Question

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 at an instant when its linear velocity is $v$ and angular velocity is $\omega$ will be:

• A. $\sqrt{(a+r\alpha {)}^2+(r{\omega }^2{)}^2}$
• B. $\frac{ar}{R}$
• C. $\sqrt{r^2{\alpha }^2+r^2{\omega }^2}$
• D. $r\alpha$

Answer 1

The correct option is A $\sqrt{(a+r\alpha {)}^2+(r{\omega }^2{)}^2}$

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 $r\alpha$.

Also, every point on the disc has a forward acceleration of a. Thus we have total acceleration in the forward direction as $a+r\alpha$.

Thus we get the resultant acceleration of point P as $\sqrt{(a+r\alpha {)}^2+(r{\omega }^2{)}^2}$.