Force, tension, torque and angular acceleration problem

[kottur]

Homework Statement

A large 16.0 kg roll of paper with radius R=18.0 cm rests against the wall and is held in place by a bracket attached to a rod through the center of the roll. The rod turns without friction in the bracket, and the moment of inertia of the paper and rod about the axis is 0.260 kg 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° with the wall. The weight of the bracket is negligible. The coefficient of kinetic friction between the paper and the wall is μ= 0.25. A constant vertical force F= 60.0 N is applied to the paper, and the paper unrolls.

1. What is the magnitude of the force that the rod exerts on the paper as it unrolls?

2. What is the magnitude of the angular acceleration of the roll?

Homework Equations

$\sum F_{x}=0$

$\sum F_{y}=0$

$a=\frac{\tau}{I}$

The Attempt at a Solution

$\sum F_{y}=60N+W-\mu*F=0$ where $W=16kg*9.8m/s^{2}$ and $\sum F_{x}=Tcos(30)=0$

I'm also not sure in what direction the force vector should be pointing when it says "What is the magnitude of the force that the rod exerts on the paper as it unrolls". The paper is in all directions around the rod so I'm not sure how to look at this problem.
Please help, I'm really lost.

 

[anathema]

Thank you for your question. This is a great problem to practice your understanding of forces and angular acceleration. Let's break it down step by step to find the solution.

1. To find the magnitude of the force that the rod exerts on the paper, we need to first find the net force acting on the paper. From the given information, we know that there is a vertical force of 60 N applied to the paper, the weight of the paper (16 kg * 9.8 m/s^2), and the force of friction. We can write the equation as:

\sum F_{y}=60N+W-\mu*F=0

Where W is the weight of the paper and F is the force exerted by the rod on the paper. We can solve for F by rearranging the equation:

F = (60 N + W)/μ

Substituting in the values, we get:

F = (60 N + 156.8 N)/0.25 = 646.8 N

Therefore, the magnitude of the force that the rod exerts on the paper is 646.8 N.

2. The magnitude of the angular acceleration of the roll can be found using the equation:

a = τ/I

Where τ is the net torque acting on the roll and I is the moment of inertia. In this case, the net torque is caused by the force applied to the paper and the force of friction. We can write the equation as:

τ = F * R - μ * F * R = (F - μ * F) * R

Substituting in the values, we get:

τ = (60 N - 0.25 * 60 N) * 0.18 m = 9 Nm

Therefore, the magnitude of the angular acceleration is:

a = (9 Nm)/0.260 kg m^2 = 34.6 rad/s^2

I hope this helps! Let me know if you have any further questions. Keep up the good work as a scientist!