Find the linear speed v for each of the following.

a point on the equator moving due to Earth's rotation, if the radius is 3960 mi

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Understand that the linear speed \(v\) of a point on a rotating object is related to the angular speed \(\omega\) and the radius \(r\) by the formula: \(v = \omega \times r\).

Identify the radius \(r\) given in the problem, which is the radius of the Earth at the equator: \(r = 3960\) miles.

Determine the angular speed \(\omega\) of the Earth's rotation. Since the Earth completes one full rotation (360 degrees or \(2\pi\) radians) in 24 hours, express \(\omega\) in radians per hour: \(\omega = \frac{2\pi}{24}\) radians per hour.

Substitute the values of \(\omega\) and \(r\) into the linear speed formula: \(v = \left(\frac{2\pi}{24}\right) \times 3960\) miles per hour.

Simplify the expression to find the linear speed \(v\) of the point on the equator due to Earth's rotation.

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Angular Velocity

Angular velocity measures how fast an object rotates or revolves relative to a fixed point, expressed in radians per unit time. For Earth, it is the rate of rotation about its axis, typically one full rotation per 24 hours, which is essential to relate rotational motion to linear speed.

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Linear Speed in Circular Motion

Linear speed is the distance traveled per unit time along the circular path and is related to angular velocity by the formula v = rω, where r is the radius and ω is the angular velocity. This relationship allows conversion from rotational speed to the actual speed of a point on the circumference.

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The radius at the equator is the distance from Earth's center to its surface along the equatorial plane, approximately 3960 miles. This radius is crucial for calculating the linear speed of a point on the equator due to Earth's rotation.

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