Question

Please Answer it

A string is wrapped around a uniform solid cylinder of radius r, as shown in (Figure 1). The cylinder can rotate freely about its axis. The loose end of the string is attached to a block. The block and cylinder each have mass m. Find the magnitude alpha of the angular acceleration of the cylinder as the block descends. Express your answer in terms of the cylinder's radius r and the magnitude of the acceleration due to gravity g.

Answer 1

at Mo

From the figure, we can write,

from Newton's second law

now, we know that,

$a = r\alpha$

then we can write,

$mg =m r\alpha + T$

now, we know that torque is given by,

$\tau = Tr$

where r is perpendicular distance.

also, Torque is given by,

$\tau = I \alpha$

therefore,

$\tau = I \alpha = Tr$

$T = \frac{I \alpha}{r} = \frac{1}{2}Mr^2 \times \frac{\alpha}{r}$

$T = \frac{\alpha Mr}{2}$

now putting above equation in, 1

$mg =m r\alpha + \frac{mr \alpha}{2}$

$mg = \frac{3mr \alpha}{2}$

${\color{Red} \alpha= \frac{2g}{3r}}$