Explanation:
Linear acceleration of a rotating particle is given as:
$\overrightarrow{\text{a}}=\overrightarrow{\text{r}}\times\overrightarrow{\alpha}$
(b) The sphere is rotating about a diameter; therefore, the position vector of the particles on the diameter is zero. Thus, linear acceleration of the particle is zero.
(c) and (d): All the particles of the body have the same angular velocity. All the particle on the surface have different linear speeds that depend on the position of the particle from the axis of rotation.
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$\text{A}\omega^2$
$\frac{\text{A}\omega^2}{2}$
$\frac{\text{A}\omega^2}{\sqrt{2}}$
$\text{Zero}.$
$(A)$ $ \overrightarrow{\mathrm{p}}_1^{\prime}=\mathrm{a}_1 \hat{\mathrm{i}}+\mathrm{b}_1 \hat{\mathrm{j}}+\mathrm{c}_1 \hat{\mathrm{k}} $
$ \overrightarrow{\mathrm{p}}_2^{\prime}=\mathrm{a}_2 \hat{\mathrm{i}}+\mathrm{b}_2 \hat{\mathrm{j}}$
$(B)$ $ \overrightarrow{\mathrm{p}}_1^{\prime}=\mathrm{c}_1 \hat{\mathrm{k}} $
$ \overrightarrow{\mathrm{p}}_2^{\prime}=\mathrm{c}_2 \hat{\mathrm{k}}$
$(C)$ $ \overrightarrow{\mathrm{p}}_1^{\prime}=\mathrm{a}_1 \hat{\mathrm{i}}+\mathrm{b}_1 \hat{\mathrm{j}}+\mathrm{c}_1 \hat{\mathrm{k}} $
$ \overrightarrow{\mathrm{p}}_2=\mathrm{a}_2 \hat{\mathrm{i}}+\mathrm{b}_2 \hat{\mathrm{j}}-\mathrm{c}_1 \hat{\mathrm{k}}$
$(D)$ $ \vec{p}_1^{\prime}=a_1 \hat{i}+b_1 \hat{j} $
$ \overrightarrow{\mathrm{p}}_2^{\prime}=a_2 \hat{\mathrm{i}}+b_1 \hat{\mathrm{j}}$