Question
Two circular discs of radius each 10 cm are joined at their centres by a rod of length 30 cm and mass 600 gm as shown in figure.
If the mass of each disc is 600 gm and applied torque between two discs is $43 \times 10^{5}$ dyne cm , the angular acceleration of the discs about the given axis $AB$ is $\_\_\_\_$ $rad/s^{2}$.
Options
Solution
$\alpha = \frac{\tau}{I}$
$${I = \frac{1}{4}{mR}^{2} + {mR}^{2} + \frac{1}{4}{mR}^{2} + m(2R)^{2} + \frac{m(3R)^{2}}{12} + m\left( \frac{R}{2} \right)^{2} }{= \left( \frac{3}{2} + 4 + 1 \right){mR}^{2} = \frac{13}{2}{mR}^{2} = \frac{13}{2} \times 600 \times 10^{2} = 39 \times 10^{4} }{\alpha = \frac{43 \times 10^{5}}{39 \times 10^{4}}rad/s^{2} = \frac{430}{39}rad/s^{2} \approx 11rad/s^{2}}$$
Create a free account to view solution
View Solution Free