Problem 44 - Physics for Scientists and Engineers with Modern Physics

Question

(II) A bucket of mass \(2.00 \mathrm{~kg}\) is whirled in a vertical circle of radius \(1.10 \mathrm{~m} .\) At the lowest point of its motion the tension in the rope supporting the bucket is \(25.0 \mathrm{~N}\). \((a)\) Find the speed of the bucket. \((b)\) How fast must the bucket move at the top of the circle so that the rope does not go slack?

Step-by-Step Solution

Answer

(a) The speed at the lowest point is approximately 1.72 m/s. (b) The minimum speed at the top to avoid slackness is approximately 3.28 m/s.

Step 1: Analyze Forces at the Lowest Point

At the lowest point, two forces act on the bucket: the gravitational force (weight) acting downward and the tension in the rope acting upward. Use Newton's second law to find the net force acting in the radial direction: \[ F_{net} = T - mg \]where \( T = 25.0 \space \mathrm{N} \) is the tension, \( m = 2.00 \space \mathrm{kg} \) is the mass, and \( g = 9.81 \space \mathrm{m/s^2} \) is the acceleration due to gravity. The net force provides the centripetal force required for circular motion:

Step 2: Write the Equation for Centripetal Force

The centripetal force \( F_{c} \) is given by:\[ F_{c} = \, \frac{mv^2}{r} \]At the lowest point, \( T - mg = \frac{mv^2}{r} \). Substituting the known values:\[ 25.0 \, - \, 2.00 \, \times \, 9.81 = \, \frac{2.00 \, \times \, v^2}{1.10} \]Solve for \( v^2 \).

Step 3: Solve for Speed at the Lowest Point

First, calculate the net force:\[ 25.0 \, N \, - \, 2.00 \, \times \, 9.81 \, N \, = \, 5.38 \, N \]Then solve for \( v^2 \):\[ 5.38 = \, \frac{2.00 \, \times \, v^2}{1.10} \]Now solve for \( v \):\[ v^2 = \, \frac{5.38 \, \times \, 1.10}{2.00} \approx 2.959 \]\[ v \approx \sqrt{2.959} \approx 1.72 \, \mathrm{m/s} \]

Step 4: Analyze Forces at the Top Point

At the top of the circle, for the rope not to go slack, the centripetal force should be equal to or greater than the gravitational force. The tension can be zero in this situation if the gravitational force provides the necessary centripetal force:\[ F_{c} = mg = \frac{mv^2}{r} \]

Step 5: Solve for Minimum Speed at the Top

Set \( mg = \frac{mv^2}{r} \) and solve for \( v \):\[ g = \frac{v^2}{r} \Rightarrow v^2 = gr \Rightarrow v = \sqrt{gr} \]Substitute the values:\[ v = \sqrt{9.81 \, \times \, 1.10} \approx \sqrt{10.791} \approx 3.28 \, \mathrm{m/s} \]

Key Concepts

Centripetal Force

Centripetal force is crucial for any object undergoing circular motion. It acts to keep the object moving along its circular path. Imagine revolving a bucket tied to a rope; the rope provides the centripetal force. This force always points towards the circle's center, maintaining the bucket's circular trajectory.

For calculating this, the formula is:

• \[ F_c = \frac{mv^2}{r} \]

In this scenario:

• \( F_c \) is centripetal force required,
• \( m \) is the mass of the object,
• \( v \) is the velocity or speed,
• \( r \) is the radius of the circular path.

When analyzing a problem, understanding centripetal force is the first step, as it informs how fast the object needs to move to sustain its path.

Newton's Second Law

Newton’s Second Law is the backbone for understanding how forces impact motion. It states that the acceleration of an object is proportional to the net force acting on it and inversely proportional to its mass. The law is represented as:

• \[ F_{net} = ma \]

Here:

• \( F_{net} \) is the total force acting on an object,
• \( m \) is the mass of the object,
• \( a \) is the acceleration.

In circular motion, this law helps in understanding how gravitational pull and tension contribute to the centripetal force. For example, at the lowest point in our bucket exercise, the net force comprises tension in the rope and the weight of the bucket. Calculating these gives a clear picture of how fast the object moves and strains the rope.

Vertical Circular Motion

Vertical circular motion combines gravity’s effects with circular motion dynamics. As an object moves in a vertical circle, the forces acting on it change based on the position. The consistent gravitational force affects how centripetal force and tension are calculated and perceived.

At the lowest point of a vertical motion, tension is maximal because it supports both the gravitational force and the centripetal force. As our bucket travels upwards to the highest point:

• The tension can drop to zero if gravity alone provides necessary centripetal force.
• The minimum speed at the top is critical to prevent the rope from slackening.

For maintaining this minimum speed, calculate it using:

• \[ v = \sqrt{gr} \]

where \( g \) is the acceleration due to gravity and \( r \) is the circle's radius. Recognizing these interactions in vertical circular motion aids in predicting movements and ensuring the system remains intact.