Question

A playground merry-go-round has radius 2.40 m and moment of inertia 2100 kg \(\cdot\) m\(^2\) about a vertical axle through its center, and it turns with negligible friction. (a) A child applies an 18.0-N force tangentially to the edge of the merry-go-round for 15.0 s. If the merry-go-round is initially at rest, what is its angular speed after this 15.0-s interval? (b) How much work did the child do on the merry-go-round? (c) What is the average power supplied by the child?

Short answer

(a) 0.309 rad/s, (b) 100.1 J, (c) 6.67 W.

Step-by-step solution

Calculate the Torque

Torque (\(\tau\)) is calculated using the formula \(\tau = r \cdot F\), where \(r = 2.40 \text{ m}\) is the radius and \(F = 18.0 \text{ N}\) is the force applied tangentially. Substituting the values:\[\tau = 2.40 \cdot 18.0 = 43.2 \text{ Nm}\]

Use Torque to Find Angular Acceleration

Angular acceleration (\(\alpha\)) is found using the formula \(\alpha = \frac{\tau}{I}\), where \(I = 2100 \text{ kg} \cdot \text{m}^2\) is the moment of inertia. Substituting the values:\[\alpha = \frac{43.2}{2100} = 0.02057 \text{ rad/s}^2\]

Calculate Final Angular Speed

The final angular speed (\(\omega\)) can be found by using the equation \(\omega = \omega_0 + \alpha t\), where the initial angular speed \(\omega_0 = 0\) (since it starts from rest), \(\alpha = 0.02057 \text{ rad/s}^2\), and \(t = 15.0 \text{ s}\). Substituting the values:\[\omega = 0 + 0.02057 \times 15.0 = 0.30855 \text{ rad/s}\]

Calculate Work Done

The work done (\(W\)) can be calculated using the formula \(W = \tau \cdot \theta\), where \(\theta\) (rotational displacement) is found using \(\theta = \omega_0 t + \frac{1}{2} \alpha t^2\). Since \(\omega_0 = 0\), substituting the values:\[\theta = 0 + \frac{1}{2} \times 0.02057 \times (15)^2 = 2.31555 \text{ rad}\]Then, work done:\[W = 43.2 \cdot 2.31555 = 100.071 \text{ J}\]

Calculate Average Power Supplied

Average power (\(P\)) is calculated using \(P = \frac{W}{t}\). Using the work done \(W = 100.071 \text{ J}\) and \(t = 15.0 \text{ s}\), the power is:\[P = \frac{100.071}{15.0} = 6.6714 \text{ W}\]

Key concepts

Moment of Inertia

The concept of moment of inertia is an essential element when dealing with angular mechanics problems. Think of it as the rotational equivalent of mass in linear motion. It represents how difficult it is to change the rotational speed of an object. The moment of inertia depends on both the mass of the object and how the mass is distributed in relation to the axis of rotation.
For an object like a merry-go-round, the moment of inertia \(I\) is expressed in units of kg \(\cdot\) m\(^2\). In our example, the merry-go-round has a moment of inertia of 2100 kg \(\cdot\) m\(^2\).\

• Factors affecting moment of inertia include the shape and axis of rotation.
• More distributed mass from the axis means a larger moment of inertia.

Understanding this helps to recognize why an object with a larger moment of inertia is harder to start or stop spinning.

Torque Calculation

Torque is a measure of the rotational force applied to an object. It can be thought of as a twist that causes an object to spin around an axis. Torque is given by the formula \(\tau = r \cdot F\), where \(r\) is the radius, and \(F\) is the force applied.
In the exercise, a force of 18.0 N is applied tangentially to the edge of the merry-go-round with a radius of 2.40 m. Substituting these values, the torque \(\tau\) can be calculated as:\[ \tau = 2.40 \cdot 18.0 = 43.2 \text{ Nm} \].

• The direction of torque is crucial because it determines the way an object rotates.
• Positive torque usually indicates counterclockwise rotation while negative torque indicates clockwise rotation.

Understanding how torque is calculated and applied helps you comprehend the forces influencing rotational motion.

Angular Acceleration

Angular acceleration represents how quickly an object's rotational speed changes over time. It is a measure of how the angular velocity changes due to applied torque. In our example, the angular acceleration \(\alpha\) is calculated by the formula \(\alpha = \frac{\tau}{I}\), where \(\tau\) is torque and \(I\) is the moment of inertia.
With torque \(\tau = 43.2\) Nm and moment of inertia \(I = 2100\) kg \(\cdot\) m\(^2\), the angular acceleration is:\[ \alpha = \frac{43.2}{2100} = 0.02057 \text{ rad/s}^2 \].

• Higher torque results in higher angular acceleration for a given moment of inertia.
• Objects with higher moments of inertia require more torque to achieve the same angular acceleration compared to objects with lower moments of inertia.

Comprehending angular acceleration provides insights into how rotational forces affect the speed at which an object spins.