Problem 46 - Physics for Scientists and Engineers with Modern Physics

Question

(II) Use dimensional analysis (Section \(1-7\) ) to obtain the form for the centripetal acceleration, \(a_{\mathrm{R}}=v^{2} / r\)

Step-by-Step Solution

Answer

Using dimensional analysis, the form for centripetal acceleration is \(a_{\mathrm{R}} = \frac{v^2}{r}\).

Step 1: Identify the dimensions of the relevant quantities

Centripetal acceleration \(a_{\mathrm{R}}\) has the dimension of acceleration, which is \([L][T]^{-2}\). Velocity \(v\) has the dimension \([L][T]^{-1}\). Radius \(r\) is a length with dimension \([L]\).

Step 2: Set up the dimensional equation

Assume \(a_{\mathrm{R}}\) can be expressed as a function of \(v\) and \(r\), i.e., \(a_{\mathrm{R}} = k \cdot v^m \cdot r^n\), where \(k\) is a dimensionless constant, \(m\) and \(n\) are unknown exponents to be determined.

Step 3: Equate dimensions on both sides

Equate the dimensions of both sides: the left side has dimensions \([L][T]^{-2}\), and the right side's dimensions are \([L][T]^{-1}]^m\cdot[L]^n = [L]^m[T]^{-m}[L]^n = [L]^{m+n}[T]^{-m}\).

Step 4: Solve for the exponents

To match dimensions, for length, \(m+n=1\) and for time, \(-m=-2\). Solving these, \(m=2\) and \(n=-1\).

Step 5: Write the formula

Using the values for \(m\) and \(n\), the derived formula is \(a_{\mathrm{R}} = k \cdot v^2 \cdot r^{-1}\). Since \(k\) is dimensionless, we assume it to be 1, giving the formula \(a_{\mathrm{R}} = \frac{v^2}{r}\).

Key Concepts

Centripetal Acceleration

Centripetal acceleration is a key concept when studying objects in rotational motion, like a car taking a turn or a planet orbiting a star. This type of acceleration keeps an object moving in a circular path by constantly changing the direction of its velocity toward the center of the circle. While the speed is constant in uniform circular motion, the direction of velocity changes, resulting in acceleration.

The formula to calculate centripetal acceleration is given by:
\[ a_{\mathrm{R}} = \frac{v^2}{r} \]where:

• \(a_{\mathrm{R}}\) is the centripetal acceleration
• \(v\) is the velocity
• \(r\) is the radius of the circular path

Dimensional analysis aids in deriving this formula by ensuring that the same physical units apply on both sides of the equation. This method applies logical reasoning to ensure that mathematical expressions have meaningful physical significance. A strong grasp of dimensional analysis fortifies your understanding of how physical quantities relate to one another.

Velocity Dimensions

Velocity is a vector quantity that represents the rate of change of an object's position. It has both a magnitude, known as speed, and a direction. Velocity gives us more comprehensive information than speed alone.

Due to its nature, velocity has specific dimensions. These are expressed in terms of fundamental physical quantities. The dimensions of velocity are:

• Length: \([L]\), because velocity is related to distance
• Time: \([T]^{-1}\), as velocity measures distance per unit time

Combining these, the dimensional formula for velocity is \([L][T]^{-1}\). Reliably utilizing dimensional analysis can check the correctness of equations involving velocity. For instance, in the centripetal acceleration formula, velocity is squared, maintaining balance on both sides of the equation.

Radius Dimensions

The radius in a circle is a crucial component in calculations of circular motion. The radius is the distance from the center of the circle to any point on its circumference. It influences both the speed and the centripetal force experienced by an object in circular motion.

The dimension of radius is straightforward since it is a measure of length. Expressed dimensionally, it is:

• Length: \([L]\)

In using dimensional analysis, radius combines with other dimensions to yield the desired units of acceleration. In the formula \(a_{\mathrm{R}} = \frac{v^2}{r}\), the radius affects the centripetal acceleration inversely. This inverse relationship indicates that as the radius increases, the required acceleration for maintaining the same velocity in a larger circle decreases, highlighting the efficiency of larger curves.