Problem 40 - College Physics

Question

To start her lawn mower, Julie pulls on a cord that is wrapped around a pulley. The pulley has a moment of inertia about its central axis of \(I=0.550 \mathrm{~kg} \cdot \mathrm{m}^{2}\) and a radius of \(5.00 \mathrm{~cm}\). There is an equivalent frictional torque impeding her pull of \(\tau_{\mathrm{f}}=0.430 \mathrm{~m} \cdot \mathrm{N}\). To accelerate the pulley at \(\alpha=4.55 \mathrm{rad} / \mathrm{s}^{2},\) (a) how much torque does Julie need to apply to the pulley? (b) How much tension must the rope exert?

Step-by-Step Solution

Answer

(a) Julie needs to apply a torque of 2.9325 Nm. (b) The tension must be 58.65 N.

Step 1: Calculate Total Torque Required

To find the total torque required to accelerate the pulley, we use the equation of motion for rotation: \( \tau_{total} = I \cdot \alpha \), where \( I = 0.550 \mathrm{~kg} \cdot \mathrm{m}^{2} \) and \( \alpha = 4.55 \mathrm{rad/s^2} \). Plug in these values: \( \tau_{total} = 0.550 \cdot 4.55 = 2.5025 \mathrm{~Nm} \).

Step 2: Calculate Torque Julie Needs to Apply

Julie needs to apply a torque that overcomes the frictional torque and provides the necessary acceleration. The formula is \( \tau_{Julie} = \tau_{total} + \tau_{f} \). We know \( \tau_{f} = 0.430 \mathrm{~Nm} \). Thus, \( \tau_{Julie} = 2.5025 + 0.430 = 2.9325 \mathrm{~Nm} \).

Step 3: Calculate Tension in the Rope

The tension in the rope can be found by using the relationship between torque and force. Torque due to tension is \( \tau_{T} = F \cdot r \), which gives \( F = \frac{\tau_{T}}{r} \), with \( r = 0.05 \mathrm{~m} \) (radius converted from cm to m). Since this tension provides the needed torque applied by Julie, \( \tau_{T} = \tau_{Julie} = 2.9325 \mathrm{~Nm} \). Therefore, \( F = \frac{2.9325}{0.05} = 58.65 \mathrm{~N} \).

Key Concepts

Torque Calculation

When Julie pulls the cord to start her lawn mower, she needs to exert a force that creates a rotational motion in the pulley. This rotational force is known as torque. Torque is the rotational equivalent of linear force and is calculated using the formula \( \tau = F \cdot r \), where \( F \) is the force applied and \( r \) is the radius of the pulley. In rotational motion, the unit of torque is Newton-meter (Nm). This measure tells us how much twisting force is being applied by the rope at a certain distance from the axis of rotation.

In this scenario, the total torque required also accounts for both the torque needed to accelerate the pulley and to overcome any opposing forces like friction. By using the equation \( \tau_{total} = I \cdot \alpha \), we know how much torque is needed to achieve the desired angular acceleration. Julie needs to apply a bit more torque to handle any resistance that may impede her pull, which is known as frictional torque. Combining these gives us the total torque that Julie needs: \( \tau_{Julie} = \tau_{total} + \tau_{f} \).

Moment of Inertia

The moment of inertia is a key concept in understanding rotational motion. It can be thought of as a measure of how much resistance an object has against changes in its rotational speed. The moment of inertia, denoted by \(I\), depends on both the mass of the object and how that mass is distributed about the axis of rotation.

For example, a heavier and larger object will generally have a greater moment of inertia compared to a lighter one. In Julie's situation, the pulley has a given moment of inertia of \(0.550 \mathrm{~kg} \cdot \mathrm{m}^{2}\). This value is crucial because it directly affects how much torque she needs to apply to get the desired acceleration \(\alpha\). Understanding this helps in predicting how an object will respond to the applied forces and can be calculated as \(I \cdot \alpha\) for the total torque necessity.

Frictional Torque

Friction is a force that resists the motion of one surface over another, and when it comes to rotational motion, frictional torque plays a similar role in opposing the spinning of an object. It acts like a hurdle that needs to be overcome by exerting an additional rotational force, otherwise known as torque.

In the case of Julie's lawn mower, there is an equivalent frictional torque, denoted as \(\tau_{f}\), of \(0.430 \mathrm{~Nm}\). This value is an oppositional force that reduces the effectiveness of her pull. For Julie to successfully accelerate the pulley, she must apply more than just the minimum torque needed to overcome inertia; she must also overcome this frictional torque. By adding \(\tau_{f}\) to \(\tau_{total}\), the sum tells us the entire amount of torque Julie must apply to achieve the required effect. Understanding frictional torque is essential when calculating how much effort is needed to maintain or change the motion in systems involving spinning objects.