Question

(a) What is the centripetal acceleration of an object on the Earth's equator due to the rotation of the Earth? (b) A \(25.0-\mathrm{kg}\) object hangs from a spring scale at the equator. If the free-fall acceleration due only to the Earth's gravity is \(9.80 \mathrm{~m} / \mathrm{s}^{2}\), what is the reading on the spring scale?

Short answer

For part (a) we find that the centripetal acceleration of an object on the Earth's equator due to the Earth's rotation is approximately \(0.034 m/s^2\). For part (b) the spring scale reading, after considering the net impact of both gravitational and centripetal forces, is about \(245 N\).

Step-by-step solution

Calculate the Earth's Rotational Speed

The velocity of the rotation on the equator, \(v\), can be calculated by using the formula \(v = 2\pi r/T\), where \(r\) is the Earth's radius and \(T\) is the period of rotation (24 hours or 86400 seconds). The Earth's radius can be approximated as 6.37 x \(10^6\) m.

Find the Centripetal Acceleration

Once we have the velocity, we can substitute its value into the formula for centripetal acceleration: \(a_c = \frac{v^2}{r}\). This will give us the centripetal acceleration at the Earth's equator due to its rotation.

Calculate the Effect on the 25.0 kg Object

The effect of this acceleration on the reading of the spring scale can be found by finding the force exerted by this mass. It should be noted that the net effect of centripetal acceleration and gravitational acceleration acting on the object at the equator should be considered. As the object moves in a circular path due to Earth's rotation, the net force will be the gravitational force minus the centrifugal force (mass * centripetal acceleration). So, by using the formula \(F= m(a_g - a_c)\) we can figure out the new weight of the object where \(a_g\) is the acceleration due to gravity and \(a_c\) is the centripetal acceleration.

Key concepts

Earth's rotation and its effects

The Earth rotates on its axis every 24 hours, which means any point on the equator travels in a circular path. This rotation causes objects at the equator to experience centripetal acceleration, which slightly reduces the effect of gravity. The velocity of a point on the equator is given by the formula \(v = \frac{2\pi r}{T}\), where \(r\) is Earth's radius (about 6.37 x \(10^6\) m) and \(T\) is the rotation period (86400 seconds). This rotation creates a tiny but significant effect on objects' apparent weight.

Understanding Earth's rotation helps us see that we're all moving in circular paths, even if we're standing still on the ground.

Gravitational acceleration on Earth

Gravitational acceleration is the force that pulls objects toward the center of the Earth. It's typically measured as \(9.80 \mathrm{~m/s^2}\). However, at the equator, this gravitational pull is slightly offset by the Earth's rotation, which introduces centripetal acceleration.

When calculating the apparent weight of an object, both gravitational and centripetal accelerations need to be considered. For an object at the equator, the net effect is a reduced gravitational pull because centripetal force slightly opposes gravitational force. The formula \(F = m(a_g - a_c)\) helps calculate the apparent weight, where \(a_g\) is gravitational acceleration and \(a_c\) is centripetal acceleration.

Circular motion and centripetal acceleration

Circular motion occurs when an object moves along a circular path. For this motion to happen, a centripetal force is needed, which points towards the center of the circle. This force results in centripetal acceleration, calculated as \(a_c = \frac{v^2}{r}\).

In the context of Earth's rotation, every point on the equator follows a circular path due to this rotational motion. The force maintaining this path is the centripetal force, which slightly counteracts gravity. This is why the weight of an object is measured a bit less at the equator compared to the poles.

Solving physics problems involving rotation

When tackling physics problems related to Earth's rotation, it's essential to consider both gravitational and centripetal accelerations. Here’s a simple approach:

• Calculate the rotation speed using \(v = \frac{2\pi r}{T}\).
• Determine centripetal acceleration \(a_c = \frac{v^2}{r}\).
• Use \(F = m(a_g - a_c)\) to find the net force or apparent weight.

By following these steps, you can solve complex problems by breaking them down into simpler calculations. Understanding each element of the motion helps in intuitively grasping the problem, making it easier to find solutions.