Problem 69 - University Physics with Modern Physics

Question

A large 16.0 -kg roll of paper with radius \(R=18.0 \mathrm{cm}\) rests against the wall and is held in place by a bracket attached to a rod through the center of the roll (Fig. P10.69). The rod turns without friction in the bracket, and the moment of inertia of the paper and rod about the axis is 0.260 \(\mathrm{kg} \cdot \mathrm{m}^{2} .\) The other end of the bracket is attached by a frictionless hinge to the wall such that the bracket makes an angle of \(30.0^{\circ}\) with the wall. The weight of the bracket is negligible. The coefficient of kinetic friction between the paper and the wall is \(\mu_{\mathrm{k}}=0.25 .\) A constant vertical force \(F=60.0 \mathrm{N}\) is applied to the paper, and the paper unrolls. (a) What is the magnitude of the force that the rod exerts on the paper as it unrolls? (b) What is the magnitude of the angular acceleration of the roll?

Step-by-Step Solution

Answer

(a) 99.2 N, (b) 14.4 rad/s².

Step 1: Identify the Forces Acting on the Paper Roll

First, identify and understand all the forces acting on the paper roll. There is a vertical force of 60 N ( F ) applied downward, the normal force exerted by the wall on the paper roll, and the frictional force opposing the motion. The frictional force can be given by \( f = \mu_k \times N \) where \( \mu_k \) is the coefficient of kinetic friction and \( N \) is the normal force.

Step 2: Calculate the Torque Due to the Applied Force

The torque caused by the applied force can be calculated using the formula: \( \tau = F \times R \), where \( F = 60 \text{ N} \) and \( R = 18 \text{ cm} = 0.18 \text{ m} \). Substitute the values to find the torque: \( \tau = 60 \times 0.18 = 10.8 \text{ Nm} \).

Step 3: Calculate the Frictional Force

Use the frictional force equation, \( f = \mu_k \times N \). Since there is no vertical acceleration, set the vertical force equal to the friction and normal force relation: \( f = \mu_k \times (mg) \), where \( m = 16 \text{ kg} \) is the mass of the roll. Calculate the normal force: \( f = 0.25 \times 16 \times 9.8 = 39.2 \text{ N} \).

Step 4: Calculate the Torque due to Friction

The torque due to friction is: \( \tau_f = f \times R \). Substituting the values: \( \tau_f = 39.2 \times 0.18 = 7.056 \text{ Nm} \).

Step 5: Apply Newton's Second Law for Rotation

Calculate the net torque exerted on the paper by subtracting the torque due to friction from the torque due to the applied force: \( \tau_{net} = \tau - \tau_f = 10.8 - 7.056 = 3.744 \text{ Nm} \). Using \( \tau_{net} = I \cdot \alpha \), where \( I = 0.260 \text{ kg} \cdot \text{m}^2 \) is the moment of inertia, solve for the angular acceleration (\( \alpha \)): \( \alpha = \frac{\tau_{net}}{I} = \frac{3.744}{0.260} = 14.4 \text{ rad/s}^2 \).

Step 6: Determine the Magnitude of the Force Exerted by the Rod

The force exerted by the rod on the paper is essentially the sum of the vertical force and the component of the frictional force in the vertical direction. Since the rod must counteract the rotation induced by the wall's friction and the vertical force, calculate the force as \( F_{rod} = F + f = 60 + 39.2 = 99.2 \text{ N} \).

Key Concepts

Torque Calculation

Torque is what makes an object rotate around an axis, akin to how force makes an object move. When calculating torque, consider both the magnitude of the force and the distance from the pivot point, known as the radius. The formula for torque is given by \[ \tau = F \times R \] where \( F \) is the force applied and \( R \) is the radius. In the exercise, a force of 60 N was applied at the radius of 18 cm (or 0.18 m), resulting in torque of: \[ \tau = 60 \times 0.18 = 10.8 \text{ Nm} \] This torque will try to unroll the paper along its circumference. Always remember that torque depends on the lever arm length, so even a small force can produce substantial torque if it's applied far from the pivot.

• Torque is measured in Newton-meters (Nm).
• Torque can cause clockwise or counterclockwise rotation.
• Higher torque means more rotational power.

Kinetic Friction

Kinetic friction occurs when two surfaces slide past one another. In this context, it acts to slow down the paper’s unrolling motion. It can be quantified using: \[ f = \mu_k \times N \] where \( \mu_k \) is the coefficient of kinetic friction and \( N \) is the normal force, which is the perpendicular force exerting on the object. For the paper roll against the wall, the frictional force opposing the motion was calculated as 39.2 N using the given coefficient of kinetic friction, 0.25.

• Kinetic friction always acts opposite to the direction of motion.
• The coefficient \( \mu_k \) is dimensionless and specific to the material interaction.
• This force affects how easily the paper unrolls.

Moment of Inertia

Moment of inertia can be thought of as rotational mass—how resistant an object is to changes in its rotation. It depends on both the mass of the object and how that mass is distributed relative to the axis of rotation. The formula used here is: \[ I = 0.260 \ \text{kg} \cdot \text{m}^2 \] for the paper roll. A larger moment of inertia means the object is harder to start or stop rotating. You can think of it as the rotational equivalent of mass in linear motion.

• Calculated in \( \text{kg} \cdot \text{m}^2 \).
• Important for understanding angular motion dynamics.
• Affects rotational speed and angular acceleration.

Angular Acceleration

Angular acceleration measures how quickly an object speeds up or slows down its rotation. It's similar to linear acceleration but in a rotational context. Using the net torque calculated and the moment of inertia: \[ \alpha = \frac{\tau_{net}}{I} \] For this problem, the net torque is 3.744 Nm after considering friction, and the moment of inertia is 0.260 kg·m², giving an angular acceleration of: \[ \alpha = \frac{3.744}{0.260} = 14.4 \text{ rad/s}^2 \] This tells us how quickly the paper starts to spin faster as it unrolls. Knowing angular acceleration is essential for predicting future states of rotational motion.

• Measured in radians per square second (rad/s²).
• Indicates change in rotational speed.
• Higher angular acceleration means quicker spinning.