Question

Consider an object moving according to the position function \(\mathbf{r}(t)=a \cos \omega t \mathbf{i}+a \sin \omega t \mathbf{j}\) If the angular velocity \(\omega\) is halved, by what factor is \(a_{\mathrm{N}}\) changed?

Short answer

The normal acceleration \(a_N\) is reduced by a factor of 1/4 when the angular velocity \(\omega\) is halved.

Step-by-step solution

Understanding the Position Function

Consider the given position function \(\mathbf{r}(t) = a \cos(\omega t) \mathbf{i} + a \sin(\omega t) \mathbf{j}\). This represents uniform circular motion in a plane with radius 'a' and angular speed '\omega'.

Computing the Velocity

The velocity is the derivative of the position function. Differentiate \(\mathbf{r}(t)\) with respect to \(t\). This gives \(\mathbf{v}(t) = -a\omega \sin(\omega t) \mathbf{i} + a\omega \cos(\omega t) \mathbf{j}\).

Computing the Acceleration

Acceleration is the derivative of the velocity function. Differentiate \(\mathbf{v}(t)\) with respect to \(t\). This gives \(\mathbf{a}(t) = -a\omega^2 \cos(\omega t) \mathbf{i} - a\omega^2 \sin(\omega t) \mathbf{j}\). By taking the magnitude of acceleration \(a(t)\), we get \(|a(t)| = a\omega^2\). This is the centripetal or radial acceleration.

Computing the Normal Acceleration

The normal acceleration \(a_N\) (towards the center of the circle) for circular motion is equal to the magnitude of acceleration \(a(t)\). Accordingly, we get \(a_N = a\omega^2\). As we're interested in the factor by which \(a_N\) changes when \(\omega\) is halved, we need to find \(a_N\) for half of \(\omega\). If we replace \(\omega\) with \(\omega/2\) in the equation, we get \(a_N = a(\omega/2)^2 = a\omega^2/4\). Comparing this to \(a_N = a\omega^2\), we see that \(a_N\) is reduced by 1/4th. The factor by which \(a_N\) is changed when \(\omega\) is halved is 1/4.