Question

A ball of radius $R=10.0cm$ rolls without slipping on a horizontal plane so that its centre moves with constant acceleration $a=2.50cm/s^2;t=2.00s$ after the beginning of motion its position corresponds to that shown in Fig. Find:
(a) the velocities of the points $A,B$ and $O$
(b) the accelerations of these points.

Answer 1

$a=2.5cm/s^2,R=10cm$

After $t=2sec$

velocity of centre of mass

$v=at=2.5\times 2=5m/s^2$

As the constant acceleration of it centre of mass is $2.5m/s^2$, the angular acceleration of ball is

$\alpha =\frac{a}{R}=\frac{2.5}{10}$

$⟹\alpha =0.25rad/s^2$

So,after 2 second its angular velocity $\omega =\alpha t=0.25\times 2=0.5rad/s$

Figure(b) shows velocity configuration of point A,B,O after 2 second,Now as the motion is pure rotational we have $\omega R=v$

Velocity of point B is $=\sqrt{v^2+(\omega R{)}^2}=\sqrt{2}v=5\sqrt{2}m/s$

Velocity of point A is $=v+\omega R=2v=10m/s$

Velocity od point O is $=v-\omega R=v-v=0$

Figure(c) shows configurations of acceleration at all these point a centriodal acceleration ${\omega }^{2}R$ acts

${\omega }^{2}R=(0.5{)}^2\times 10=2.5m/s^2$

Acceleration of point A $=\sqrt{a^2+({\omega }^{2}R{)}^2}=2.5\sqrt{2}m/s^2$

Acceleration of point B $=a-{\omega }^{2}R=2.5-2.5=0$

Acceleration of point O $=\sqrt{a^2+({\omega }^{2}R{)}^2}=2.5\sqrt{2}m/s^2$