Question
Two identical discs of mass $$m$$ and radius $$r$$ are arranged as shown in the figure. If $$\alpha $$ is the angular acceleration of the lower disc and $${a_{cm}}$$ is acceleration of centre of mass of the lower disc, then relation between $${a_{cm}},\alpha $$ and $$r$$ is
Two identical discs of mass $$m$$ and radius $$r$$ are arranged as shown in the figure. If $$\alpha $$ is the angular acceleration of the lower disc and $${a_{cm}}$$ is acceleration of centre of mass of the lower disc, then relation between $${a_{cm}},\alpha $$ and $$r$$ is
A.
$${a_{cm}} = \frac{\alpha }{r}$$
B.
$${a_{cm}} = 2\alpha r$$
C.
$${a_{cm}} = \alpha r$$
D.
None of these
Answer :
$${a_{cm}} = 2\alpha r$$
Solution :
$$\eqalign{ & Tr = \frac{{m{r^2}}}{2}{\alpha _1}\,......\left( {\text{1}} \right) \cr & Tr = \frac{{m{r^2}}}{2}\alpha \,......\left( {\text{2}} \right) \cr & {\alpha _1} = \alpha \,......\left( {\text{3}} \right) \cr} $$
Acceleration of point $$b$$ = acceleration of point $$a$$
$$r{\alpha _1} = {a_{cm}} - r\alpha \,......\left( 4 \right)$$
Hence, $$2r\alpha = {a_{cm}}$$
$$\eqalign{ & Tr = \frac{{m{r^2}}}{2}{\alpha _1}\,......\left( {\text{1}} \right) \cr & Tr = \frac{{m{r^2}}}{2}\alpha \,......\left( {\text{2}} \right) \cr & {\alpha _1} = \alpha \,......\left( {\text{3}} \right) \cr} $$
Acceleration of point $$b$$ = acceleration of point $$a$$
$$r{\alpha _1} = {a_{cm}} - r\alpha \,......\left( 4 \right)$$
Hence, $$2r\alpha = {a_{cm}}$$
