A uniform disc of mass M and radius R is mounted on a fixed horizontal axis. A block of mass m hangs from a massless string that is wrapped around the rim of the disc. The magnitude of the acceleration of the falling block ( m ) is

Question

A uniform disc of mass M and radius R is mounted on a fixed horizontal axis. A block of mass m hangs from a massless string that is wrapped around the rim of the disc. The magnitude of the acceleration of the falling block ( m ) is (A) (2 M/M + 2 m) g (B) (2 m/M + 2 m) g (C) (M + 2 m/2 M ) g (D) (2 M + m/2 M ) g

Tardigrade Admin · Accepted Answer

When pulley has a finite mass M and radius R , then tension in two segments of the string are different. <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-1wuklqawwvbc.png" /> Here, ma=mg-T a=(m/m + (M)2)g=(2 m/2 m + M)g

Q. A uniform disc of mass $M$ and radius $R$ is mounted on a fixed horizontal axis. A block of mass $m$ hangs from a massless string that is wrapped around the rim of the disc. The magnitude of the acceleration of the falling block ( $m$ ) is

$\frac{2 M}{M + 2 m} g$

$\frac{2 m}{M + 2 m} g$

$\frac{M + 2 m}{2 M} g$

$\frac{2 M + m}{2 M} g$

When pulley has a finite mass $M$ and radius $R$ , then tension in two segments of the string are different.

Here, $ma = m g - T$
$a = \frac{m}{m + \frac{M}{2}} g = \frac{2 m}{2 m + M} g$

Q. A uniform disc of mass M and radius R is mounted on a fixed horizontal axis. A block of mass m hangs from a massless string that is wrapped around the rim of the disc. The magnitude of the acceleration of the falling block ( m ) is

M+2m2M​g

M+2m2m​g

2MM+2m​g

2M2M+m​g