Problem 35 - Physics for Scientists and Engineers with Modern Physics

Question

(I) A child sitting \(1.20 \mathrm{~m}\) from the center of a merry-goaround moves with a speed of \(1.30 \mathrm{~m} / \mathrm{s} .\) Calculate \((a)\) the centripetal acceleration of the child and \((b)\) the net horizontal force exerted on the child \((\) mass \(=22.5 \mathrm{~kg})\)

Step-by-Step Solution

Answer

(a) Centripetal acceleration is 1.41 m/s². (b) Net horizontal force is 31.73 N.

Step 1: Understand the Formula for Centripetal Acceleration

Centripetal acceleration is given by the formula \( a_c = \frac{v^2}{r} \), where \( v \) is the speed and \( r \) is the radius of the circular path.

Step 2: Substitute Given Values for Part (a)

Using the provided values, speed \( v = 1.30 \, \text{m/s} \) and radius \( r = 1.20 \, \text{m} \), substitute these into the formula for centripetal acceleration: \( a_c = \frac{(1.30)^2}{1.20} \).

Step 3: Calculate Centripetal Acceleration

Carry out the calculation: \( a_c = \frac{1.69}{1.20} \approx 1.41 \, \text{m/s}^2 \). So, the centripetal acceleration of the child is \( 1.41 \, \text{m/s}^2 \).

Step 4: Use the Formula for Net Horizontal Force

The net horizontal force (centripetal force) can be calculated using \( F_c = m \times a_c \), where \( m \) is the mass of the child, and \( a_c \) is the centripetal acceleration.

Step 5: Substitute Given Values for Part (b)

With mass \( m = 22.5 \, \text{kg} \) and \( a_c = 1.41 \, \text{m/s}^2 \), substitute into the formula: \( F_c = 22.5 \times 1.41 \).

Step 6: Calculate Net Horizontal Force

Compute the force: \( F_c = 31.725 \, \text{N} \). Therefore, the net horizontal force exerted on the child is approximately \( 31.73 \, \text{N} \).

Key Concepts

Centripetal Acceleration

Centripetal acceleration is a crucial concept in understanding how objects move in a circular path. Put simply, it's the acceleration that points towards the center of the circle along which the object is traveling. This is what keeps the object moving in a circle instead of flying off in a straight line.
To calculate centripetal acceleration, you can use the formula:

• \(a_c = \frac{v^2}{r}\)

Here, \(v\) is the speed of the object in meters per second, and \(r\) is the radius of the circular path in meters.
This means that if you know how fast something is moving and how big the circle is, you can find out the acceleration keeping it in that circle.
So, when a child moves on a merry-go-round, this acceleration is what holds them in the circular path, stopping them from moving straight outwards.

Horizontal Force

In the context of circular motion, horizontal force—or centripetal force—is the force directed towards the center of the circle. It's what keeps an object moving in its circular path.
In simpler terms, if you imagine swinging a ball around on a string, the tension in the string is the centripetal force pulling the ball to the center.
We can calculate this force using the formula:

• \(F_c = m \times a_c\)

where \(m\) is the mass of the object, and \(a_c\) is the centripetal acceleration.
For a child on a merry-go-round, this force comes into play with the weight of the child and the speed of the spin, holding them securely on the ride. This is why a stronger force is needed to keep heavier objects, like a bigger child, in motion.

Circular Motion

Circular motion is when an object moves along the circumference of a circle. This type of motion is common in everyday life, like a car turning around a bend or a child on a merry-go-round.
Key aspects include:

• Constant Speed, Changing Direction: Even if an object moves at a steady speed, its direction changes continually.
• Inward Force: An inward force, known as centripetal force, is required to keep it moving in a circle.

Because the path is round, the direction of the velocity changes constantly, requiring this inward push.
Moving in a circle is more than just moving fast; it's about balancing speed and radius to maintain the path. For instance, a slight increase in speed or a reduction in radius would require a greater force to maintain the circular motion. This is why understanding these components is vital for different applications, like amusement park rides to ensure safety while having fun.