Problem 43 - University Physics with Modern Physics

Question

A machine part consists of a thin 40.0 -cm-long bar with small 1.15 -kg masses fastened by screws to its ends. The screws can support a maximum force of 75.0 \(\mathrm{N}\) without pulling out. This bar rotates about an axis perpendicular to it at its center. (a) As the bar is turning at a constant rate on a horizontal, frictionless surface, what is the maximum speed the masses can have without pulling out the screws? (b) Suppose the machine is redesigned so that the bar turns at a constant rate in a vertical circle. Will one of the screws be more likely to pull out when the mass is at the top of the circle or at the bottom? Use a free-body diagram to see why. (c) Using the result of part (b), what is the greatest speed the masses can have without pulling a screw?

Step-by-Step Solution

Answer

(a) 3.61 m/s; (b) More likely to pull out at the bottom; (c) 3.42 m/s.

Step 1: Understand the Problem and Given Data

We have a thin bar of length 40.0 cm with two masses of 1.15 kg each attached at its ends. The maximum force the screws can withstand before pulling out is 75.0 N. The bar rotates about its center, and we need to find the maximum speed at which these masses can rotate without exceeding the force limit.

Step 2: Convert Units and Determine Radius

The bar length is given as 40.0 cm, which can be converted to meters: 0.40 m. Since the bar rotates about its center, each mass is 0.20 meters from the axis of rotation. This distance (0.20 m) is the radius of rotation.

Step 3: Use Centripetal Force Formula

The centripetal force \( F_c \) required for circular motion is given by \( F_c = \frac{mv^2}{r} \), where \( m \) is the mass, \( v \) is the speed, and \( r \) is the radius. We know the maximum \( F_c = 75.0 \) N, \( m = 1.15 \) kg, and \( r = 0.20 \) m. Plug these into the formula and solve for \( v \).

Step 4: Compute for Maximum Speed

Set \( 75.0 = \frac{1.15v^2}{0.20} \). Solve for \( v^2 \):\[75.0 \times 0.20 = 1.15v^2\]\[v^2 = \frac{75.0 \times 0.20}{1.15}\]\[v^2 = 13.043\]\[ v = \sqrt{13.043} \approx 3.61 \text{ m/s}\]The maximum speed of the masses is approximately 3.61 m/s.

Step 5: Consider Force Positions in Vertical Circular Motion

When the setup rotates vertically, the centripetal force required changes with the position of the masses. At the top of the circle, gravity assists in providing the centripetal force, whereas at the bottom of the circle, gravity acts against it. Therefore, there will be more tension on the screw at the bottom since the required centripetal force is greater.

Step 6: Find the Greatest Speed with the Screws at the Bottom

At the bottom of the circle, the total force (centripetal force plus gravitational force) cannot exceed the maximum screw force of 75 N. Circle formula considering the component of gravity:\[ m\frac{v^2}{r} + mg = 75 \]Plug the numbers (\( m = 1.15 \), \( g \approx 9.81 \), \( r = 0.20 \)) into this equation:\[ \frac{1.15v^2}{0.20} + 1.15 \times 9.81 = 75.0 \]Solve for \( v^2 \):\[ 1.15v^2 = (75.0 - 1.15 \times 9.81) \times 0.20 \]\[ v^2 \approx 11.72 \]\[ v = \sqrt{11.72} \approx 3.42 \text{ m/s} \]The greatest speed with the masses in a vertical circle is approximately 3.42 m/s without pulling a screw.

Key Concepts

Centripetal Force

When anything moves along a circular path, a special force is always at play, known as centripetal force. This force is what keeps bodies moving in that curved path rather than shooting off in a straight line. Centripetal force is always directed towards the center of the circle, which is crucial for maintaining circular motion.
Let's break it down with the formula: for any object moving in a circle of radius \( r \) with mass \( m \) and speed \( v \), the centripetal force \( F_c \) is expressed as:

• \( F_c = \frac{mv^2}{r} \)

This formula shows that greater speeds or larger masses need more force to keep moving in a circle. There's also a clear relationship between the force and the radius of the rotation. If the radius is smaller, the force needed increases for the same speed and mass. In our problem, we used this concept to figure out the maximum speed our masses could have without overwhelming the screws with more force than they can handle.

Rotational Dynamics

Rotational dynamics is all about understanding how and why objects rotate, considering the forces that affect these rotations. Just like how linear dynamics deals with straight-line motion, rotational dynamics is concerned with spinning bodies.
For our exercise, the bar rotates about an axis through its center. The masses on each end of the bar create a rotational effect due to centripetal forces acting on them, ensuring they move in a circular path.
In a more general sense, rotational dynamics encompasses concepts such as torque, angular velocity, and angular momentum. In this scenario, we mainly consider forces leading to rotational motion, such as our centripetal force, and the constraints like the maximum force the screws can withstand. Understanding these concepts is vital because they dictate how and if parts will stay intact under rotational stress.

Equilibrium of Forces

Equilibrium of forces is a principle stating that when the net force acting on an object is zero, the object remains at rest or in uniform motion. In the context of rotating systems, it means the forces acting on the body don't unbalance its circular motion.
For our rotating bar, forces must be balanced to prevent the screws from pulling out. In horizontal circular motion, the centripetal force is the main focus. But when rotated vertically, gravity also influences rotation. At the top of the circle, gravity helps centripetal force, while at the bottom, it requires additional force to maintain equilibrium.

• At the bottom, total force = centripetal force + gravitational force.

Thus, equilibrium of forces is a critical consideration in ensuring the bar and masses move smoothly without any part failing due to excessive force. In our case, understanding the interplay of these forces helped determine safe operating conditions for vertical rotations.