Question

If they start together at $t=0$ at what time does each reach the bottom?

Answer 1

Torque on about point of contact

$=Mg.AB$

$MgR\sin {30}^o$

Again if $\alpha$ is its angular acceleration the torque for relation is

$=I_A^{\alpha }$

Where $I_A=$Moment of Inertia about point of contact $A$

$=I_c+M{R}^2$$\left(T_c\right)=$moment of inertia about COM.

For solid Sphere $I_c=\frac{2}{5}M{R}^2$

so, ${\left(I_A\right)}_{Sphere}=\frac{7}{5}M{R}^2$

For disc $I_c=M{r}^2$ so, ${\left(I_a\right)}_{disk}=2M{R}^2$

For hoop $I_c=\frac{1}{2}M{R}^2$

so, ${\left(I_A\right)}_{Hoop}=\frac{3}{2}M{R}^2$

Hence for sphere

${\alpha }_{sphere}=\frac{MgR\sin {30}^o}{\frac{7}{5}M{R}^2}=\frac{5g}{14R}$

and for Hoop

${\alpha }_{Hoop}=\frac{MgR\sin {30}^o}{\frac{3}{2}M{R}^2}=\frac{g}{3R}$

For Disc

${\alpha }_{disc}=\frac{MgR\sin {30}^o}{M{R}^2}=\frac{g}{4R}$

Now in one complete rotation any object of circular cross section transverse a linear distance equal to its circumference of circular cross-section

$=2\pi R$

$=2\times \frac{22}{7}\times 0.1m=\frac{4.4}{7}m$

To cover $42m$ it has to rotate by,

$\frac{2}{\frac{4.4}{7}}=3$

In $3$ rotation it will cover an angular distance

$\theta =3\times 2\pi =6\pi$ rod.

Time taken by sphere to cover the distance is

$t_{sphere}=\sqrt{\frac{2\theta }{{\alpha }_{sphere}}}$

$\sqrt{\frac{2\times 6\pi \times 14\times R}{5g}}$

$\simeq 1.03$

and time taken by Hoop

$t_{Hoop}=\sqrt{\frac{2\theta }{{\alpha }_{hoop}}}=\sqrt{\frac{2\times 6\pi \times 3\times 4}{g}}$

$\simeq 1.07sec$

and the time taken by disc,

$t_{disc}=\sqrt{\frac{2\theta }{{\alpha }_{disc}}}=\sqrt{\frac{2\times 6\pi \times 3\times 4}{g}}$

$\simeq 1.074sec$