A disc of radius \(R\) rolls on a horizontal ground with linear acceleration \(a\) and angular acceleration…

A disc of radius \(R\) rolls on a horizontal ground with linear acceleration \(a\) and angular acceleration \(\alpha\) as shown in figure. The magnitude of acceleration of point \(P\) as shown in the figure at an instant when its linear velocity is \(v\) and angular velocity is \(\omega\) will be
  1. \(\sqrt{(a+r \alpha)^{2}+\left(r \omega^{2}\right)^{2}}\)
  2. \(\frac{\mathrm{ar}}{R}\)
  3. \(\sqrt{r^{2} \alpha^{2}+r^{2} \omega^{4}}\)
  4. \(r \alpha\)

Solution

We have two components of acceleration of point P- one in radial direction which is \(r \omega^{2}\) and the other in tangential direction which is \(r a\). Also, every point on the disc has a forward acceleration of a. Thus we have total acceleration in the forward direction as a \(+\mathrm{r} \alpha\).
Thus we get the resultant acceleration of point \(\mathrm{P}\) as \(\sqrt{(\mathrm{a}+\mathrm{ra})^{2}+\left(\mathrm{r} \omega^{2}\right)^{2}}\).

Asked in: JEE Mains - Rotational Motion - Chapter Test

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