotating about its center is given by the formula:\[I = \frac{1}{2} m r^2\]Substitute the given values, where the mass \(m = 250 \, \text{kg}\) and the radius \(r = 1.50 \, \text{m}\):\[I = \frac{1}{2} \times 250 \times (1.50)^2 = 281.25 \, \text{kg m}^2\]

Step 3: Use energy methods to calculate work required

The work done on the disk is equal to its final rotational kinetic energy, since it starts from rest:\[K = \frac{1}{2} I \omega_f^2\]Substitute the earlier results:\[K = \frac{1}{2} \times 281.25 \times \left(\frac{2\pi}{3}\right)^2 \]Calculate:\[K \approx 1032.04 \, \text{J (joules)}\]

Step 4: Relate work and force to solve for force

The work done is also equal to the force applied times the distance moved along the edge, which is the arc length for one full turn while the force is applied. Since the merry-go-round turns at a constant acceleration from rest to this speed, we can use:\[W = F \cdot s\]where \(s\) is the arc length, and in this scenario the total arc length is the circumference multiplied by the average rotations:\[s = \omega_{avg} \times \text{Time} \times 2\pi r\]The average angular speed \(\omega_{avg} = \frac{\omega_f}{2} = \frac{\pi}{3}\) and time is \(5 \, \text{s}\):\[s = \left(\frac{2\pi}{3}\right) \times 5 \times 1.50 = 5\pi \, \text{m}\]Rearrange for force:\[F = \frac{W}{s} = \frac{1032.04}{5\pi}\]Calculate:\[F \approx 65.64 \, \text{N}\]

Step 5: Final Answer and Verification

Thus, the force you need to apply tangentially to accelerate the merry-go-round to the desired speed in 5 seconds is approximately \(65.64 \, \text{N}\). You can verify the correctness by ensuring that the computed force, when used in conjunction with the radius and moment of inertia, yields the desired kinetic energy at \(5 \, \text{s}\).

Key Concepts

Moment of Inertia

The moment of inertia is a crucial concept in understanding rotational motion. It quantifies how an object's mass is distributed relative to its axis of rotation. Just as mass measures resistance to linear acceleration, moment of inertia measures resistance to rotational acceleration.

In our example with the merry-go-round, considered as a disk, the moment of inertia is given by the formula:

• \( I = \frac{1}{2} m r^2 \)

Where *\( m \)* is the mass and *\( r \)* is the radius. For the exercise, substituting the values \( m = 250 \, \text{kg} \) and \( r = 1.50 \, \text{m} \), gives:

• \( I = \frac{1}{2} \times 250 \times (1.50)^2 = 281.25 \, \text{kg m}^2 \)

This calculation shows that the mass and radius exponentially affect the moment of inertia, thereby emphasizing that larger, heavier objects are harder to set spinning.

For students, this implies that understanding the moment of inertia helps predict how much force you will need to rotate a given object. It is a measure of an object's resistance to change in its rotational state.

Rotational Kinetic Energy

Rotational kinetic energy is the energy due to the rotation of an object and is part of its total kinetic energy. For a rotating disk, the energy due to its rotation is calculated similarly to linear kinetic energy, using the formula:

• \( K = \frac{1}{2} I \omega^2 \)

Where *\( I \)* is the moment of inertia, and *\( \omega \)* is the angular velocity. In the merry-go-round example, the final angular velocity \( \omega_f = \frac{2\pi}{3} \, \text{rad/s} \), when inserted into the formula, appears as:

• \( K = \frac{1}{2} \times 281.25 \times \left(\frac{2\pi}{3}\right)^2 \)
• \( K \approx 1032.04 \, \text{J} \)

This calculation shows how the rotational kinetic energy forms as a result of the motion, indicating the energy required to reach a certain speed.

Understanding rotational kinetic energy allows students to connect rotational motion to concepts they already know about linear motion, showing that energy principles apply universally across different types of movement.

Tangential Force

A tangential force is one that acts perpendicular to the radius of a circular path, creating rotational motion. In our merry-go-round scenario, applying this kind of force causes the disk to spin faster over time. This is deeply connected to the concepts of torque and angular acceleration.

The work done by this tangential force is expressed as:

• \( W = F \cdot s \)

Where *\( F \)* is the force and *\( s \)* is the distance moved along the circumference, given by \( s = \omega_{avg} \times \text{Time} \times 2\pi r \). The average angular speed *\( \omega_{avg} = \frac{\omega_f}{2} \)* helps compute the total arc length of movement:

• \( s = \left(\frac{2\pi}{3}\right) \times 5 \times 1.50 = 5\pi \, \text{m} \)

The tangential force required is then found by rearranging the work equation:

• \( F = \frac{W}{s} = \frac{1032.04}{5\pi} \)
• \( F \approx 65.64 \, \text{N} \)

This highlights that the tangential force depends on how quickly and how far you want to rotate the object. A firm grasp of tangential forces helps students understand how to apply real-world actions like pushing, to achieve desired mechanical effects.